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About The Author
Dan Fleisch is a Professor in the Department of Physics at
Wittenberg University, where he specializes in
electromagnetics and space physics. He is the author of the
internationally best-selling book A Student’s Guide to
Maxwell’s Equations, published by Cambridge University
Press in January 2008. This book, currently in its 11th printing,
has been translated into Japanese, Korean, and Chinese. Dr.
Fleisch’s latest book, A Student’s Guide to Vectors and
Tensors, was published by Cambridge Press in September of 2011. Fleisch is also the co-author
with the late Prof. John Kraus of The Ohio State University of the McGraw-Hill textbook
Electromagnetics with Applications.
Prof. Fleisch has published technical articles in the IEEE Transactions, The Journal of Atmospheric
and Terrestrial Physics, and Microwave Journal, and has presented more than a dozen
professional papers on topics related to high-speed microwave instrumentation and radar cross-
section measurement. He has been a regular contributor of science commentary to PBS station
WYSO of Yellow Springs, and in 2006 he appeared in the documentary "The Dayton
Codebreakers" shown on Public Television.
Prof. Fleisch was named Outstanding Faculty Member at the Wittenberg Greek scholarship awards
in 2000, and in 2002 he won the Omicron Delta Kappa award for Excellence in Teaching. In 2003
and 2005 he was recognized for Faculty Excellence and Innovation by the Southwestern Ohio
Council for Higher Education (SOCHE), and in 2004 he received Wittenberg’s Distinguished
Teaching Award, the university’s highest faculty award.
In November, 2010 Prof. Fleisch was named the Ohio Professor of the Year by the Carnegie
Foundation and the Council for the Advancement and Support of Education.
Fleisch received his B.S. in Physics from Georgetown University in 1974 and his M.S. and Ph.D. in
Space Physics and Astronomy from Rice University in 1976 and 1980, respectively.
To contact Dan Fleisch, send e-mail to
A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Matrix Algebra
Review
Alternative forms of
the EM tensor
About Dan Fleisch
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Home > Chapter 1 Problems
Problem 1.3
Imagine that the y-axis points north and the x-axis points east.
(a) If you travel a distance r = 22 km in a straight line from the origin in
a direction 35 degrees south of west, what is your position in Cartesian
(x,y) coordinates?
(b) If you travel 6 miles due south from the origin and then turn west
and travel 2 miles, how far from the origin and in what direction is your
final position?
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution A
Go
Full solution for part (a):
Continue to Problem 1.4
A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press
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Home > Chapter 1 Problems
Problem 1.3
Imagine that the y-axis points north and the x-axis points east.
(a) If you travel a distance r = 22 km in a straight line from the origin in
a direction 35 degrees south of west, what is your position in Cartesian
(x,y) coordinates?
(b) If you travel 6 miles due south from the origin and then turn west
and travel 2 miles, how far from the origin and in what direction is your
final position?
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution B
Go
Full solution for part (b):
From a sketch such as Figure P1.3b, you can see that point P has coor-
dinates x = -2 miles and y = -6 miles. Thus
and
where 180° has been added to the result of the arctan operation since
the denominator was negative.
Continue to Problem 1.4
A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press
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Home > Chapter 1 Problems
Problem 1.4
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full solution for part all parts (a through c):
Continue to Problem 1.5
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Home > Chapter 1 Problems
Problem 1.5
Cylindrical coordinates
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution A
Go
Full solution for part (a):
Continue to Problem 1.6
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Home > Chapter 1 Problems
Problem 1.5
Cylindrical coordinates
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution B
Go
Full solution for part (b):
Continue to Problem 1.6
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Home > Chapter 1 Problems
Problem 1.6
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution A
Go
Full solution for part (a):
Continue to Problem 1.6
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Home > Chapter 1 Problems
Problem 1.6
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution B
Go
Full solution for part (b):
Continue to Problem 1.7
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Home > Chapter 1 Problems
Problem 1.7
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution A
Go
Full solution for part (a):
Continue to Problem 1.8
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Home > Chapter 1 Problems
Problem 1.7
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution B
Go
Full solution for part (b):
Continue to Problem 1.8
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Home > Chapter 1 Problems
Problem 1.8
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution A
Go
Full solution for part (a):
Continue to Problem 1.9
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Home > Chapter 1 Problems
Problem 1.8
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution B
Go
Full solution for part (b):
Continue to Problem 1.9
A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press
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Home > Chapter 1 Problems
Problem 1.9
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full solution:
Continue to Problem 1.10
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Home > Chapter 1 Problems
Problem 1.10
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full solution:
To find the unit basis vectors, use
Thus
A Student's Guide to Vectors and Tensors © 2012 Cambridge University Press
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Home > Chapter 2 Problems
Problem 2.1
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full solution:
The definition of the scalar product in Cartesian coordinates is
which in this case gives
or
and
so
and
Continue to Problem 2.2
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Home > Chapter 2 Problems
Problem 2.2
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SOLUTION!
Full Solution
Go
Full solution:
or
Thus in this case
Continue to Problem 2.3
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Home > Chapter 2 Problems
Problem 2.3
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
so
or
Likewise, for the vector product
so
or
Continue to Problem 2.4
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Home > Chapter 2 Problems
Problem 2.4
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
which is
So
Continue to Problem 2.5
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Home > Chapter 2 Problems
Problem 2.5
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Using the BAC-CAB rule,
Inserting values for the vectors that appear in the dot products gives
So
Continue to Problem 2.6
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Home > Chapter 2 Problems
Problem 2.6
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Using the rules of partial differentiation, the partial derivative with re-
spect to x is
and the partial derivative with respect to y is
Continue to Problem 2.7
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Home > Chapter 2 Problems
Problem 2.7
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 2.8
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Home > Chapter 2 Problems
Problem 2.8
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 2.9
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Home > Chapter 2 Problems
Problem 2.9
What is the curl of the vector field given in the previous problem?
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 2.10
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Home > Chapter 2 Problems
Problem 2.10
Find the Laplacian of the function given in Problem 2.6.
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Taking the first partial derivatives makes this
Taking the second partial derivatives makes this
Continue to Problem 2.11
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Problem 2.11
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SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 2.12
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Home > Chapter 2 Problems
Problem 2.12
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
At the optimum angle of 90°, for a force of 25 Newtons and a distance
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Home > Chapter 3 Problems
Problem 3.1
Solve the box-on-a-ramp problem (that is, find the acceleration of
the box) for the frictionless case using a Cartesian coordinate system for
which the y-axis points vertically upward and the x-axis points horizon-
tally to the right.
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
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Continue to Problem 3.2
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Problem 3.2
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SOLUTION!
Full Solution
Go
Full Solution:
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Problem 3.3
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SOLUTION!
Full Solution
Go
Full Solution:
Begin by making a free-body diagram such as Fig. P3.3 showing all the
forces acting on the box along with the coordinate axes. The equations
for the sum of the forces in the x- and y-directions are
and the y-components are
This means that Newton's 2nd law for this case is
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To find the velocity after the deliverywoman has pushed the box for 2
meters, use the constant-acceleration kinematic equation
So, for example, if the mass of the box is 1 kg, the speed of the box
after the deliverywoman has pushed with a force of 10 Newtons over a
distance of 2 meters will be
Continue to Problem 3.4
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Home > Chapter 3 Problems
Problem 3.4
If the hammer thrower shown on the cover of this book wishes to
launch a hammer of mass 7.26 kg on a cable of length 1.22 meters with a
speed of 22 m/s, what is the magnitude of the centripetal force he must
supply?
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
and the magnitude of the centripetal acceleration is related to the ve-
locity and radius of curvature by
thus
Plugging in values gives
Continue to Problem 3.5
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Home > Chapter 3 Problems
Problem 3.5
Imagine a Formula 1 car going around a curve with radius of 10
meters while slowing down from a speed of 180 mph to a speed of 120
mph in 2 seconds. What are the magnitude and direction of the car's
acceleration at the instant the car's speed is 150 mph?
Give me a HINT or show me THE FULL
SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 3.6
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Problem 3.6
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SOLUTION!
Full Solution
Go
Full Solution:
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Continue to Problem 3.7
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Problem 3.7
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SOLUTION!
Full Solution
Go
Full Solution:
Thus
Taking the partial derivatives gives
Thus
Continue to Problem 3.8
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Home > Chapter 3 Problems
Problem 3.8
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SOLUTION!
Full Solution
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Full Solution:
and in cylindrical coordinates, the gradient is given by
Taking the partial derivatives makes this
and
Continue to Problem 3.9
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Home > Chapter 3 Problems
Problem 3.9
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SOLUTION!
Full Solution
Go
Full Solution:
Continue to Problem 3.10
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Problem 3.10
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SOLUTION!
Full Solution
Go
Full Solution:
which in this case
or
Thus
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Problem 4.1
Write the inverse transformation matrix for a 70° rotation of the
2D Cartesian coordinate axes and the direct transformation matrix for
the rotation of a vector through an angle of 70° degrees. Show that the
product of these two transformation matrices is the identity matrix.
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which is the identity matrix.
Continue to Problem 4.2
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Problem 4.2
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Plugging in values gives
or
which means
Continue to Problem 4.3
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Home > Chapter 4 Problems
Problem 4.3
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and
and
Remember that these are the components of the rotated unit vectors in
the original (non-rotated) coordinate system.
Continue to Problem 4.4
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Problem 4.4
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Full Solution:
Continue to Problem 4.5
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Home > Chapter 4 Problems
Problem 4.5
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Plugging in values gives
and
This is consistent with the results of Problems 4.2 and 4.4.
Continue to Problem 4.6
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Problem 4.6
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and
which in this case is
and
Solving the first of these two simultaneous equations for A1 gives
and substituting this into the second equation gives
Thus
and
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Another approach to finding the contravariant components is to use the
matrix method and Cramer's Rule. To do this, write
in matrix form as
or
Now use Cramer's Rule to find A1 and A2:
and
in agreement with the results of the substitution method.
Continue to Problem 4.7
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Home > Chapter 4 Problems
Problem 4.7
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Continue to Problem 4.8
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Problem 4.8
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which may be written as
or
You can now use Cramer's Rule:
and
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Continue to Problem 4.9
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Home > Chapter 4 Problems
Problem 4.9
Use the substitution method and the elimination method to solve the
two simultaneous equations that result from vector Equation 4.27.
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Home > Chapter 4 Problems
Problem 4.10
Show that the elements of the Cartesian-to-polar transformation
matrix are the components of the basis vectors tangent to the original
(Cartesian) coordinate axes.
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Home > Chapter 5 Problems
Problem 5.1
Show that the process of subtracting one tensor from another results
in a quantity that is also a tensor.
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Continue to Problem 5.2
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Home > Chapter 5 Problems
Problem 5.2
Find the elements of the metric tensor for spherical coordinates by
forming the dot products of the relevant basis vectors.
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You can determine the elements of the metric tensor by finding the scalar products of these basis
vectors:
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Continue to Problem 5.3
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Home > Chapter 5 Problems
Problem 5.3
Show how the derivatives given after Equation 5.16 lead to the ele-
ments of the metric tensor for spherical polar coordinates (Equation 5.17).
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Full Solution:
The steps in getting from Equation 5.16 to Equation 5.17 can be found
in the solution for Problem 5.2. But in case you skipped that problem,
here it is again. The relevant derivatives are
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Continue to Problem 5.4
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Home > Chapter 5 Problems
Problem 5.4
Use the scale factors for spherical polar coordinates to verify the
expressions given in Chapter 2 for the gradient, divergence, curl, and
Laplacian in spherical coordinates.
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in agreement with Equation 2.34 in the text.
For the divergence, inserting the spherical polar scale factors gives
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in agreement with Equation 2.39 in the text.
The curl is given by
Inserting the scale factors gives
which simplified to
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or
in agreement with Equation 2.42 in the text.
The Laplacian can be found as
which is
in agreement with Equation 2.54 in the text.
Continue to Problem 5.5
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Home > Chapter 5 Problems
Problem 5.5
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SOLUTION!
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Full Solution:
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Continue to Problem 5.6
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Home > Chapter 5 Problems
Problem 5.6
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SOLUTION!
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Full Solution:
For diagonal matrices, the inverse matrix is also diagonal, and each (di-
agonal) element of the inverse matrix is the inverse of the corresponding
element of the original matrix. Since
the inverse matrix is
Continue to Problem 5.7
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Home > Chapter 5 Problems
Problem 5.7
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SOLUTION!
Full Solution
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Full Solution:
Continue to Problem 5.8
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Home > Chapter 5 Problems
Problem 5.8
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SOLUTION!
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Full Solution:
Continue to Problem 5.9
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Home > Chapter 5 Problems
Problem 5.9
What are the Christoffel symbols for the 2-D spherical surface of
Problem 5.8?
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Full Solution:
To find Christoffel symbols, use
and from the solution for Problem 5.8, you know that
Continue to Problem 5.10
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Home > Chapter 5 Problems
Problem 5.10
Show that the covariant derivative of the metric tensor equals zero.
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The covariant derivative of a second-rank tensor is
Applying this to the metric tensor gives
Based on Equation 5.23 in the text, the two Christoffel symbols in this
expression are related to the metric tensor as
in which the dummy index is written as m since the letter k is being
used as the index with respect to which the covariant derivative is being
taken. Thus
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Home > Chapter 6 Problems
Problem 6.1
Find the inertia tensor for a cubical arrangement of eight identical masses with the origin of
coordinates at one of the corners and the co-ordinate axes along the edges of the cube.
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Continue to Problem 6.2
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Home > Chapter 6 Problems
Problem 6.2
How would the moment of inertia tensor of Problem 6.1 change if one of the eight masses is
removed?
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Continue to Problem 6.3
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Home > Chapter 6 Problems
Problem 6.3
Find the moment of inertia tensor for the arrangement of masses of
Problem 6.2 if the coordinate system is rotated by 20 degrees about one
of the coordinate axes (do this by finding the locations of the masses in
the rotated coordinate system).
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A sketch such as Figure P6.3(a) will help you determine the impact
of rotating the coordinate system. In this case, the Cartesian axes are
rotated 20° about the z-axis, so the x and y coordinates of the masses will
change (excepting the mass at the origin). One approach to determining
the values of x' and y' for each mass is to make a sketch such as Figure
P6.3(b), which is the view you would have if you looked directly down the
z-axis (toward the origin). To make it a bit easier to determine the x' and
y' values for each mass, "picture has been tilted" in Figure 6.3(c) so that
the x' axis is horizontal and the y' axis is vertical. Applying geometry
to a figure such as this allows you can determine the coordinates of each
of the masses in the primed coordinate system.
For example, the x' coordinate of masses 2 and 6 may be seen from
the figure to equal s sin 20° = 0.342s, and the y' coordinate of masses 2
and 6 is to equal s cos 20° = 0.940s.
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An alternative approach is to use a rotation matrix to convert the
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Home > Chapter 6 Problems
Problem 6.4
Use the similarity-transform approach to verify the moment of iner-
tia tensor you found in Problem 6.3.
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Home > Chapter 6 Problems
Problem 6.5
Show how the vector wave equation results from taking the curl of
both sides of Faraday's Law and inserting the curl of the magnetic field
from the Ampere-Maxwell Law.
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Continue to Problem 6.6
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Home > Chapter 6 Problems
Problem 6.6
If an observer in one coordinate system measures an electric field of
5 volts per meter in the z-direction and zero magnetic field, what electric
and magnetic fields would be measured by a second observer moving at
1/4 the speed of light along the x-axis?
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Full Solution:
Continue to Problem 6.7
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Home > Chapter 6 Problems
Problem 6.7
If an observer in one coordinate system measures a magnetic field
of 1.5 Tesla in the z-direction and zero electric field, what electric and
magnetic fields would be measured by a second observer moving at 1/4
the speed of light along the x-axis?
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SOLUTION!
Full Solution
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Full Solution:
Continue to Problem 6.8
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Home > Chapter 6 Problems
Problem 6.8
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Continue to Problem 6.9
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Problem 6.9
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Problem 6.10
Find the Ricci tensor and scalar for the 2-sphere of Section 6.3.
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1
EM Tensor Comparison
This document compares the electromagnetic field tensor presented in AStudent’s Guide to Vectors and Tensors (SGVT) to the equivalent tensorin several other texts. It is not meant to provide additional informationabout this tensor, but rather to help you understand why the tensor mayappear different when you encounter it elsewhere. This initial documentwill compare the EM field tensor as presented in SGVT to the versionspresented in Griffiths’ Introduction to Electrodynamics, Jackson’s Clas-sical Electrodynamics, and Shadowitz’s The Electromagnetic Field ; I’lladd the versions of other texts based on readers’ suggestions.
The electromagnetic field tensor presented in Section 6.2 of SGVT hasthe following form:
Fαβ =
⎛⎜⎜⎝
0 −Ex/c −Ey/c −Ez/c
Ex/c 0 −Bz By
Ey/c Bz 0 −Bx
Ez/c −By Bx 0
⎞⎟⎟⎠ .
Notice that this is the contravariant form of the EM field tensor (in bothindices), so the first thing to check when making comparisons betweentexts is that you’re using the same form with respect to contravariantand covariant indices.
In Griffiths’ Introduction to Electrodynamics, the doubly contravariantEM field tensor is presented as
Fµν =
⎛⎜⎜⎝
0 Ex/c Ey/c Ez/c
−Ex/c 0 Bz −By
−Ey/c −Bz 0 Bx
−Ez/c By −Bx 0
⎞⎟⎟⎠
2 EM Tensor Comparison
which differs by a minus sign from the version presented in SGVT. Thisdifference is attributable to the fact that Griffiths uses the sign conven-tion of placing a minus sign on the time component and a plus sign onthe spatial components of four-vectors, and the opposite convention isused in SGVT.
In Jackson’s Classical Electrodynamics, the doubly contravariant EMfield tensor is given as
Fαβ =
⎛⎜⎜⎝
0 −Ex −Ey −Ez
Ex 0 −Bz By
Ey Bz 0 −Bx
Ez −By Bx 0
⎞⎟⎟⎠
In this case, the sign convention is the same as that used in SGVT,but the electric-field terms do not contain the factor of 1/c included inthe SGVT version. The reason for this difference is that Jackson usesGaussian rather than SI units for all chapters after Chapter 10 of thelatest (third) edition of his book (that’s what the little “G” means afterthe chapter headings on the top of the even pages). You can understandthe impact of those units by looking at Table 3 in the Appendix of theJackson text. There he indicates that you should make the followingsubstitutions to convert his equations between systems:
�E in SI →�E√4πε0
in Gaussian system
�B in SI →√
mu0
4π�B in Gaussian system
�J in SI → √4πε0 �J in Gaussian system
If you make the appropriate conversions in the EM field tensor (as wellas in the expression relating the derivatives of the field-tensor elementsto the four-current), you can get to Maxwell’s Equations in the samemanner described in Section 6.2 of SGVT.
EM Tensor Comparison 3
In Shadowitz’s The Electromagnetic Field, the doubly contravariant EMfield tensor is written as
F JK =
⎛⎜⎜⎝
0 −cBz −cBy −Ex
−cBz 0 cBx −Ey
cBy −cBx 0 −Ez
Ex Ey Ez 0
⎞⎟⎟⎠
This appears quite different from the other versions of this tensor dis-cussed above, but notice that a large part of this difference comes aboutbecause the electric-field terms appear in the fourth row and columnrather than in the first row and column as in the previous versions.The reason for this difference is that Shadowitz writes four-vectors withthe time component in the last position (after the spatial components)rather than in the first position. Additionally, the EM field tensor pre-sented by Shadowitz is essentially c times the EM field tensor in SGVT,so Ex/c in SGVT becomes Ex in Shadowitz, and Bx in SGVT becomescBx in Shadowitz.
To compensate for this difference when deriving Maxwell’s Equationsfrom the EM field tensor, Shadowitz includes a factor of
√µ0/ε0 in front
of his four-current (J) term:
∂Fij
∂xj=
õ0
ε0Ji
so if you pull the multiplicative factor of c out of Shadowitz’s EM tensorand divide it into the four-current term, this becomes
1c
õ0
ε0Ji =
√µ0ε0
õ0
ε0Ji
=
õ2
0ε0ε0
Ji = µ0Ji
where the fact that c = 1/√
µ0ε0 has been used, making the final ex-pression consistent with the four-current term in SGVT.
References
Griffiths, D., Introduction to Electrodynamics, Benjamin-Cummings 1999.
Jackson, J.D., Classical Electrodynamics, John Wiley 1999.
Shadowitz, A., The Electromagnetic Field, McGraw-Hill 1975.
1
Matrix Algebra Review
Introduction
This is the matrix review for A Student’s Guide to Vectors and Tensors(SGVT). It is not meant to be a thorough introduction to the theory andpractice of matrix algebra, but rather a review of the matrix conceptsyou should understand when learning about vectors and tensors.
A matrix is defined as an array of numbers. Some authors restrict theterm “matrix” to refer only to square arrays of real numbers, but in com-mon usage a matrix is a rectangular (m x n) array of complex numbersor “elements,” where “m” represents the number of rows, “n” representsthe number of columns, and “complex” means that the numbers may bepurely real, purely imaginary, or mixed. Thus
A =[
3 −21 0
]B =
⎡⎢⎢⎢⎢⎣
80−215
⎤⎥⎥⎥⎥⎦ C =
[9 15 3 −8 12 −2
]
and
D =
⎡⎢⎢⎣
5 − 2i 3i
0 15−4i 2 + 8i
12 i − 1
⎤⎥⎥⎦ E =
⎡⎣ 1 0 0
0 1 00 0 1
⎤⎦ F =
[0 00 0
]
are all matrices. In these examples, matrix A has “dimensions” or “or-der” (rows x columns) of 2x2, B of 5x1, C of 1x6, D of 4x2, E of 3x3,and F of 2x2.
2 Matrix Algebra Review
If both m and n for a matrix are equal to one, that matrix is calleda scalar (because it’s just a single value), and if either (but not both)of m and n are one, that matrix may be called a vector. So in theexamples shown above, B and C are vectors; in some texts you’ll seeB called a column vector and C called a row vector. A square matrix(that is, a matrix with m=n) with ones on the diagonal and zeroes in alloff-diagonal elements (such as matrix E above) is called the “unit” or“identity” matrix, and a matrix with all elements equal to zero is calledthe “null” or “zero” matrix.
Two matrices are said to be equal only if they have the same numberof rows as well as the same number of columns and if every element inone matrix is equal in value to the corresponding element in the othermatrix. Matrices are usually denoted using uppercase letters or bold font,often surrounded by square brackets (such as [A]), and the elements areoften written using lowercase letters with subscripts. So you may see theelements of matrix [A] written as aij, although some authors use Aij or[A]ij to refer to the elements of A (but be careful to note that Aij mayalso refer to the “Matrix of Cofactors” of matrix A, which you can readabout in Section E of this document).
So does terminology such as “row vector” mean that matrices, vectors,and tensors are the same thing? Not really. It’s certainly true that thevectors and tensors in SGVT are often represented using matrices, butit’s important to remember what those matrices stand for. Those arraysof values represent the components of vectors and tensors, and thosecomponents have meaning only when associated with the basis vectorsof a particular coordinate system. Since the basis vectors are not alwaysshown, it’s convenient to think of the matrix as representing the vectoror tensor itself, and that’s fine as long as you remember that the actualvector or tensor has existence independent of any coordinate system.
A) Matrix addition, multiplication by a scalar, and subtraction
Matrices may be added only if both the row dimension (m) and theircolumn dimension (n) are equal (such matrices are said to be “of thesame order”). The addition is accomplished simply by adding each el-ement of one matrix to the corresponding element of the other matrix.For example:
[5 2−3 0
]+
[ −3 10 4
]=
[2 3−3 4
]
Matrix Algebra Review 3
or in general[a11 a12
a21 a22
]+
[b11 b12
b21 b22
]=
[a11 + b11 a12 + b12
a21 + b21 a22 + b22
].
Note that the result is a matrix of the same order as the matrices beingadded. Note also that matrix addition is commutative, so [A]+[B]=[B]+[A].
Multiplication of a matrix by a scalar is straightforward; you simplymultiply each element of the matrix by the scalar. Thus
3A = 3[
3 −21 0
]=
[9 −63 0
]
and generally
kA = k
[a11 a12
a21 a22
]=
[ka11 ka12
ka21 ka22
]
You can use the rules for addition of matrices and scalar multiplicationto see that subtraction of matrices is accomplished simply by subtractingthe corresponding elements. Thus [A]-[B]=[A]+(-1)[B]. So if
A =[
a11 a12
a21 a22
]
and
B =[
b11 b12
b21 b22
]
then
A−B =[
a11 + (−1)b11 a12 + (−1)b12
a21 + (−1)b21 a22 + (−1)b22
]=
[a11 − b11 a12 − b12
a21 − b21 a22 − b22
].
Just as for addition, subtraction of matrices only works for matrices ofthe same order.
B) Matrix multiplication
There are several different ways to multiply matrices; the most commonand most relevant to the vector and tensor concepts in SGVT is tomultiply two matrices (call them A and B) by multiplying the elementsof each row in A by the elements of each column of B and then summingthe results. So if matrix A is given by[
a11 a12 a13
a21 a22 a23
]
4 Matrix Algebra Review
and matrix B is given by⎡⎣ b11 b12 b13
b21 b22 b23
b31 b32 b33
⎤⎦
then the product A times B is given by
[a11 a12 a13
a21 a22 a23
]×
⎡⎣ b11 b12 b13
b21 b22 b23
b31 b32 b33
⎤⎦
=[
a11b11 + a12b21 + a13b31 a11b12 + a12b22 + a13b32 a11b13 + a12b23 + a13b33
a21b11 + a22b21 + a23b31 a21b12 + a22b22 + a23b32 a21b13 + a22b23 + a23b33
].
This result is achieved by multiplying the elements of the first row ofA by the elements of the first column of B, summing the results, andplacing the sum in the first row and first column of the result matrix,as shown in Figure 1.1.
a11
a12
a13
( ) ( ) ( )
b11
( ) ( )
b21
( ) ( )
b31
( ) ( )
a11
b11
+ a12
b21
+ a13
b31
( ) ( )
( ) ( ) ( )x =
Multiply first rowby first column {
Sum products
Place result in first rowand first column
Figure 1.1 Multiplying first row by first column
The next step is to multiply the elements of the first row of A by theelements of the second column of B, summing the products, and placingthe sum in the first row and second column of the result matrix, asshown in Figure 1.2.
After multiplying the first row of A by the third column of B and placingthe sum in the first row and third column of the result matrix, thesame procedure is done with the second row of A - those elements aremultiplied by the first column of B, summed, and placed in the second
Matrix Algebra Review 5
a11
a12
a13
( ) ( ) ( )
( ) b12
( )
( ) b22
( )
( ) b32
( )
( ) a11
b12
+ a12
b22
+ a13
b32
( )
( ) ( ) ( )x =
Multiply first rowby second column
{
Sum products
Place result in first rowand second column
Figure 1.2 Multiplying first row by second column
( ) ( ) ( )
a21
a22
a23
b11
( ) ( )
b21
( ) ( )
b31
( ) ( )a
21b
11+ a
22b
21+ a
23b
31( ) ( )x =
{
Sum products
Multiply second rowby first column
Place result in second rowand first column
( ) ( ) ( )
Figure 1.3 Multiplying second row by first column
row and first column of the result matrix, as shown in Figure 1.3.
Notice that the matrix that results from multipling a 2x3 matrix (A) bya 3x3 matrix (B) is a 2x3 matrix - the result has the number of rows ofthe first matrix and the number of columns of the second matrix. Youshould also note that this type of matrix multiplication only works whenthe number of columns of the first matrix (3 in the case of A) equals thenumber of rows of the second matrix (also 3 in the case of B).
To understand why matrices are multiplied in this way, consider thesorting of five types of toy marbles (Bowlers, Cat’s Eyes, Steelies, Aggies,and Commons) into four sizes of package (Small, Medium, Large, andExtra Large). The number of each type of marble in each size package
6 Matrix Algebra Review
is shown in the following table:
Bowlers Cat’s Eyes Steelies Aggies Commons
Small package 1 3 2 5 4Medium package 2 5 3 7 6Large package 5 10 5 12 10Extra Large package 10 15 12 16 14
This array of numbers can be put into a matrix - call it “P” the packagematrix:
P =
⎡⎢⎢⎣
1 3 2 5 42 5 3 7 65 10 5 12 1010 15 12 16 14
⎤⎥⎥⎦
Now imagine three containers (Cans, Boxes, and Crates) with differentnumbers of packages in each type:
Small Pkgs Medium Pkgs Large Pkgs Extra-Large Pkgs
Can 10 8 5 2Box 15 12 7 3Crate 40 25 15 10
Put this array into a matrix called “C” the container matrix:
C =
⎡⎣ 10 8 5 2
15 12 7 340 25 15 10
⎤⎦
If you wish to find the number of each type of marble in each type ofcontainer, you could do it like this:
Bowlers per Can = 10 small packages × 1 Bowler per small package+ 8 medium packages × 2 Bowlers per medium package+ 5 large packages × 5 Bowlers per large package+ 2 extra-large packages × 10 Bowlers per extra-large package= (10)(1)+(8)(2)+(5)(5)+(2)(10)=71 Bowlers per Can
Matrix Algebra Review 7
Likewise, you can find the number of Cat’s Eyes per can using
Cat’s Eyes per Can=10 small packages×3 Cat’s Eyes per small package+ 8 medium packages × 5 Cat’s Eyes per medium package+ 5 large packages × 10 Cat’s Eyes per large package+ 2 extra-large packages × 15 Cat’s Eyes per extra-large package= (10)(3)+(8)(5)+(5)(10)+(2)(15)= 150 Cat’s Eyes per Can
And if you wished to find the number of Bowlers per Box, you could use
Bowlers per Box = 15 small packages × 1 Bowler per small package+ 12 medium packages × 2 Bowlers per medium package+ 7 large packages × 5 Bowlers per large package+ 3 extra-large packages × 10 Bowlers per extra-large package= (15)(1)+(12)(2)+(7)(5)+(3)(10)=104 Bowlers per Box
If you compare the numbers in these calculations to the values in the Pand C matrices, you’ll see that in each case you’re multiplying the rowelements of C by the column elements of P, which is exactly how matrixmultiplication works:
CP =
⎡⎣ 10 8 5 2
15 12 7 340 25 15 10
⎤⎦
⎡⎢⎢⎣
1 3 2 5 42 5 3 7 65 10 5 12 1010 15 12 16 14
⎤⎥⎥⎦ =
⎡⎣ 71 150 93 198 166
104 220 137 291 244265 545 350 715 600
⎤⎦
So the product matrix shows you how many of each type of marble arein each type of container:
Bowlers Cat’s Eyes Steelies Aggies Commons
Can 71 150 93 198 166Box 104 220 137 291 244Crate 265 545 350 715 600
In addition to showing why the rows of the first matrix are multi-plied by the columns of the second matrix, this example also illustratesthe point made earlier: when you multiply two matrices, the number ofcolumns in the first matrix must equal the number of rows in the sec-ond matrix. So in this case, multiplication works only if you multiply Cby P, not P by C. The larger point is that matrix multiplication is not
8 Matrix Algebra Review
commutative, so even if you’re able to multiply two matrices in eitherorder (which you can do, for example, with square matrices), the answeris not the same. So in matrix world, AB is in general not equal to BA.
Another difference between matrix multiplication and multiplicationof numbers is that it’s possible to get zero as the result of multiplyingtwo matrices, even if neither matrix is zero. For example,[
2 10 0
]×
[1 0−2 0
]=
[0 00 0
].
If you study vectors and tensors, you’re likely to come across the mul-tiplication of a row vector times a column vector or vice versa, and youshould understand that such products use the same process describedabove for matrix multiplication, although that may not be immediatelyobvious when you look at the result. For example, multiplying a rowvector A by a column vector B gives
[a11 a12 a13
] ×⎡⎣ b11
b21
b31
⎤⎦ = [a11b11 + a12b21 + a13b31] .
This scalar result (essentially the dot product between A and B) comesfrom multiplying the first (and only) row of A by the first (and only)column of B and adding the sums. Likewise, if A is a column vector andB a row vector, the product AB is
⎡⎣ a11
a21
a31
⎤⎦ × [
b11 b12 b13
]=
⎡⎣ a11b11 a11b12 a11b13
a21b11 a21b12 a21b13
a31b11 a31b12 a31b13
⎤⎦
Once again, the same rule of matrix multiplication has been applied.The first row of A (just a11 in this case) is multiplied by the first columnof B (just b11 in this case), and since there are no other elements in thefirst row of A or the first column of B, there is nothing to add, andthe result (a11b11) is written in the first row and column of the resultmatrix. Then the first row of A (again, only a11) is multiplied by thesecond row of B (only b12), and the result is written in the first row,second column of the result matrix. After doing the same for the firstrow of A and the third column of B, you then multiply the second rowof A (which is just a21 in this case) by the first column of B (again justb11) and write the result in the second row, first column of the resultmatrix. So although you get a scalar when you multiply a row vector by acolumn vector (sometimes called the “inner product” of the matrics) and
Matrix Algebra Review 9
a matrix when you multiply a column vector by a row vector (sometimescalled the “outer product” of the matrices), the process is exactly thesame in both cases.
Although matrix multiplication is not commutative (so AB is notnecessarily equal to BA), matrix multiplication is the associative anddistributive over addition, so for matrices A, B, and C
(AB)C = A(BC)
and
A(B + C) = AB + AC
as long as you remember not to reverse the order of any of the products.You may be wondering if there’s ever a need to multiply each element
of one matrix by the corresponding element of an equal-size matrix.There certainly is (for example, when applying a two-dimensional fil-ter funtion to an image). This process is called “element-by-element”or “entrywise” matrix multiplication and the result is sometimes calledthe “Hadamard product.” In MATLAB, such multiplication is denotedA.*B, where the decimal point before the multiplication symbol signifieselement-by-element multiplication.
C) Transpose and trace of a matrix
The transpose of a matrix is accomplished by interchanging the rowswith the columns of that matrix. This is usually denoted by placing thesuperscript “T” after the matrix name, so the transpose of matrix A iswritten as AT (MATLAB uses A′). So if the matrix A has elements
A =[
a11 a12 a13 a14
a21 a22 a23 a24
]
then the transpose of A is given by
AT =
⎡⎢⎢⎣
a11 a21
a12 a22
a13 a23
a14 a24
⎤⎥⎥⎦ .
So (Aij)T = Aji; notice that the indices have been switched. The trans-pose of the product of two matrices is equal to the transpose of eachmatrix multiplied in reverse order, so
(AB)T = BT AT .
10 Matrix Algebra Review
so long as the dimensions of the matrices allow such products to beformed.
For square matrices, the trace of the matrix is given by the sum of thediagonal elements. The trace is usually denoted by “Tr” so the trace ofmatrix A is written as Tr(A). So if the matrix A is given by
A =
⎡⎢⎢⎣
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44
⎤⎥⎥⎦
the trace of A is
Tr(A) = a11 + a22 + a33 + a44 =∑
aii.
D) Determinant, minors, and cofactors of a matrix
The determinant of a square matrix is a scalar calculated by multiplying,adding, and subtracting various elements of the matrix. If A is a 2x2matrix
A =[
a11 a12
a21 a22
]
the determinant of A (denoted using vertical bars on each side of A) isfound by cross-multiplying the upper-left element times the lower-rightelement and then subtracting the product of the lower-left element timesthe upper-right element:
|A| =∣∣∣∣ a11 a12
a21 a22
∣∣∣∣ = a11a22 − a21a12.
Hence if A is
A =[
4 −23 1
]
the determinant of A is
|A| =∣∣∣∣ 4 −2
3 1
∣∣∣∣ = (4)(1) − (3)(−2) = 4 − (−6) = 10.
To find the determinant of higher-order matrices, you must use the“minors” and “cofactors” of the determinant of the matrix. The minorfor each element of the determinant of a square matrix is found by
Matrix Algebra Review 11
eliminating the entire row and the entire column in which the elementappears and then writing the remaining elements as a new determinant,with row and column dimensions reduced by one (so the minor of eachelement of a 3x3 determinant is a 2x2 determinant). So for the 3x3determinant of matrix A
|A| =
∣∣∣∣∣∣a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣the minor of element a11 is∣∣∣∣∣∣
− − −− a22 a23
− a32 a33
∣∣∣∣∣∣ =∣∣∣∣ a22 a23
a32 a33
∣∣∣∣obtained by elminating the first row and column of the determinant ofmatrix A (since the element for which this minor is taken (a11) is inthe first row and the first column). Likewise, the minor for element a12
is found by eliminating the row and column in which a12 appears (firstrow and second column):∣∣∣∣∣∣
− − −a21 − a23
a31 − a33
∣∣∣∣∣∣ =∣∣∣∣ a21 a23
a31 a33
∣∣∣∣ .
For element a22, the minor is∣∣∣∣∣∣a11 − a13
− − −a31 − a33
∣∣∣∣∣∣ =∣∣∣∣ a11 a13
a31 a33
∣∣∣∣ .
Before you can use minors to find the determinant of a higher-ordermatrix, you have turn each minor into a “cofactor.” The cofactor of anelement is just the minor of that element multiplied by either +1 or -1.To know which of these to use, just determine whether the sum of theelement’s row index and its column index is even or odd. If even (as itis, for example, for the element in the first row and first column, since1+1=2 which is even), the cofactor equals the minor of that elementmutiplied by +1. If odd (for example, for the element in the first rowand second column, since 1+2=3 which is odd) the cofactor equals theminor of that element multiplied by -1. So for element aij with row indexof i and column index of j,
Cofactor of (aij) = (−1)(i+j)Minor of (aij).
12 Matrix Algebra Review
So for the minors shown above, the cofactors are
Cofactor of (a11) = (−1)(1+1)
∣∣∣∣ a22 a23
a32 a33
∣∣∣∣ = (1)(a22a33 − a32a23)
Cofactor of (a12) = (−1)(1+2)
∣∣∣∣ a21 a23
a31 a33
∣∣∣∣ = (−1)(a21a33 − a31a23)
and
Cofactor of (a22) = (−1)(2+2)
∣∣∣∣ a11 a13
a31 a33
∣∣∣∣ = (1)(a11a33 − a31a13).
With cofactors in hand, the determinant of a 3x3 matrix is straight-forward to find. Simply pick one row or column, and multiply each ofthe elements of that row or column by its cofactor and sum the results.So, for example, if you choose to use the first row of the 3x3 matrix A,the determinant is given by
|A| = a11(Cofactor of a11)+a12(Cofactor of a12)+a13(Cofactor of a13).
or
|A| = a11(1)(a22a33−a32a23)+a12(−1)(a21a33−a31a23)+a13(1)(a21a32−a31a22).
The value of the determinant would have been the same if you had useda different row (or any of the columns) of matrix A, as long as youmultiply each element in your selected row or column by its cofactorand sum the results.
The same approach can be used to find the determinant of 4x4 andhigher-order matrices; you just have to expand each of the determinantsusing elements and cofactors until you get down to the 2x2 level. Ofcourse, the calculation becomes a bit tedious, so it’s worth your time tolearn to find determinants using a program like MATLAB.
Some useful characteristics of determinants are that the determinantof the transpose of a matrix equals the determinant of the matrix (so|AT | = |A|) and the determinant of the product of two matrices is thesame as the determinant of the reverse product (so |AB| = |BA|) pro-vided that the dimensions of A and B allow both of these products tobe made.
You may also find it useful to know that if a matrix has two identicalrows or columns, or if one row or column is an integer multiple of an-other row or column, the determinant of that matrix must be zero.
Matrix Algebra Review 13
E) Inverse of a matrix
As indicated in Section C of this document, matrix multiplication isquite different from ordinary multiplication of numbers. One example ofthis is that the product of two matrices may be zero even if both of thematrices are non-zero (so you can’t simply divide both sides of a matrixequation such as AB = 0 by A to get B = 0). Additionally, the matrixequation AB = AC does not mean that matrix B equals matrix C (sodividing both sides of this equation by A does not work for matrices).
Differences such as this suggest that matrix division has little in com-mon with ordinary division of numbers. About the closest you can getto a process similar to division for matrices comes about by consideringthe matrix equation
AX = B
where A, X, and B are all matrices. If you wish to find matrix X fromthis equation, here’s one thing you definitely cannot do:
AX
A=
B
A
X =B
A
because this type of division does not work for matrices.What you can do to find X is this: you can try to find a matrix A−1
(called the “inverse” or “reciprocal” of A) with the following property:
A−1A = I
where I represents the identity matrix (ones on the diagonal and zeroeseverywhere else). Now you can find X rather easily:
AX = B
A−1(AX) = A−1B
(A−1A)X = A−1B
IX = A−1B
X = A−1B
since IX = X (multiplying the identity matrix times any matrix doesnot change that matrix).
So although you haven’t really divided matrix B by matrix A, you’veused the matrix equivalent of multiplying by the reciprocal of a numberto achieve the same result as dividing.
14 Matrix Algebra Review
The question is, does every matrix have an inverse (that is, anothermatrix with which it multiplies to give the identiry matrix), and if so,how do you find it? The first part of this question is easy to answer:any matrix that has a non-zero determinant (|A| �= 0) has an inverse.Matrices that have no inverse ((|A| = 0) are called “singular” matrices,and matrices that have an inverse are called “non-singular” matrices.
But if a certain matrix has an inverse, how do you find that inverse?There are several ways to go about this, but one approach is to usethe concepts of cofactors, transpose, and determinant described above.Those concepts appear in the following equation for the inverse of matrixA:
A−1 =(Matrix of cofactors of A)T
|A|
in which the “Matrix of cofactors of A” is a matrix in which each elementof A is replaced by that element’s cofactor.
Here’s how that works for a 3x3 matrix. If matrix A is the usual
A =
⎡⎣ a11 a12 a13
a21 a22 a23
a31 a32 a33
⎤⎦
then the matrix of cofactors (MoC) for A is
MoC (A) =
⎡⎣ a22a33 − a32a23 −(a21a33 − a31a23) a21a32 − a31a22
−(a12a33 − a32a13) a11a33 − a31a13 −(a11a32 − a31a12)a12a23 − a22a13 −(a11a23 − a21a13) a11a22 − a21a12
⎤⎦ .
The equation for the inverse of A shown above requires that you takethe transpose of this matrix, which is
[MoC (A)]T =
⎡⎣ a22a33 − a32a23 −(a12a33 − a32a13) a12a23 − a22a13
−(a21a33 − a31a23) a11a33 − a31a13 −(a11a23 − a21a13)a21a32 − a31a22 −(a11a32 − a31a12) a11a22 − a21a12
⎤⎦ .
Dividing this matrix by the determinant of A provides the inverse of A.Fortunately, the inverse of non-singular matrices is easily found using
most scientific calculators or computers, usually by defining a matrix A
and then raising A to the power of minus one.For a diagonal matrix, the inverse is simply another diagonal matrix
in which each diagonal element is the reciprocal of the corresponding
Matrix Algebra Review 15
element in the original matrix. So⎡⎣ a11 0 0
0 a22 00 0 a33
⎤⎦−1
=
⎡⎢⎣
1a11
0 00 1
a220
0 0 1a33
⎤⎥⎦ .
F) Simultaneous linear equations and Cramer’s Rule
The relationship between matrices and simultaneous linear equationscan be understood by considering equations such as
2x + 5y − z = 12
−3x − 3y + z = −1
x + y = 1.
Using the rules of matrix multiplication, this system of equations can bewritten as a single matrix equaton:⎡
⎣ 2 5 −1−3 −3 11 1 0
⎤⎦ ×
⎡⎣ x
y
z
⎤⎦ =
⎡⎣ 12
−11
⎤⎦ .
In general, three linear equations in three unknowns (x1, x2, x3) can bewritten as
a11x1 + a12x2 + a13x3 = b1
a21x1 + a22x2 + a23x3 = b2
a31x1 + a32x2 + a33x3 = b3
or ⎡⎣ a11 a12 a13
a21 a22 a23
a31 a32 a33
⎤⎦ ×
⎡⎣ x1
x2
x3
⎤⎦ =
⎡⎣ b1
b2
b3
⎤⎦ .
If you define the matrices
A =
⎡⎣ a11 a12 a13
a21 a22 a23
a31 a32 a33
⎤⎦
and
x =
⎡⎣ x1
x2
x3
⎤⎦
16 Matrix Algebra Review
and
b =
⎡⎣ b1
b2
b3
⎤⎦
then the system of equations can be written as a single matrix equation:
Ax = b
which can sometimes be solved using the inverse of matrix A:
x = A−1b.
The reason that the word “sometimes” appears in the previous sentenceis that there are several conditions which will prevent this approach fromsucceeding. For example, the system of equations may be inconsistent -that is, there may be no set of values for x1, x2, x3 that satisfy all threeequations. In that case, you will not be able to find the inverse of A
because it will be a singular matrix. And if matrix b equals zero, youhave a system of homogeneous linear equations, which means there willbe only the trivial solution (x1 = x2 = x3 = 0) if A is non-singular oran infinite number of solutions if A is singular.
In cases for which A is non-singular and b does not equal zero, youcan find the values of x1, x2, and x3 by finding the inverse of matrixA and multiplying that inverse by matrix b, or you can use Cramer’sRule. In that approach, the first unknown (x1 in this case) is found byreplacing the values in the first column of the coefficient matrix (A) withthe elements of matrix b and dividing the determinant of that matrix bythe determinant of A. Here’s how that looks:
x1 =
∣∣∣∣∣∣b1 a12 a13
b2 a22 a23
b3 a32 a33
∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣
.
Likewise, to find the second unknonwn (x2 in this case), replace the
Matrix Algebra Review 17
values in the second column of A with the elements of b:
x2 =
∣∣∣∣∣∣a11 b1 a13
a21 b2 a23
a31 b3 a33
∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣and to find x3 use
x3 =
∣∣∣∣∣∣a11 a12 b1
a21 a22 b2
a31 a32 b3
∣∣∣∣∣∣∣∣∣∣∣∣a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣
.
Thus for the equations given at the start of this section,
A =
⎡⎣ 2 5 −1
−3 −3 11 1 0
⎤⎦
x =
⎡⎣ x1
x2
x3
⎤⎦ =
⎡⎣ x
y
z
⎤⎦
and
b =
⎡⎣ b1
b2
b3
⎤⎦ =
⎡⎣ 12
−11
⎤⎦ .
Hence
x = x1 =
∣∣∣∣∣∣12 5 −1−1 −3 11 1 0
∣∣∣∣∣∣∣∣∣∣∣∣2 5 −1−3 −3 11 1 0
∣∣∣∣∣∣
18 Matrix Algebra Review
so
x =12[(−3)(0) − (1)(1)] + 5(−1)[(−1)(0) − (1)(1)] − 1[(−1)(1) − (1)(−3)]2[(−3)(0) − (1)(1)] + 5(−1)[(−3)(0) − (1)(1)] − 1[(−3)(1) − (1)(−3)]
=−12 + 5 − 2−2 + 5 − 0
= −3.
Proceeding in the same way for y and z gives
y = x2 =
∣∣∣∣∣∣2 12 −1−3 −1 11 1 0
∣∣∣∣∣∣∣∣∣∣∣∣2 5 −1−3 −3 11 1 0
∣∣∣∣∣∣so
y =−2 + 12 + 2−2 + 5 − 0
= 4
and
z = x3 =
∣∣∣∣∣∣2 5 12−3 −3 −11 1 1
∣∣∣∣∣∣∣∣∣∣∣∣2 5 −1−3 −3 11 1 0
∣∣∣∣∣∣so
z =−4 + 10 + 0−2 + 5 − 0
= 2.
G) Matrix diagonalization using eigenvectors and eigenvalues
In the matrix equation discussed in the previous section
Ax = b
the matrix A operates on matrix x to produce matrix b. If x and b
are vectors, this equation represents the transformation by matrix A
of vector x into vector b. In some cases, the operation of A on thecomponents of x produces the components of a vector b that is a scaled(but not rotated) version of x. In such cases, the equation becomes
Ax = λx
Matrix Algebra Review 19
where λ represents a scalar multiplier (and scalar multipliers can changethe length but not the direction of a vector).
Any vector x which satisfy this equation for a matrix A is called an“eigenvector” of matrix A, and the scalar λ is called the “eigenvalue”associated with that eigenvector.
The eigenvalues of a matrix can be very useful in finding a diagonalversion of that matrix, so you may wish to understand how to find theeigenvalues of a given matrix. To do that, write the previous equationas
Ax − λx = 0
which, since Ix = x, can be written as
Ax − λ(Ix) = 0
or
(A − λI)x = 0.
which means that either x = 0 (which is a the trivial case) or
|A − λI| = 0.
This equation is called the “characteristic equation” for matrix A, andfor a 3x3 matrix it looks like this:∣∣∣∣∣∣
a11 a12 a13
a21 a22 a23
a31 a32 a33
∣∣∣∣∣∣ − λ
∣∣∣∣∣∣1 0 00 1 00 0 1
∣∣∣∣∣∣ = 0
or ∣∣∣∣∣∣a11 − λ a12 a13
a21 a22 − λ a23
a31 a32 a33 − λ
∣∣∣∣∣∣ = 0.
This expands to
(a11 − λ)[(a22 − λ)(a33 − λ) − a32a23]
+ a12(−1)[a21(a33 − λ) − a31a23]
+ a13[a21a32 − a31(a22 − λ)] = 0.
Finding the roots of this polynomial provides the eigenvalues (λ) formatrix A, and substituting those values back into the matrix equationAx = λx allows you to find eigenvectors corresponding to each eigen-value. The process of finding the roots is less daunting than it may
20 Matrix Algebra Review
appear, as you can see by considering the following example. For the3x3 matrix A given by
A =
⎡⎣ 4 −2 −2
−7 5 85 −1 −4
⎤⎦
the characteristic equation is∣∣∣∣∣∣4 − λ −2 −2−7 5 − λ 85 −1 −4 − λ
∣∣∣∣∣∣ = 0
or
(4 − λ)[(5 − λ)(−4 − λ) − (−1)(8)]
− 2(−1)[(−7)(−4 − λ) − (5)(8)]
− 2[(−7)(−1) − (5)(5 − λ)] = 0.
Multiplying through and subtracting within the square brackets makesthis
(4 − λ)(λ2 − λ − 12) + 2(7λ − 12) − 2(5λ − 18) = 0
or
−λ3 + 5λ2 + 12λ − 36 = 0.
Finding the roots of a polynomial like this is probably best left to acomputer, but if you’re lucky enough to have a polynomial with integerroots, you know that each root must be a factor of the term not involvingλ (36 in this case). So (+/-) 2,3,4,6,9,12,and 18 are possibilities, and itturns out that +2 works just fine:
−(2)3 + 5(22) + 12(2) − 36 = −8 + 20 + 24 − 36 = 0.
So you know that one root of the characteristic equation (and hence oneeigenvalue) must be +2. That means you can divide a factor of (λ − 2)out of the equation and try to see other roots in the remainder. Thatdivision yields this:
−λ3 + 5λ2 + 12λ − 36(λ − 2)
= −λ2 + 3λ + 18.
The roots remaining polynomial on the right-hand side of this equationare +6 and -3, so you now have
−λ3 + 5λ2 + 12λ − 36 = (λ − 2)(6 − λ)(λ + 3) = 0.
Matrix Algebra Review 21
So matrix A has three distinct eigenvalues with values +6, -3, and +2;these are the factors by which matrix A scales its eigenvectors. Youcould find the eigenvectors of A by plugging each of the eigenvaluesback into the characteristic equation for A, but as long as you can findN distinct eigenvalues for an NxN matrix, you can be sure that A canbe diagonalized simply by constructing a new diagonal matrix with theeigenvalues as the diagonal elements. So in this case, the diagonal matrix(call it A′) associated with matrix A is
A′ =
⎡⎣ 6 0 0
0 −3 00 0 2
⎤⎦ .
To see why this is true, consider the operation of matrix A on each ofits three eigenvectors (call them e, f , and g):
Ae = λ1e
Af = λ2f
Ag = λ3g
Now imagine a matrix E whose columns are made up of the eigenvectorsof matrix A:
E =
⎡⎣ e1 f1 g1
e2 f2 g2
e3 f3 g3
⎤⎦
where the components of eigenvector e are (e1, e2, e3), the componentsof eigenvector f are (f1, f2, f3), and the components of eigenvector g are(g1, g2, g3). Mutiplying matrix A (the original matrix) by E (the matrixmade up of the eigenvectors of A), you get
AE =
⎡⎣ a11 a12 a13
a21 a22 a23
a31 a32 a33
⎤⎦ ×
⎡⎣ e1 f1 g1
e2 f2 g2
e3 f3 g3
⎤⎦
which is
AE =
⎡⎣ a11e1 + a12e2 + a13e3 a11f1 + a12f2 + a13f3 a11g1 + a12g2 + a13g3
a21e1 + a22e2 + a23e3 a21f1 + a22f2 + a23f3 a21g1 + a22g2 + a23g3
a31e1 + a32e2 + a33e3 a31f1 + a32f2 + a33f3 a31g1 + a32g2 + a33g3
⎤⎦
The columns of this AE matrix are the result of multiplying A by eachof the eigenvectors. But you know from the definition of eigenvectors
22 Matrix Algebra Review
and eigenvalues that
Ae = λ1e = λ1
⎡⎣ e1
e2
e3
⎤⎦ =
⎡⎣ λ1e1
λ1e2
λ1e3
⎤⎦
and
Af = λ2f = λ2
⎡⎣ f1
f2
f3
⎤⎦ =
⎡⎣ λ2f1
λ2f2
λ2f3
⎤⎦
and
Ag = λ3g = λ3
⎡⎣ g1
g2
g3
⎤⎦ =
⎡⎣ λ3g1
λ3g2
λ3g3
⎤⎦ .
This means that the product AE can be written
AE =
⎡⎣ λ1e1 λ2f1 λ3g1
λ1e2 λ2f2 λ3g2
λ1e3 λ2f3 λ3g3
⎤⎦ .
But the matrix on the right-hand side can also be written like this:⎡⎣ λ1e1 λ2f1 λ3g1
λ1e2 λ2f2 λ3g2
λ1e3 λ2f3 λ3g3
⎤⎦ =
⎡⎣ e1 f1 g1
e2 f2 g2
e3 f3 g3
⎤⎦ ×
⎡⎣ λ1 0 0
0 λ2 00 0 λ3
⎤⎦
This means that you can write
AE = E
⎡⎣ λ1 0 0
0 λ2 00 0 λ3
⎤⎦
and multiplying both sides by the inverse of matrix E (E−1) gives
E−1AE = E−1E
⎡⎣ λ1 0 0
0 λ2 00 0 λ3
⎤⎦ =
⎡⎣ λ1 0 0
0 λ2 00 0 λ3
⎤⎦ .
If you tracked the discussion of the similarity transform in Section 6.1,you’ll recognize the expression E−1AE as the similarity transform of ma-trix A to a coordinate system with basis vectors that are the columnsof matrix E. Those columns are the eigenvectors of matrix A, and thematrix that results from the similarity transform (call it A′) is diagonaland has the eigenvalues of A as its diagonal elements.
Matrix Algebra Review 23
References
Pettofrezzo, A., Matrices and Transformations, Prentice-Hall 1966.
Riley, K., Hobson, M.P., and Bence, S.J. Mathematical Methods forPhysics and Enginnering, Cambridge 2006.
Schwartz, J.T., Introduction to Matrices and Vectors, McGraw-Hill 1961.
Errata and Clarifications (September 2015)
1) On page 18, the radial line from the origin to point P should be labeled “r”rather than “ρ”.
2) On page 19, the last of the three equations of Eq. 1.14 should say z = k.
3) On page 24 in the second line of Problem 1.9, the word “spherical” shouldbe replaced by the word “cylindrical”. Hence the phrase should be “...a steadycurrent I is given in cylindrical coordinates by the expression....”
4) On the top of page 35 in the second line of the right-hand side of the equation
for ~A× ( ~B × ~C), the term AxCy should be AxCx.
5) On the top of page 48 in the right-hand side of the equation for ~∇◦ ~A, thereshould be no vector symbols over the components Ax, Ay, and Az.
6) On page 57, the paragraph beginning with ”And how does the difference...”should read ”And how does the difference between a function’s value at a pointand the average value at neighboring points relate to the divergence of the gra-dient of that function? To understand that, think about a point which is a localmaximum of the function - at such at point, the function’s value is greater thanthe surrounding average. Likewise, at a point which is a local minimum, thefunction’s value is less than the surrounding average. This is the reason youmay find the Laplacian described as a ”concavity detector” - it finds points atwhich the value of the function sticks above or falls below the average value ofthe surrounding points.”
7) On page 60 in Problem 2.3, in two places vertical-line symbols are printedas slanting lines (this occurs after the letter “B”, once before “cos” and oncebefore “sin”). Here is the correct text:
2.3 Show that ~A◦ ~B = AxBx+AyBy +AzBz = | ~A|| ~B| cos(θ) and that | ~A◦ ~B| =| ~A|| ~B| sin(θ).
8) On page 68, the sentence beginning “According to this equation...” shouldsay that the y-component of the box’s acceleration is proportional to (ratherthan equal to) the difference between the magnitude of the normal force andthe y-component of the gravitational force.
9) On page 71, 77, and 80 the word “accelerate” is misspelled.
10) On page 87, the units of the permittivity of free space should be C2/(Nm2).
11) On page 88, the equation on the second line should contain the term(keqr2
)
1
rather than(keqr
).
12) On page 90, the equation for the divergence of the magnetic field near thebottom of the page (after Eq. 3.36) should read
~∇ ◦ ~B =1
r
∂Bφ∂φ
=1
r
∂
∂φ
(µ0I
2πr
)
13) On page 114 in the right portion of Figure 4.13, the basis vector along they axis labeled ~e 2 should be ~e2. Here’s a corrected version of the figure:
14) Clarification: On page 126, the paragraph beginning with “And here’s theimportant insight...” should make it clear that the columns of the transforma-tion matrix are the components of the basis vectors.
15) On the bottom of page 129 in the right side of Equation 4.62, Ai should beAj .
16) In the statement for Problem 4.1 on page 130, the word “indirect” shouldbe “direct.”
17) In the statements for Problems 4.6 and 4.8 on page 131, the vector compo-nents A1, A2, A1 and A2 should not have vector arrows.
18) On pages 133 to 136, in Equations 5.1 through 5.9 and the text in-between,the prime marks should appear before the superscripts. So on these pages, ∂xi
′
2
should be ∂x′i.
19) On page 134, the statement beginning with “And just as in Equation 5.1...”should read “And just as in Equation 5.1, the elements of the inverse transfor-mation matrix also represent the basis vectors tangent to the original coordinateaxes.”
20) On page 135, the statement beginning with “In this case, the weighting fac-
tors...” should read “In this case, the weighting factors ( ∂xj
∂xi′ ) are the elementsof the direct transformation matrix from the unprimed to the primed coordinatesystems...”
21) On page 140, in the sentence “...or using covariant components and dualbasis vectors (~ei)...” the subscript ‘i‘ should be a superscript. So the correctedsentence reads “...or using covariant components and dual basis vectors (~e i)...”.
22) On page 141, the final equation in the second block of equations (after thesentence “A third option is to use contravariant components on one side of thedot and covariant components on the other:”) should say = dxidxi.
23) On the top of page 142 in the second line of the equation for | ~A, the expres-sion
√AiAj should be
√AiAi.
24) On page 148, in the final equation on the page, the subscript 1 in the basisvector ~e1 should be an i. So the corrected equation is
∂ ~A
∂x1=∂(A1 ~e1 +A2 ~e2 +A3 ~e3)
∂x1
=∂(Ai~ei)
∂x1
=∂Ai
∂x1~ei +Ai
∂~ei∂x1
25) On page 149, in the fourth line from the top of the page, the word “differ-entiation” is misspelled as “defferentiation”.
26) On page 151, the word “of” is missing from the sentence beginning “Nowit’s just a matter of pulling out...”
27) On page 153, in the expression for Γ122, the multiplying factor for the third
term should be 12g
31.
28) On page 155 in the expression for Aφ;φ on the right side of the equation the
second term should contain Ar instead of Aφand the third term should contain
3
Aφ instead of Ar. The correct equation is:
Aφ;φ =∂Aφ
∂φ+ArΓφrφ +AφΓφφφ +AzΓφzφ
=∂Aφ
∂φ+Ar
(1
r
)+ 0 + 0
29) On page 155, the final equation should be
∂ ~A
∂φ= (
∂Ar
∂φ− rAφ)~er + (
∂Aφ
∂φ+
1
rAr)~eφ +
∂Az
∂φ~ez
30) On page 176, in the expression for the square of the distance covered by awavefront of the light wave in the unprimed coordinate system, the right sideshould be (ct)2, and in the primed coordinate system, the right side should be(c′t′)2.
31) On page 182 in the matrix at the top of the page, at the bottom row ofthe second column, the expression (By)(−γ) should be (−By)(γ), and in the
expression for~~F ′, the element in the third column of the first row should be
−γ(Ey/c− βBz).
32) On page 183 near the end of the second paragraph, the second-last sentenceshould ask“So does the electric field exist or not?”
33) On page 190, in the expression
Γθφφ = (1
2)− gθθ ∂gφφ
∂θ
the minus sign should be in front of the (1/2) factor. So the correct expressionis
Γθφφ = −(1
2)gθθ
∂gφφ∂θ
34) On page 191, in the second equation from the top, which is
Rθφθθ =∂Γθφθ∂θ−∂Γθφθ∂φ
+ ΓθφθΓθθθ + ΓφφθΓ
θφθ − ΓθφθΓ
θθθ − ΓφφθΓ
θφθ
the second term after the equals sign should have θ instead of φ in the denomi-nator. The correct expression is
Rθφθθ =∂Γθφθ∂θ−∂Γθφθ∂θ
+ ΓθφθΓθθθ + ΓφφθΓ
θφθ − ΓθφθΓ
θθθ − ΓφφθΓ
θφθ
35) On page 193, the statement for Problem 6.3 should say“Find the momentof inertia tensor for the arrangement of masses in Problem 6.1....”
4
