A website that has programs that will do most operations in this course (an online calculator for matrices)

http://www.math.odu.edu/~bogacki/cgi-bin/lat.cgi

3.3 of a Subspace

This is a hypercube

Dimension - Hypercubes

DimensionA basis of a subspace consists of the fewest number of

vectors needed to span the subspace.

The basis of a subspace V in Rn consist of the same number of vectors.

The number of vectors in the basis of a subspace V is called the dimension of V.

Standard basis

The vectors

Form as standard basis for Rn

How to find a basis for the imageFind all of the columns that are independent.There are 2 ways this is generally done.1) If the matrix is simple do by inspection.(use number sense to determine which columns are

independent).2) Put the matrix in reduced row echelon form.

Every column that has a leading one in this form is independent. The columns (in the un-simplified form) form a basis for the image.

Note: the number of vectors in the basis of the image is the dimension of the image

Recall this problem:Find the basis of the kernel and the

image of

Find the dimension of the image and the kernel.

Find the basis of the image by inspection

Consider the columns1) The first column is in the basis for the image because it is non-zero2) The second column is not in the basis for the image because it is a

multiple of column 13) The third column is not in the basis because it is the zero vector4) The fourth column is in the basis because it is not a multiple of

column 1 (the only vector that is in our basis so far)5) The 5th column is not in the basis because it is the sum of column 1

and 4

Answer: columns 1 and 4 form a basis of the image

The dimension of the image is 2

Find the basis of the kernel

Set Ax = 0Solve and find all solutions

Find rref(A)1 2 0 0 10 0 0 1 10 0 0 0 0 0 0 0 0 0

x2, x3 and x5 are free variables

Set x2 = r, x3 = s, x5 = t

x1 = -2r –tx2 = rx3 = sx4 = -tx5 = t

r

-21000

+ s

00100

+

-100-11

The three column vectors are a basis for the kernel

The dimension of the kernel is 3

Problem 6

Find the basis of the kernel (null space) and image (column Space) of the matrix

Problem 6 Solution

Problem 10Find the basis of the kernel (null space) and image (column Space) of the matrix

Solve by inspection

Problem 10 Solution

What is the relationship between the dimension of the image and the

dimension of the kernel?

The Fundamental Theorem of Linear Algebra

For an nxm matrix

Dim(Im(A)) + Dim (Ker(A)) = m

For an nxn matrix

Of course for a square matrix the dim(Im(A)) + Dim(ker(A)) = n

Why is this true?

Problem 16Find the basis of the kernel (null space) and image (column Space) of the matrix

Problem 16Find the basis of the kernel (null space) and image (column Space) of the matrix

Columns 1,2 and 4 appear to be independent,Hence they form a basis of the image.

Due to the fundamental thm of LA we need To find only 1 vector to form a basis of the Kernel. Find 1 vector that results in zero.

-3 -2 1 0

For an nxn matrix A, the following statements are equivalent

1 A is invertible2. The linear system Ax=b has a unique solution3. rref(A) = I4. The rank of (A) = n5. Im(A) = Rn

6.Ker(A) = 07. The column vectors of A form a basis of Rn

8. The columns vectors of A span Rn

9. The column vectors of A are linearly independent.10. Det A ≠ 0

Problem 29

29 Solution

The dimension of this subspace is 2

Problem 39

Problem 39 Solution

Problem 34

Problem 34 Solution

I found a new technique for working in higher dimensions…When ever I run into a hypercube I give it Ritalin - Mr. Whitehead

A Mathematician (M) and an Engineer (E) attend a lecture by a Physicist. The topic concerns Kulza-Klein theories involving physical processes that occur in spaces with dimensions of 9, 12 and even higher dimensions. The M is sitting, clearly enjoying the lecture, while the E is frowning and looking generally confused and puzzled. By the end the E has a terrible headache. At the end, the M comments about the wonderful lecture. E: "How do you understand this stuff?" M: "I just visualize the process" E: "How can you POSSIBLY visualize something that occurs in 9-dimensional space?" M: "Easy, first visualize it in N-dimensional space, then let N go to 9

Homework p.133 1-29 odd, 37