Correction Homework 10

7/30/2019 Correction Homework 10

http://slidepdf.com/reader/full/correction-homework-10 1/7

CORRECTION OF HOMEWORK 8

1. Page 149, Exercises 12 and 13

In these exercises, one considers a function D : Mn×n(F) → F, defined on then× n matrices with entry in a field F, with values in F. (You can consider F = R

for concreteness, this is unimportant here). We also assume that D has the samemultiplicative property as the determinant, namely

D(AB) = D(A)D(B) (1.1)

whenever A and B are n × n matrices. The goal of these questions is to see that

D is (under some nondegeneracy conditions) the determinant function. This givesanother way to define the determinant.

Show that either D(A) = 0 for all A or D(I n) = 1We want to prove that one of two possibilities occur. To prove this, we suppose

that one does not occur, and try to see that the second has to occur.1 Hence wesuppose that the first possibility is not true.

So we have two information on D, namely (1.1) and

there exists a matrix M such that D(M) = 0

and we want to prove that

D(I n) = 1 (1.2)

For the moment, the second information seems mysterious, so we focus on (1.1)

and what we want to prove. We have an information relating D and the multiplica-tion of matrices, and we want to prove something about I n. But I n also has somerelation to the multiplication. Namely that for any matrix M , there holds that

M = I n ×M = M × I n.

Now, we want to use our information, so we take D of the identity above and using(1.1), we get

D(M ) = D(I n ×M ) = D(I )D(M ).

This is almost what we want. We just want to make sure that we do not divideby 0. But this is precisely how we can use our second information. Namely we usethe equality above (which is true for all matrix M ) when M = M. In this case,D(M) = 0, and dividing by D(M), we get that D(I n) = 1

If D(I n) = 1, prove that if M is invertible, D(M ) = 0.We want to prove something for all invertible matrices. So we let M be such

a matrix, and we try to prove the result for this particular M . If we can do itwithout making any additional assumption on M , then the proof works for all

Date : Thursday, April 16th 2008.1This has the advantage to give us another information, besides (1.1), namely that the first

possibility does not occur.

1



matrices2 What do we know about M . A priori, we know exactly that there existsanother matrix N = M −1 such that

I n = MN = N M.This look encouraging, since we have a relation between a multiplication (so wecan use our first hypothesis (1.1)) and I n (so we can use our second hypothesisD(I n) = 1). Applying D to the equality above, we get that

1 = D(I n) = D(MN ) = D(M )D(N )

and now, we see that for the product of D(M )D(N ) to be nonzero, both termsmust be nonzero3. Thus D(M ) = 0, and since we have made no assumption on M ,this works for all invertible matrices. This ends the proof.

Now, we make the additional Hypothesis that

n = 2

D(I 2) = D(J )(1.3)

where J is the matrix

J =

0 11 0

.

In particular the analysis above shows that D(I 2) = 1.4

At this point, we have introduced one new object, namely J . Since we only haveinformation about the multiplicative properties of D, it is interesting to see whatthe multiplicative properties of J are. A simple computation shows that

J

a bc d

=

c da b

and

a bc d

J =

b ad c

. (1.4)

Hence, multiplication on the left by J exchange the rows of a matrix, while multi-plication on the right exchange the columns.

Besides, since we are considering products, we can compute J 2 = I 2, and hence,taking D on both sides of this equality, we get that 1 = D(I 2) = D(J )2, and using(1.3), we conclude that

D(J ) = −1. (1.5)

These facts will be useful later on.

Prove that

D(0) = 0. (1.6)

Again, we ask ourselves whether 0 has some special multiplicative property. Andindeed, we know that for every matrix M , there holds that

0 = 0×M,

and applying D to both sides of this equality and using (1.1), we get that, for allmatrices M ,

D(0) = D(M )D(0).

2if we need to make additional assumptions, then we have at least a partial result (case 1). Tofinish the proof, we can then try to prove the result with the additional information that (case 1)

does not hold.3we even have the more precise information that D(M −1) = 1/D(M ).4I think that all the results here would hold for any n if we replace (1.3) by D(I n) = D(E τ )

for one transposition τ ∈ Sn, where E τ is the matrix with (i, j)-entry δiτ (j). For simplicity, we

stick to the n = 2 case.



And from this we conclude that either D(0) = 0, in which case we are done, orD(0) = 0, but then, we can divide by D(M ) in the equality above to get

D(M ) = 1for all matrices M . But this is not true for M = J by (1.3).5 Consequently D(0) = 0and the proof of (1.6) is complete

Prove that

if A2 = 0, then D(A) = 0. (1.7)

We still only know three things, namely (1.1), (1.3) and (1.6). But the multi-plicative property combines nicely with latter to give that in our case when A2 = 0,

0 = D(0) = D(A2) = D(A)D(A) = D(A)2

hence D(A) = 0, and (1.7) is proved.

Prove that

D(B) = −D(A) (1.8)

whenever B is obtained from A by exchanging the rows or the columns.Now, this question is not apparently, linked to any multiplication property, so

we must examine more precisely what we know, which now consist of (1.1), (1.2),(1.3), (1.6) and (1.7). And all these are just consequences of our hypothesis, i.e(1.1) and (1.3). Now, there is one thing that we haven’t used so far, it is the precisestatement of (1.3), which in particular involves a special matrix J . Then, (in casewe did not do that before) we start to examine J and its multiplicative properties,and we rapidly see that (1.4) and (1.5) hold. But this is exactly what we need toanswer the question since if B is obtained from A by interchanging the rows, wesee by (1.4) that B = JA and, using (1.1) and (1.5)

D(B) = D(JA) = D(J )D(A) = −D(A).

In the case when B is obtained by interchanging the columns, we get that B = AJ

and we conclude similarly.

Prove that

D(A) = 0 (1.9)

if one row or one column of A is 0.

Now, we have accumulated some results, (1.1)–(1.8), and it is not apparent whatto use. So to get some idea, let us deal with the rows, and try to simplify thequestion. Using the action of J (1.4), we see that

J

a b0 0

=

0 0a b

and taking D of both sides, we see that it suffices to prove the result when thesecond row is 0.6 But, actually, examining the equality above, we see that since thesecond row is 0, it was not important what the second column of J was and theequality above holds with many other matrices. Indeed, for all c, d, we get that

0 c1 d

=

a b0 0

=

0 0a b

.

5another way to conclude without using that D(I − 2) = 1 is to say that if D(0) = 0, then allmatrices have the same image (1) under D, which again is ruled out by (1.3)

6Multiplying on the right by J , we also see that we can assume that a < b, although we won’tuse this fact.



And then, we can try some special choices for c or d.A first obvious choice leads to the result, since taking c = d = 0 we get a matrix

J such that

(J )2 =

0 01 0

2= 0.

Hence using (1.7), we see that D(J ) = 0, and consequently, using (1.1)

0 = D

J

a b0 0

= D

0 0a b

This proves the result for the case when a row is 0, and the case when a column istreated similarly.

Another special choice for J could be c = 0 and d = 1. Then

J J =

0 01 1

J =

0 01 1

and taking D of this equality, we get that D(J ) = −D(J ). Hence D(J ) = 0, andwe conclude similarly.

Prove that D(A) = 0 whenever A is singular

To prove this fact, we need to express the information A is singular by somethingmore amenable to analysis.

Let

A =

a bc d

be a singular matrix. Since A is a square matrix, this is equivalent to saying thatthe system AX = 0 has a nontrivial solution X = (α, β )T with α = 0 or β = 0, i.e.that

a b

c dα

β = 0

0 . (1.10)

But this looks encouraging: we have the product of two matrices that give a newmatrix with a column which is 0. If we were dealing with square matrices, we couldapply (1.1) and (1.9) and conclude. So let us try to make our column matrices intosquare matrices. The first trial gives

AB =

a bc d

α γ β δ

=

0 aγ + bδ 0 cγ + dδ

= C

and taking D of this equality gives D(A)D(B) = D(C ) = 0 after using (1.1) and(1.9). To conclude, we just need to be able to say that D(B) = 0. And using theresult of the second question, we see that this is the case if B is invertible. Hence,we just need to choose γ , δ such that B is invertible. But det B = αδ − βγ , so wesee that choosing δ = α and γ = −β we get an invertible matrix since (α, β ) wasnot trivial.

To sum up, using (1.10), we get that

AB =

a bc d

α −β β α

=

0 −aβ + bα0 −cβ + dα

and evaluating D of both side and using (1.1) and (1.9), we get that

D(AB) = D(A)D(B) = 0,



and since det B = α2 + β 2 > 0, B is invertible, so D(B) = 0, and consequentlyD(A) = 0. Since we have made no assumption on A (except that it is noninvertible),we get that for all noninvertible matrices A, D(A) = 0 and the proof is complete.

2. Page 155, Exercises 1,2,3,7

This section deals with computation of some determinants. Remember that tocompute determinants, one can

(1) Use the multilinearity with respect to rows or columns(2) Add a multiple of a row to another row, or a multiple of a column to another

column without changing the value of the determinant.(3) Switch two rows or two columns. This multiplies the determinant by −1.(4) Expand with respect to a row or a column.(5) Use the fact that if two rows or columns are linearly dependent (in particular

if they are the same, or one i 0!), the determinant is 0.(6) Use the formula for a 2× 2 determinant:

det

a bc d

= ad− bc.

Note that this formula follows from item 4) and the fact that for a 1 × 1determinant, det(a) = a.

We refer to the book for more precise formulas.

2.1. Problem 1. Compute the following determinant:

det A = det

0 a b−a 0 c−b −c 0

.

Here we have three parameters on which we have no information (except that theybelong to a field). To get some intuition, let us first examine some particular cases.Looking at the matrix, it is easy to see that if a = 0, then the determinant is 0.Similarly if b = 0 or c = 0. So we can assume that abc = 0 and that all scalars areinvertible. But in this case, let us try to come back to the case when (say) a = 0.

Using item 2) in the list of properties of the determinant above, we decide todelete the entry a by adding −a/b times the third column to the second. This gives

det A = det

0 0 b−a −ac/b c−b −c 0

.

To get back to a matrix of the original form, but with a = 0, we also need to cancelthe first entry in the second row. To use the symmetry of the matrix, we proceed

as before, but with rows, that is we add −a/b times the third row to the secondrow. This gives

det A = det

0 0 b

0 0 c−b −c 0

and the two first column of this matrix are obviously linearly dependent (and thisis indeed a matrix of the same shape as A, but with a = 0). Thus in all cases,det A = 0.



2.2. Problem 2. Compute the determinant of the Vandermonde matrix

V (a,b,c) = det1 a a2

1 b b2

1 c c2

.

Using item 2) in the list of properties of the determinant, we use the first row todelete the other entries in the first column. This gives

V (a,b,c) = det

1 a a2

0 b− a b2 − a2

0 c− a c2 − a2

.

Now, we use the fact that b2 − a2 = (b− a)(b + a), and similarly for c2 − a2, anditem 1) (the multilinearity) above to factor (b−a) out of the second row and (c−a)out of the third one. This gives

V (a,b,c) = (b− a)(c− a)det

1 a a2

0 1 b + a0 1 c + a

.

Expanding along the first column (item 4)) and using the formula for a 2 × 2determinant (item 6)), we finally get that V (a,b,c) = (c− b)(c− a)(b− a).

2.3. Problem 3. Find all the elements in S3, the symmetric group of 3elements.

We need to find all the permutations of the set X = {1, 2, 3}.First, we can find the permutations that send 1 to 1 (thus, we consider σ such

a permutation, and we assume that σ(1) = 1). For such a permutation, either the

image of 2 is 2, in which case the image of 3 is 3, and we get the identity transfor-mation (no shuffling), (1, 2, 3), or the image of 2 is 3, and we get the transpositionwhich exchange 2 and 3, τ 23 = (1, 3, 2).

Now, we can find the permutations that send 1 to 2, and we look at the imageof 2. It is either 1 or 3. In the first case, we get (2, 1, 3) = τ 12, and in the latter,we get (2, 3, 1) = τ 13 ◦ τ 12.

Finally, the permutations that send 1 to 3 either send 2 to 2, in which case weget τ 13 = (3, 2, 1), or send 2 to 1, and we obtain (3, 1, 2) = τ 12 ◦ τ 13. Finally, we getall the permutations of S3:

Permutation signId 1τ 23 −1

τ 12 −1τ 13 −1(1, 3, 2) 1(3, 1, 2) 1

In particular, one can check that S3 has cardinal 6 = 3! and that for any twopermutations σ1, σ2, there holds that

Sign(σ1 ◦ σ2) = Sign(σ1)Sign(σ2).



2.4. Problem 7. An n × n matrix, M = (mij)1≤i,j≤n is called triangular if

mij = 0 when i < j. Prove that if M is triangular,

det M = m11m22 . . . mnn,that is to say, the determinant of M is the product of the diagonal entries

of M .To prove this, we first observe that this is true for a 2× 2 matrix, using item 6)

in the list above:

det

a 0b c

= ac.

And for a general triangular matrix, we see that to compute the determinant, it isnatural to expand with respect to the first row (item 4)). But this gives

det A = det

a11 0X A2

= a11 det A2, (2.1)

and if A was a triangular matrix, A2 is again a triangular matrix, but of smaller

size, (n− 1)× (n− 1).So, suppose that the formula is true for all matrices of size n, and let

A = (aij)1≤i,j≤n+1

be a matrix of size n + 1. Since A is triangular, we can define a new n× n matrix,A2 = (mij)1≤i,j≤n such that A has the shape in (2.1), that is

mij = ai+1,j+1

with this formula, we see that if i < j, then i+1 < j +1 and hence mij = ai+1,j+1 =0, so that A2 is also triangular. Using the equality in (2.1), and the fact that theformula is true for n× n matrices (hence for A2), we get

det A = a11 det A2 = a11a22a33 . . . an+1,n+1.

This is the formula for (n + 1) × (n + 1) matrices. Consequently, the formula is

true for triangular matrices of all size (it is true for matrices of size 2 × 2, hence formatrices of size 3× 3, hence for matrices of size 4× 4, . . . ).

Brown University

E-mail address: [email protected]

Documents

Correction Homework 10