Chapter 5. Matrices. 5.1 Inverses, Part 1

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1 Chapter 5 Matrices The classification result at the end of the previous chapter says that any finite-dimensional vector space looks like a space of column vectors. In the next couple of chapters we re going to take this further by showing that rewriting vectors as column vectors turns linear transformations into matrices. Before we can do this, we need to collect some information and set up some notation about matrices. Some of this stuff you will have already seen, but maybe not in this generality. I am not going to remind you about how and when you can multiply matrices, or all of the basic properties of matrix multiplication, such as associativity: ABC = ABC whenever this makes sense. Instead, we ll concentrate on the notion of an inverse and the determinant for square matrices. 5.1 Inverses, Part 1 Recall that for a positive integer n, the identity matrix I n is the n n square matrix with 1s on the main diagonal and 0s elsewhere. Eg., I 1 = 1, I 2 =, I 3 = 0 1 0, etc The identity matrix gets its name because given any n n matrix A with entries in a field K, we have AI n = I n A = A, so I n acts as an identity element for the set of n n matrices under multiplication. If the context is clear, then we sometimes drop the n and write I for the identity matrix. In analogy with the usual sort of multiplication eg. for numbers in a field, we say that an n n matrix A with entries in K is invertible if there exists a n n matrix B with AB = I = BA. Lemma 5.1. If A is invertible, then the inverse of A is unique. Hence we are safe writing A 1 for the inverse of an invertible matrix A. Proof. Suppose B and C are two inverses of A, so AB = I = BA and AC = I = CA. Then B = IB = CAB = CAB = CI = C. 29

2 Examples 5.2. Here are some basic examples: i The matrix has inverse. 1 4 ii A general 2 2 matrix A = γ δ only if αδ βγ 0 K. If αδ βγ 0, then A 1 = 1 αδ βγ you can verify by direct calculation with entries in a field K is invertible if and δ β, which γ α iii Any n n matrix A with entries in K can be viewed as a linear transformation from the space K n to itself. Then A is invertible if and only if it is an isomorphism when viewed as a linear transformation. 5.2 Determinants The function αδ βγ on the entries of a 2 2 matrix A = given in Example γ δ 5.2ii above is called the determinant. This important function can be defined for any square matrix. In order to do that, we first need to recall a couple of very basic things about symmetric groups. Definition 5.3. Given a positive integer n, the symmetric group S n is the set of bijections from the set {1,2,...,n} to itself. It has the following basic properties among others: i The order of S n is n!. ii The elements of S n can be written as products of disjoint cycles, for example: σ = is the element of S 6 which sends 1 5, 5 4, 4 1, 2 3, 3 2, 6 6. iii There is a map sgn : S n {1, 1} called signature, defined as follows: for σ S n, write σ as a product of disjoint cycles. Let l 1,...,l r be the lengths of the cycles. Then sgnσ = r i=1 1l i 1. With this in hand, we can define determinants. Definition 5.4. Let A = α ij be an n n matrix with entries in K. Then we define the determinant of A, denoted deta, as follows: deta := σ S n sgnσ n α σj,j. The function det maps matrices with entries in K to elements of K. 30

3 Examples 5.5. The definition of the determinant is not easy to get to grips with without doing some examples. Here are a couple: i Let s check that the two definitions coincide for 2 2 matrices. Let A = γ δ as before. Then, in the notation of the definition above, we have α 11 = α, α 12 = β, α 21 = γ and α 22 = δ. Further, we are interested in the symmetric group S 2 = {12,12}. Then the formula in Definition 5.4, together with the definition of the signature sgn, gives deta = sgn12α 11 α 22 +sgn12α 21 α 12 = αδ+ 1 1 γβ = αδ βγ, as we had before. ii Working over the real numbers, set A = 1 2 π e Then we re in the case n = 3, so we have to consider So. S 3 = {123, 123, 132, 231, 123, 132} with signatures deta = α 11 α 22 α 33 α 21 α 12 α 33 α 31 α 22 α 13 α 11 α 32 α 23 +α 21 α 32 α 13 +α 31 α 12 α 23 = π e 1 = 3 8 3π 2 6e 1. We now give an extremely important result about determinants: the determinant function is multiplicative. Lemma 5.6. IfAandB aren nmatricesoverafieldk, thendetab = detadetb. Remark. For such a fundamental result, the proof is rather tricky to get at. You can, in theory, jump straight in and try to manipulate the various determinants involved, but this is very longwinded. There s a shorter proof in the appendix to this chapter, but it is not examinable. For your piece of mind, you might at least want to check the result directly for 2 2 matrices and if you re brave 3 3. In any case, you should certainly remember the result! Corollary 5.7. If A is invertible, then deta 0 and deta 1 = deta 1. 31

4 Proof. If A 1 exists, then we have by Lemma 5.6, which gives the result. 1 = deti = detaa 1 = detadeta 1, This result shows that the determinant is intimately related to the existence of the inverse for a matrix. There are lots and lots of different ways of calculating determinants. One that you may be familiar with is the Laplace expansion or cofactor expansion: Definition 5.8. Let A = α ij be an n n matrix with entries in K. For each 1 i,j n we can obtain an n 1 n 1 matrix by deleting the i th row and j th column of A. The determinant of this matrix is called the i,j-minor of A, denoted M ij. The i,j- cofactor of A is then defined to be C ij := 1 i+j M ij. Then the Laplace expansion along the i th row is given by the formula: deta = α ij C ij. Similarly, the Laplace expansion along the j th column is: deta = α ij C ij. i=1 Note that the formula in each case looks very similar; however, in the row expansion formula i is fixed and the js are changing, whereas in the column expansion formula j is fixed and the is are changing. It is an exercise to show that the different Laplace expansions all agree, and that they agree with the definition of the determinant. The point of the Laplace expansion is that it reduces the problem of finding an n n determinant to the problem of finding several n 1 n 1 determinants. Therefore, in principle, you can find any determinant by applying this formula recursively until you get down to matrices which are small enough to deal with easily. Example 5.9. As usual, an example will help. Let s do the matrix 1 2 π A = 0 1 e from Example 5.5 again. Looking at Laplace s formula, it is often sensible to choose a row or column which has 0s in it, because that reduces the number of calculations. Let s expand along the second row, so the formula becomes: deta = α 2j C 2j. 32

5 Since α 21 = 0, we don t need to worry about the first term which is why I chose the second row. Now C 22 = π det = π so deta = C 23 = 1 5 det 5.3 Inverses Part = π 6e 1 = 3 8 3π 2 6e 1, which is the same as we had before. We know that an invertible matrix must have non-zero determinant. In this section, we show the converse is true, and write down the general form for the inverse of a matrix. Definition Suppose A is an n n matrix with entries in K. The cofactor matrix of A is the matrix whose i,j-entry is the i,j-cofactor of A, that is cofa := C ij. The adjugate matrix of A is the transpose of the cofactor matrix, that is adja := cofa has i,j-entry the j,i-cofactor of A. Theorem Let A be an n n matrix with entries in K. Then AadjA = detai = adjaa. In particular, if deta 0, then A is invertible and A 1 = 1 deta adja. Proof. The result is trivial and basically pointless if n = 1, so assume n 2. Let s write down a formula for the i,j-entry of AadjA: α ik C jk k=1 We first note that the diagonal entries of AadjA are all equal to deta, because if i = j in the above formula we get the Laplace expansion of deta along the i th row. If i j, then we want the result to be 0. The formula still looks a bit like the Laplace expansion of a determinant, and it is: it is the determinant of the matrix obtained from A by replacing the j th row with another copy of the i th row, so that α ik = α jk. So we are done if we can prove that the determinant of any matrix with two rows the same is zero. We proceed by induction on n. First note that the result is true for 2 2 matrices: det = αβ βα = 0. Now assume M is a n n matrix with two rows the same, and n 3. The value of the determinant only changes by a factor of 1 each time we swap rows so we may assume 33

6 that rows 2 and 3 of M are the same. Now expand detm along the first row. For each 1 j n, the 1,j-minor M 1j is the determinant of an n 1 n 1 matrix with the first two rows the same, so M 1j = 0 by the induction hypothesis. Hence each cofactor C 1j = 1 1+j M 1j = 0 and therefore detm = 0. We have proved what we wanted, by induction. Adding together all these ingredients, we get that A adja has diagonal entries deta and off-diagonal entries 0, so A adja = detai. A similar proof, using expansion along columns instead of rows, works to show adjaa = detai. Now, if deta 0, then we can divide through to get A 1 deta adja 1 = deta adja A = I, so A is invertible with A 1 = 1 adja, as required. deta Corollary A is invertible if and only if deta 0. Example Let s just check this all works for 2 2 matrices. Suppose A =, γ δ with deta = αδ βγ 0. Then each minor is the determinant of a 1 1 matrix, which is easy to deal with, so we get M 11 = δ, M 12 = γ, M 21 = β and M 22 = α. Adding in the correct powers of 1, we get the cofactor matrix cofa = δ γ β α. Taking the transpose, we get adja = δ β γ α. Finally, dividing by the determinant, we get A 1 1 = αδ βγ δ β γ α, which is what we had before. 5.4 Some More Important Properties of Matrices Often in algebra, one is interested in finding the simplest representation of a particular class of objects. As the course goes on, you ll see plenty of examples of this idea for matrices, and a lot of them are based on the following concept: 34

7 Definition Let A and B be n n matrices with entries in a field K. Then we say that A is similar to B if there exists an invertible matrix C with B = CAC 1. We collect some properties of similar matrices. Lemma i Similarity is an equivalence relation on the set of n n matrices over a field K. ii Similar matrices have the same determinant. Proof. i. Since A = IAI 1 for all matrices A, A is similar to itself reflexive. If B = CAC 1, then A = C 1 BC = C 1 BC 1 1, so A is similar to B implies B is similar to A symmetric. If B 1 = C 1 AC1 1 and B 2 = C 2 B 1 C2 1, then B 2 = C 2 C 1 AC1 1 C2 1 = C 2 C 1 AC 2 C 1 1 transitive. ii. detcac 1 = detcdetadetc 1 = detcdetadetc 1 = deta. One of the important things you ll do later on in the course is work out nice representatives of the equivalence classes for the similarity relation, using the Jordan Normal Form. We finish with another property of matrices which is preserved under similarity. Definition Let A = α ij be an n n matrix with entries in K. The trace of A is the sum of the diagonal entries of A, that is tra := α ii. i=1 Lemma Let A = α ij and B = β ij be n n matrices. Then: i tra+b = tra+trb; ii trab = trba; iii Similar matrices have the same trace. Proof. i. The i,i-entry of A + B is α ii + β ii, so tra + B = n i=1 α ii + β ii = n i=1 α ii+ n i=1 β ii = tra+trb. ii. The i,i-entry of AB is n α ijβ ji. The i,i-entry of BA is n β ijα ji. So trab = = i=1 α ij β ji β ji α ij i=1 = trba. iii. trcac 1 = trac 1 C = trai = tra, using part ii to swap the order of AC 1 and C. 35

8 Example Note that although similar matrices have the same trace and determinant, the converse is not true: the matrices A = and B have the same trace and the same determinant, but there does not exist C with B = CAC 1, because CAC 1 = A for all invertible C. Appendix to Chapter 5 This section is included for completeness, and the proofs contained here are not examinable at all. However, everyone should see a proof that the determinant is multiplicative at least once, so here goes! We begin by collecting some properties of the determinant function det : M n K K: Lemma i If two rows of A are equal, then deta = 0. ii IfthematrixB isobtainedfrom thematrixabymultiplyingonerowbyaconstant, then detb = λdeta. iii If the matrix B is obtained from the matrix A by adding one row to another, then detb = deta. Proof. We ll freely use the Laplace expansion of the determinant when we need it. i. A proof of this was given in the proof of Theorem 5.11 above. It relied on the fact that swapping two rows of a matrix simply changes the sign of the determinant. To see this recall that, by definition, deta = σ S n sgnσ n α σj,j. Swapping rows r and s, say, is equivalent to applying the transposition r, s to this formula. Since sgnr,s = 1, this just changes the sign of the determinant, as claimed. ii. Suppose the i th row of A = α jk is multiplied by λ to get B = β jk. Then, expanding along the i th row, all the cofactors for A and B are the same, so we get detb = β ij C ij = λα ij C ij = λ α ij C ij = λdeta, as required. iii. Suppose the i th row of A is added to the j th row to get the new j th row of B. Expand along this j th row to get: detb = β jk C jk = k=1 α jk +α ik C jk = k=1 α jk C jk + k=1 α ik C jk. The first term on the right hand side is deta, and the second term is the determinant of a matrix whose i th and j th rows are equal, which gives 0 by part i. So we are done. Properties ii and iii in the lemma above show us how elementary row operations affect the determinant. Next we recall that these operations can also be achieved using matrix multiplication. 36 k=1

9 Definition We define two types of elementary matrix: i For each λ K and each 1 i n, define the n n matrix R λ,i to be the diagonal matrix with a λ in the i th diagonal position and 1s in all other diagonal positions. ii For each 1 i,j n with i j, define the matrix R j i to be the n n matrix with 1s on the diagonal, a 1 in the i,j-position and 0s elsewhere. Lemma With the notation given above, for any matrix A M n K, we have: i For all λ K and 1 i n, R λ,i A is the matrix obtained from A by multiplying the i th row by λ. ii For all 1 i,j n with i j, R j ia is the matrix obtained from A by replacing the i th row with the row obtained by adding the i th and j th rows. iii We have detr λ,i A = λdeta for all λ and i and detr j ia = deta for all i,j. Proof. The statements in i and ii are easy to check. Part iii follows from Lemma 5.19ii and iii. We can now give a proof that invertible matrices have nonzero determinant which doesn t use the multiplicative property recall that our original proof in Lemma 5.6 used the multiplicative property Lemma A matrix A is invertible if and only if deta 0. Proof. Suppose deta 0. Then the proof of Theorem 5.11 still goes through without the multiplicative property, so A is invertible. Conversely, suppose A is invertible. Then, viewing A as a linear transformation from K n to K n, A is an isomorphism of vector spaces, so A has rank n. Since the rank of a matrix as a linear map equals its row rank, the rows of A must be linearly independent. Hence, by performing row operations on A which is just taking linear combinationsof therows, we can row reduceatothe identity matrix I n. In the language of Definition 5.20, we find a sequence of elementary matrices R 1,R 2,...,R r such that I n = R r R r 1 R 1 A. At each step, the determinant of the new matrix stays unchanged, or gets multiplied by some λ K. Since we end up with the identity matrix, we never multiply a row by 0 because that is an irreversible step, so we conclude from Lemma 5.21 that deti n = 1 = detr r R r 1 R 1 A = µdeta for some nonzero µ K. Hence deta = µ 1 0, as required. We finally need a result which will help us deal with the case where our matrices are not invertible. Lemma SupposeAisnotinvertible. ThenAB isnotinvertibleforanyb M n K. Proof. If B is not invertible either, then viewing B as a linear transformation on K n, we can find a nonzero vector v K n with Bv = 0, because B must have nontrivial kernel. But then ABv = A0 = 0, so AB has nontrivial kernel, so AB is not invertible. If B is invertible, with inverse B 1, choose nonzero v in the kernel of A. Then, since B 1 is invertible, we must have that u = B 1 v 0. But ABu = ABB 1 v = AI n v = Av = 0, so AB again has nontrivial kernel, so is not invertible. 37

10 We can now prove the result we want by applying the preceding lemmas. Theorem ForanytwomatricesA,B M n K, wehavedetab = detadetb. Proof. Suppose first that A is not invertible, then deta = 0, by Lemma Also, AB is not invertible by Lemma 5.23, so detab = 0 by Lemma 5.22 again. Therefore, we have 0 = detab = detadetb, as required. Now suppose A is invertible. Then, by the argument in the proof of Lemma 5.22, we can find a sequence of elementary matrices R 1,...,R r with R r R r 1 R 1 A = I n. Let R = R r R 1 ; then, also by the arguments in that Lemma, detrc = deta 1 detc for any C M n K. In particular, then, we have detb = deti n B = detrab = deta 1 detab. Multiplying through by deta, we get the result we want. Note that this proof works because we re working over a field K, so we can apply things we know about fields and vector spaces. In general, one has to be even more careful, but this is definitely beyond what we need to do now. 38

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