Lecture Notes: Matrix Inverse. 1 Inverse Definition

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1 Lecture Notes: Matrix Inverse Yufei Tao Department of Computer Science and Engineering Chinese University of Hong Kong Inverse Definition We use I to represent identity matrices, namely, diagonal matrices where all the elements on the main diagonal are. Definition. Let A and B be n n matrices. If AB = I, then we say that B is the inverse of A, denoted as A. For example, let A = You can verify that AB = 4,B = -4-4 /2. Hence, B = A. We state the next non-trivial but well-known result without proof: Lemma. Let A and B be n n matrices. If AB = I, then BA = I. In other words, if B is A, then A = B. The lemma has an important corollary: Corollary. Let A be an n n matrix. If A exists, then it is unique. Proof. Suppose that there were two different matrices B and C satisfying AB = I and AC = I. Then, by Lemma, we have: which is a contradiction. BA = I BAC = IC B(AC) = C BI = C B = C

2 It is important to note that some square matrices have no inverse. For example, has no inverse (you are encouraged to make an attempt to find it, and see where you will get stuck). The next lemma gives the precise condition for the existence of inverses: Lemma 2. An n n matrix A has an inverse if and only if det(a). Proof. If-Direction. If det(a), then the linear system Ax = b has a unique solution for any b. Denote by b i the i-th column ( i n) of the n n identity matrix I, and x i the solution of the system Ax i = b i. Then, we obtain A by placing x i as the i-th column of A, for each i [,n. Only-If Direction. If A exists, then given any linear system Ax = b, we have A Ax = A b, which means x = A b. This means that Ax = b has at least one solution. Furthermore, by the uniqueness of A, we know that Ax = b has a unique solution. It thus follows that the rank of A is n, namely, det(a). An n n matrix A is said to be singular if it does not have an inverse; non-singular if it does. 2 Properties of Inverse Lemma 3. Let A,B be n n non-singular matrices. Then, (AB) = B A. Proof. Hence, (AB) = B A. (AB)(B A ) = A(BB )A = AIA = AA = I. Lemma 4. Let A be an n n non-singular matrix. Then, (A T ) = (A ) T. Proof. It suffices to prove that A T (A ) T = I, which is equivalent to proving (A T (A ) T ) T = I T = I. This is true because (A T (A ) T ) T = A A = I. Lemma 5. Let A be an n n non-singular matrix. Then, det(a ) = /det(a). Proof. det(a) det(a ) = det(aa ) = det(i) =. The lemma thus follows. Lemma 6. Let A,B and C be n n matrices. If A is non-singular and AB = AC, then B = C. 2

3 Proof. From AB = AC, we have A AB = A AC, which gives B = C. Lemma 7. Let A and B be n n matrices such that AB = (where is the zero matrix). Then:. If A is non-singular, then B =. 2. If B is non-singular, then A =. 3. (Corollary) If neither A nor B is, then A and B are both singular. Proof. Statement. If A is non-singular, then we have A AB = A, which gives B =. Statement 2. If B is non-singular, then we have ABB = B, which gives A =. 3 Inverse Computation: Gauss-Jordan Elimination We will use an example to illustrate how to compute the inverse of a matrix A. Consider that A = 4 Suppose that we do not know what is A ; hence, we assume: x x 2 x 3 A = x 2 x 22 x 23 x 3 x 32 x 33 Remember that we want 4 x x 2 x 3 x 2 x 22 x 23 x 3 x 32 x 33 = This is essentially to solve three linear systems: x x 2 x 3 x 2 x 22 x 32 x 3 x 23 x 33 = = = () (2) (3) Now we can focus on solving these systems respectively using Gauss Elimination. For example, to solve the linear system (), we look at the augmented matrix: (4) 3

4 Usually, we would start back substitution from here, but now we take a different approach. In particular, we will show that (since the system has a unique solution) it is possible to get rid of back substitution, but instead, continue to use elementary row operations to make the left side of the vertical bar an identity matrix. Then, the solution of the system will present itself. Specifically: (4) 4 /2-4 /2 It is thus clear that x = 4,x 2 = /2,x 3 =. The above method is an extension of Gauss elimination, and is referred to as Gauss-Jordan elimination. Now you may proceed to solve (2) and (3) in the same way. You will then realize that the operations done to the left of the vertical line are always the same. Motivated by this, we can solve all three systems ()-(3) together in one go, as illustrated below: / /2 What is now on the right side of the bar is exactly A. It is important to observe that the above process has in fact embedded the Gauss-Jordan elimination for solving all three linear systems ()-(3). 4 Inverse Formula It is possible to give a general formula for the inverse of an n n non-singular matrix A. As before, given i,j [,n, we denote by M ij the (n ) (n ) matrix obtained from A after discarding its i-th row and j-th column. Also, define: Then we have: C ij = ( ) i+j det(m ij ). Lemma 8. A = det(a) C C 2... C n C 2 C C n C n C 2n... C nn. 4

5 We skip a proof of the lemma, but illustrate it with an example. Example. Consider once again A = 4, We have: det(a) = 2. Also: M = M 2 = M 3 = M 2 = M 22 = M 23 = M 3 = M 32 = M 33 = [ 4-2 [ 4 [ -2 [ 2-2 [ [ 2-2 [ 2 4 [ 4 [ 2, and thus C = 8, and C 2 =, and C 3 = 2, and C 2 = 2, and C 22 =, and C 23 =, and C 3 = 8, and C 32 =, and C 33 = 2 Therefore, by Lemma 8, we have: A = det(a) = 2 C C 2 C 3 C 2 C 22 C 32 C 3 C 23 C =. 4 4 /2. 5