MIT18_024s11_ChB2notes

MIT18_024s11_ChB2notes - The - inverse of a matrix We now...

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-- The inverse of a matrix We now consider the pro5lern of the existence of multiplicatiave inverses for matrices. At this point, we must take the non-commutativity of matrix.multiplication into account.Fc;ritis perfectly possible, given a matrix A, that there exists a matrix B such that A-B equals an identity matrix, without it following that B-A equals an identity matrix. r Consider the following example: Example 6. Let A and B be the matrices 0 ] 0 = - Then A-B = I2 , but B-A # I3 , as you can 1-1 check. I Definition. Let A be a k by n matrix. A matrix B of size n by k is called an inverse for A if both of the following equations hold: = Ik and = In . We shall prove that if k # n, then it is impossible for both these equations to hold. Thus only square matrices can have inverses. We also show that if the matrices aro,square and one of these equations holds, then the other equation holds as vell! Theorem 13. Let A be a matrix of size k by n. Then A has an inverse if and only if k = n = rank A. If A has an inverse, that inverse is unique.
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,Proof. Step 1. If B is an n by k matrix, we say B is a right inverse for A if A-B = Ik . We say B is a leEt inverse for A if B-A = In . QA&a.k be the rank that if A has a right inverse, then r = k; ar.d if A has a left inverse, then r = n. I of the theorem follows. Fjrst, suppose B is a right inverse for A . Then A-B = I It k * follows that the system of equations A.X = C has a solution for arbitrary C, for the vector X = B.C is one such solution, as you can check. Theorem 6 then implies that r must equal k. Second, suppose B is a left inverse for A. = In . It A-X = - 0 has only the trivial solution, for the equation = g implies that B-(A*X) = 2 , whence X = d. Nc;w the dimension of the solution space of the system = Q is n - r ; it follows that n - r = 0. * Step 2. Now let A be an n by n matrix of rank n. We show there is a matrix B such that A.B = . i i Because the rows of A are independent, the system of equations = C has a solution for arbitrary C. In particular, it has a solution when C is one of the unit coordinate vectors Ei in Vn. Let us choose Bi SG that for i = l...,n. Then if B is the n by n matrix whose successive columns are B1,. . . ,B ' the product A.B equals the matrix whose successive n columns are El, ..., E that is, A.B = I, ni . 1 Step 3. We show that if A and B are n by n matrices and . A-B = , then BmA = In. The "iEn part of the theorem follows.
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Let us note that if we apply Step 1 to the case of a square matrix \ of size n by n , it says that if such a matrix has either a right inverse or a left inverse, then its rank must be n. Now the equation A.B = In says that A has a right inverse and that B has a left inverse. Hence both A and B must have rank n. Applying Step 2 to the matrix B, we see that there is a matrix C such that B.C = I n . Now we compute The equation B*C = In now becomes B-A = In , as desired. Step 4. The computation we just made shows that if a matrix has an inverse, that inverse is unique. Indeed, we just showed that if B has an left inverse A and a ri~ht C, then A = C. k:tus state the result proved in Step 3 as a separate theorem: Theorem 14. If A and B are n
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This note was uploaded on 01/19/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024s11_ChB2notes - The - inverse of a matrix We now...

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