The Inverse of a Matrix

We now have enough tools to be able to determine whether a given square matrix is invertible or not, and - if it is - we can also compute the inverse by an efficient method.

Theorem Let A be an n x n matrix. Then the following conditions are equivalent.
(a) A is invertible
(b) A has a left inverse (that is, there is a matrix B with BA = I_n).
(c) A has a right inverse (that is, there is a matrix C with AC = I_n).
(d) The columns of A span R ⁿ .
(e) For every b in Rⁿ , the system Ax = b is consistent.
(f) The rank of A is n.
(g) The columns of A are linearly independent.
(h) The only solution to Ax = 0 is 0.
(i) The nullity of A is zero.
(j) The reduced row echelon form of A isI_n.
(k) A is a product of elementary matrices.

We have already established most of these results: for instance, we know that (d), (e) and (f) are equivalent; (g), (h), and (i)are equivalent, and (f) and (i) are equivalent, because the number of columns is n. We also know that (f) is equivalent to (j). Now, if A is invertible, then every equation Ax = b is consistent, because it has the solution x = A^-1b. This implies (by the known equivalences) that (j) holds, and from what we already know about elementary matrices, this means that PA = I_n for P a product of elementary matrices, and hence A = P^-1 is also a product of elementary matrices. On the other hand, if A is a product of elementary matrices, we know that it is invertible (since each of them is invertible). So (a) is equivalent to each of (d) through (k).

The final statements to check are about right and left invertibility. Clearly (a) implies both (b) and (c), so we only need to check that each of (b) and (c) implies (a). If (b) holds, and Ax = 0, then BAx = 0 = I_n x = x, and hence (h) holds, and we know that (a) follows. On the other hand, if (c) holds then the already-established equivalence of (b) with (a) means that C is invertible, and by the proof of uniqueness of inverses, this forces A to be the inverse to C, which means that A is invertible.

Examples

The theorem also gives us a practical way of computing A^-1: if A is invertible, then property (j) means that there is a sequence of elementary row operations, corresponding to multiplication by the product of elementary matrices P, such that PA = I_n, and hence P = A^-1. Now if we augment the matrix A by the matrix I_n, and consider the sequence of row operations on the augmented matrix [A I_n], notice that the result is the matrix [I_n P].

Algorithm for Matrix Inversion
If A is an n x n matrix, form the augmented matrix [A I_n] and compute its reduced row echelon form.
(a) If A is not invertible, then in the reduced row echelon form [R P], R is NOT the matrix I _n .
(b) If A is invertible, then the reduced row echelon form is [I_n A^-1].

Examples

Application: Rank
Let A be an m x n matrix, and let P be an invertible m x m matrix and Q an invertible n x n matrix.
Then rank(PA) = rank(A) and rank(AQ) = rank(A).

To see this, first let R denote the reduced row echelon form of PA. Then there is a product of elementary matrices, call it S, such that SPA = R. But SP is also then a product of elementary matrices (since P is) and because (SP)A = R, we see that R must also be the reduced row echelon form of A. Hence PA and A have the same rank. On the other hand, with R the reduced row echelon form of A, notice that the nullity m is the numberof zero rows of R, all of which come at the bottom. If T is a product of elementary matrices such that TA = R, thenTAQ = RQ, and the last m rows of RQ are all zero. So the nullity of AQ is at least as big as the nullity of A. The same argument again shows that the nullity of AQQ^-1 (= A) is at least as big as the nullity of AQ. But this means that the nullity of A and of AQ must be the same, and so their ranks are the same as well.

250 Lecture Index