Computing the inverse of a matrix over a field

Let A be an n \times n matrix over a field.

  1. Suppose B = [B_1 | \cdots | B_n] is a matrix with columns B_i, and let e_i denote the ith standard basis vector. Prove that B is an inverse of A if and only if B_i is a solution of the matrix equation AX = e_i for each i.
  2. Prove that A has an inverse if and only if A is row equivalent to the identity matrix.
  3. Prove that A has an inverse B if and only if [A|I] and [I|B] are row equivalent.

We begin with a lemma.

Lemma: Let A and B be matrices of dimension n \times n over a field. If AB = I, then BA = I. Proof: Thinking of A and B as linear transformations, since AB is injective, A is injective. Since we are working over a vector space of finite dimension, A is also surjective. So A is an automorphism. Thus there exists a matrix C such that CA = I, and we have C = CAB = B. So BA = I. \square

Thus, in \mathsf{Mat}_n(F), to show a matrix is invertible it suffices to show left- or right-invertible.

  1. Suppose B is an inverse of A. Then we have AB = [AB_1 | \cdots | AB_n] = [e_1 | \cdots | e_n]. In particular, B_i is a solution of the equation AX = e_i for each i. Conversely, if AB_i = e_i for each i, then AB = [AB_1 | \cdots | AB_n] = [e_1 | \cdots | e_n] = I. Thus B is an inverse of A.
  2. Suppose there is a matrix B such that BA = I. In particular, B is invertible, and so A is row equivalent to the identity matrix. Conversely, if A is row equivalent to the identity matrix, then BA = I for some invertible matrix B. By the lemma, B is an inverse of A.
  3. Suppose A is invertible with inverse B. Then B[A|I] = [BA|B] = [I|B], so that [A|I] and [I|B] are row equivalent. Now suppose [A|I] and [I|B] are row equivalent, say by an invertible matrix P. Then P[A|I] = [PA|P] = [I|B]. Thus P = B and BA = I, so that B is an inverse of A.
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