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Main result

We now prove that a left inverse of a square matrix is also a right inverse. In other words, we show the following:

Let \(A, N \in \mathbb{F}^{n\times n}\) where \(\mathbb{F}\) denotes a field. If \(NA = I\), then \(AN = I\).

Before we look at the proof, note that the above statement also establishes that a right inverse is also a left inverse because we can view \(A\) as the right inverse of \(N\) (as \(NA = I\)) and the conclusion asserts that \(A\) is a left inverse of \(N\) (as \(AN = I\)).

To prove the above statement, we first establish the claim that \(Ax = y\) has a solution for all \(y \in \mathbb{R}^n\).

The claim is not true if \(A\) does not have a left inverse. We postpone the proof of this claim to the end. Let's see how we can use this claim to prove the main result.

Take an arbitrary element in \(\mathbb{F}^n\) and call it \(y\). Then the above result tells us that there is \(x' \in \mathbb{F}\) such that \(Ax' = y\). Multiplying both sides on the left by \(N\), we get that \( N(Ax') = Ny\), giving \( (NA)x' = Ny\) by associativity of matrix multiplication. As \(NA = I\), we have \(x' = Ny\).

Hence, \(y = Ax' = A(Ny) = (AN)y\). Let \(D\) denote the product \(AN\). So \(y = Dy\). But \(y\) is arbitrary. We must have \(D = I\).

Proof of claim

Suppose that there exists \(y' \in \mathbb{F}^n\) such that \(Ax = y'\) has no solution. Now, row reduce \([A~y']\) to \([R~d]\) where \(R\) is in reduced row-echelon form.

As \(Ax = y'\) has no solution, there must be an \(i\) such that row \(i\) of \(R\) has all 0's and \(d_i \neq 0\). Hence, because \(R\) is a square matrix, not every column of \(R\) can be a pivot column. So there is at least one free variable, implying that there is a nonzero \(\tilde{x} \in N(A)\) (i.e., \(Ax' = 0_n\)). But \(\tilde{x} = I \tilde{x} = (NA)\tilde{x} = N(A\tilde{x}) = N 0_n = 0_n\), contradicting that \(\tilde{x}\) is nonzero! So the assumption that there exists \(y' \in \mathbb{F}^n\) such that \(Ax = y'\) has no solution cannot be true.

Quick Quiz


Let \(A = \begin{bmatrix} 2 & 0 \\ -1 & 0 \\ 1 & 1\end{bmatrix}\) and let \(N = \begin{bmatrix} 1 & 1 & 0\\ -1 & -1 & 1 \end{bmatrix}\). Compute the products \(NA\) and \(AN\). Do the answers contradict our main result? If not, why not?