Left and Right Inverses of Square Matrix over Field are Equal

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Theorem

Let $\Bbb F$ be a field, usually one of the standard number fields $\Q$, $\R$ or $\C$.

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $\map \MM n$ denote the matrix space of order $n$ square matrices over $\Bbb F$.


Let $\mathbf B$ be a left inverse matrix of $\mathbf A$.

Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.


Similarly, let $\mathbf B$ be a right inverse matrix of $\mathbf A$.

Then $\mathbf B$ is also a right inverse matrix of $\mathbf A$.


Proof

Consider the algebraic structure $\struct {\map \MM {m, n}, +, \circ}$, where:

$+$ denotes matrix entrywise addition
$\circ$ denotes (conventional) matrix multiplication.

From Ring of Square Matrices over Field is Ring with Unity, $\struct {\map \MM {m, n}, +, \circ}$ is a ring with unity.

Hence a fortiori $\struct {\map \MM {m, n}, +, \circ}$ is a monoid.

The result follows directly from Left Inverse and Right Inverse is Inverse.

$\blacksquare$


There is believed to be a mistake here, possibly a typo.
In particular: That's not what it actually says. What the above link says is that if $\mathbf A$ has both a right inverse matrix and a left inverse matrix, then those are equal and can be called an inverse matrix. It does not say that if $\mathbf B$ is a left inverse matrix then it is automatically a right inverse matrix. I'll sort this out when I get to exercise $1.15$.
You can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.
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Sources

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