Left or Right Inverse of Matrix is Inverse

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Theorem

Let $\mathbf A, \mathbf B$ be square matrices of order $n$ over a commutative ring with unity $\left({R, +, \circ}\right)$.

Suppose that:

$\mathbf A \mathbf B = \mathbf I_n$

where $\mathbf I_n$ is the unit matrix of order $n$.


Then $\mathbf A$ and $\mathbf B$ are invertible matrices, and furthermore:

$\mathbf B = \mathbf A^{-1}$

where $\mathbf A^{-1}$ is the inverse of $\mathbf A$.


Proof

When $1_R$ denotes the unity of $R$, we have:

\(\ds 1_R\) \(=\) \(\ds \map \det {\mathbf I_n}\) Determinant of Unit Matrix
\(\ds \) \(=\) \(\ds \map \det {\mathbf A \mathbf B}\) by assumption
\(\ds \) \(=\) \(\ds \map \det {\mathbf A} \map \det {\mathbf B}\) Determinant of Matrix Product


From Matrix is Invertible iff Determinant has Multiplicative Inverse, it follows that $\mathbf A$ and $\mathbf B$ are invertible.

Then:

\(\ds \mathbf B\) \(=\) \(\ds \mathbf I_n \mathbf B\) Unit Matrix is Unity of Ring of Square Matrices
\(\ds \) \(=\) \(\ds \paren {\mathbf A^{-1} \mathbf A} \mathbf B\) Definition of Inverse Matrix
\(\ds \) \(=\) \(\ds \mathbf A^{-1} \paren {\mathbf A \mathbf B}\) Matrix Multiplication is Associative
\(\ds \) \(=\) \(\ds \mathbf A^{-1} \mathbf I_n\) by assumption
\(\ds \) \(=\) \(\ds \mathbf A^{-1}\) Unit Matrix is Unity of Ring of Square Matrices

$\blacksquare$


Sources

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