Inverse of Transpose of Matrix is Transpose of Inverse

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Theorem

Let $\mathbf A$ be a matrix over a field.

Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.

Let $\mathbf A$ be an nonsingular matrix.


Then $\mathbf A^\intercal$ is also nonsingular and:

$\paren {\mathbf A^\intercal}^{-1} = \paren {\mathbf A^{-1} }^\intercal$

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


Proof

We have:

\(\ds \paren {\mathbf A^{-1} }^\intercal \mathbf A^\intercal\) \(=\) \(\ds \paren {\mathbf A \mathbf A^{-1} }^\intercal\) Transpose of Matrix Product
\(\ds \) \(=\) \(\ds \mathbf I^\intercal\) Definition of Inverse Matrix: $\mathbf I$ denotes Unit Matrix
\(\ds \) \(=\) \(\ds \mathbf I\) Definition of Unit Matrix

Hence $\paren {\mathbf A^{-1} }^\intercal$ is an inverse of $\mathbf A^\intercal$.

From Inverse of Square Matrix over Field is Unique:

$\paren {\mathbf A^{-1} }^\intercal = \paren {\mathbf A^\intercal}^{-1}$

$\blacksquare$


Sources

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