In this post I attempt to prove a very simple identity, that will be used in a later post.
I want to prove that for a square, non-singular matrix \(A\), we have (in the steps below, \(I\) is the identity matrix):
\begin{align} (A^T)^{-1} = (A^{-1})^{T} \end{align}
Let \((A^T)^{-1} = B\), we have
\begin{align} (A^T)^{-1} & = B \\ \implies (A^T)^{-1} A^T & = B A^T \\ \implies I & = B A^T \tag{1} \\ \implies I^T & = (BA^T)^T \\ \implies I & = AB^T \tag{2} \\ \implies A^{-1}I & = A^{-1}AB^T \\ \implies A^{-1} & = B^T \\ \implies (A^{-1})^T & = (B^T)^T \\ \implies (A^{-1})^T & = B \\ \implies (A^{-1})^T & = B = (A^T)^{-1} \end{align}
This completes the proof. I use the following identities in the labeled equations above:
- Equation (1) A square, non singular matrix multiplied by its inverse is the identity matrix
\((A^T)^{-1} A^T = I\). - Equation (2) comes from the transpose of a matrix product identity
\((BA^T)^T = AB^T\).