Exercise 3.2.4

Prove that if \(\mathbf{A}\) is an invertible square matrix, then \(\mathbf{A}^+=\mathbf{A}^{-1}\).

First we note that if \(\mathbf{A}\) is invertable then so is its transpose:

\[ \begin{aligned} \mathbf{A} \mathbf{A}^{-1} &= I\\ (\mathbf{A} \mathbf{A}^{-1})^T &= I\\ (\mathbf{A}^{-1})^T \mathbf{A}^T &= I \end{aligned} \]

So the inverse of \(\mathbf{A}^T\) is \((\mathbf{A}^T)^{-1} = (\mathbf{A}^{-1})^T\) (Sometimes written as \(\mathbf{A}^{-T}\)) So with that we can use the fact that the inverse of a product of two matrices is the product of the inverses in reverse order to find:

\[ \begin{aligned} \mathbf{A}^+ &= (\mathbf{A}^T\mathbf{A})^{-1}\mathbf{A}^T \\ &=\mathbf{A}^{-1}(\mathbf{A^T})^{-1}A^T\\ &= \mathbf{A}^{-1} \end{aligned} \]