Suppose $A$ is symmetric and thus, orthogonally diagonalizable: $A = U\Sigma U^T$, where $UU^T = I$ and $U^TU = I$. $D$ is a diagonal matrix.
Suppose $A$ is not invertible, then would it be correct to write the following?
$$A^{-1} = (U\Sigma U^T)^{-1} = U\Sigma^{-1}U^T$$
A follow-up question is about how one might express the pseudoinverse of $A$ in terms of $U$ and $\Sigma$. Please see the edit below.
Edit: Suppose $A = U\Sigma V^T$ by singular value decomposition. Now $A$ may or may not be symmetric. How can I express the Moore-Penrose pseudoinverse $A^{+}$ in terms of $U, \Sigma, V$?
Best Answer
$\newcommand{\diag}{\mathrm{diag}}$
To its most generality, assume $A \in \mathbb{C}^{m \times n}$, and let $A = U\Sigma V$ be the SVD of $A$, where $U$ and $V$ are order $m$ and $n$ unitary matrices respectively, and $$\Sigma = \diag(\diag(\mu_1, \mu_2, \ldots, \mu_r), 0, \ldots, 0), $$ and $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_r > 0$ are all (positive) singular values of $A$.
It is then straightforward to verify that $$A^+ = V^*\diag(\diag(\mu_1^{-1}, \mu_2^{-1}, \ldots, \mu_r^{-1}), 0, \ldots, 0)U^*$$ is the Moore-Penrose inverse of $A$. Here $V^*$ means the conjugate transpose of $V$.