Suppose $A = U\Sigma U^T$. If $A$ is not invertible, is it correct to write $A^{-1} = (U\Sigma U^T)^{-1}$? 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 $U, \Sigma, V$?
 A: $A$ will fail to be invertible when one of its eigenvalues is $0$. Then that will be one of the diagonal entries of $\Sigma$, which is thus itself not invertible.
Since the transformation is just changing the underlying coordinate system, rewriting the matrix this way can't change any geometric property the linear transformation.
A: Your matrix is symmetric, but not necessarily definite positive. If it isn't definite (has a null eigenvalue) it is not invertible.
In the case it IS definite, then it is invertible, and by the law $(AB)^{-1} = B^{-1}A^{-1}$; you can deduce that:
$$A^{-1} = (U\Sigma U^T)^{-1} = (U^{T})^{-1}\Sigma^{-1}U^{-1} = U\Sigma^{-1}U^T$$
If it isn't invertible/definite; then this equation does not hold, since $\Sigma^{-1}$ is undefined.
A: $\newcommand{\diag}{\mathrm{diag}}$
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 $V$.
