Calculating the matrix without using the inversion

Question

Suppose we have a non-square matrix $X$. Is it possible that using SVD we can calculate the matrix $\left(X^TX\right)^{-1}$ without its direct inversion ?

Answer 1

Yes. If $m \ge n$ and $X \in \mathbb{R}^{m \times n}$ has full rank $n$, then $X^T X \in \mathbb{R}^{n \times n}$ is nonsingular. Let $$X = U\Sigma V^T$$ be the singular value decomposition of $X$, i.e., $U \in \mathbb{R}^{m \times n}$ is orthogonal, $\Sigma \in \mathbb{R}^{n \times n}$ is diagonal with strictly positive diagonal entries, and $V \in \mathbb{R}^{n \times n}$ is orthogonal, then $$X^T X = (V \Sigma^T U^T) (U \Sigma V^T) = V \Sigma^2 V^T.$$ It follows, that $$ (X^TX)^{-1} = V \Sigma^{-2} V^T.$$