$y \in \mathbb{R}^n \implies \frac{y^T (X^TX) y}{y^Ty} \ge \sigma_n(X)^2$?

Question

I came across an expression in a book on matrix computations that is not clear to me. Let $X \in \mathbb{R}^{n \times n}$ be nonsingular. Then we have

$y \in \mathbb{R}^n \implies \frac{y^T (X^TX) y}{y^Ty} \ge \sigma_n(X)^2$.

I do not see how this is obvious? What identity is being used here?

Answer 1

If $\sigma_1(X) \ge \cdots\ge\sigma_n(X)$ are the singular values of $X$, then $\sigma_1(X)^2 \ge \cdots \ge \sigma_n(X)^2 \ge 0$ are the eigenvalues of $X^t X$. Write, as $X^tX$ is symmetric, $X^t X = Q^t \Sigma Q$, with $Q^t = Q^{-1} \in \mathrm{Mat}_n(\mathbf R)$, $\Sigma = \operatorname{diag}(\sigma_1(X)^2, \ldots,\sigma_n(X)^2)$. Then we have \begin{align*} y^t(X^tX)y &= y^tQ^t \Sigma Qy\\ &= \sum_{i=1}^n \sigma_i(X)^2(y^tQ^t)_i(Qy)_i\\ &\ge \sigma_n(X)^2 y^tQ^tQy\\ &= \sigma_n(X)^2 y^ty. \end{align*}