Can $X^TX$ have negative eigenvalues?

Question

Given a real $X_{n\times p}$ matrix where $p > n$, can the matrix $X^T X$ have any negative eigenvalues? A derivation would be appreciated!

Answer 1

Suppose $v$ is an eigenvector of $X^\top X$ corresponding to eigenvalue $\lambda$. Note that $v \neq 0$, hence $\|v\|^2 \neq 0$. Then \begin{align*} X^\top X v = \lambda v &\implies v^\top X^\top X v = \lambda v^\top v\\ &\implies (Xv)^\top (Xv) = \lambda v^\top v \\ &\implies \|Xv\|^2 = \lambda \|v\|^2 \\ &\implies \lambda = \frac{\|Xv\|^2}{\|v\|^2} \ge 0. \end{align*}

Answer 2

No, it cannot.

You can show this by contradiction: If $\lambda < 0$ was an eigenvalue with eigenvector $u$ $(\neq 0)$, you would have $$ \color{blue}{0 >} \lambda u^Tu =u^T(\lambda u) = u^TX^TXu = (Xu)^T(Xu) \color{blue}{\geq 0} \mbox{ Contradiction!}$$