# how can I prove ${\lambda _n}\left( {{X^T}AX} \right) \leqslant \left\| X \right\|_F^2{\lambda _n}\left( A \right)$ where $A$ is a PD matrix

The following inequality intuitively holds in my opinion, however I am facing hard time proving it

$${\lambda _n}\left( {{X^T}AX} \right) \leqslant \left\| X \right\|_F^2{\lambda _n}\left( A \right)$$

Keep in mind that $$\lambda_n$$ is the smallest eigenvalue of a $$n$$-by-$$n$$ square matrix. The matrix $$A$$ is a symmetric, positive definite matrix

All I know from matrix linear algebra is that

$${\lambda _n}\left( {{X^T}AX} \right) \leqslant {\lambda _1}\left( {{X^T}AX} \right) = \rho \left( {{X^T}AX} \right) \leqslant {\left\| {{X^T}AX} \right\|_F} \leqslant \left\| X \right\|_F^2{\left\| A \right\|_F}$$

where $$\rho$$ is the spectral radius.

• $A$ is symmetric? $X$ is unitary/orthogonal? Jun 16, 2022 at 8:01
• $A$ is PD and hence symmetric, $X$ can be neither! (post has been edited) @Exodd
– SAM
Jun 16, 2022 at 8:03
• If $X$ is singular, the result is easy, otherwise just characterize the least singular value of a symmetric matrix $B$ as the minimum of $v^TBv/\|v\|^2$ Jun 16, 2022 at 8:06
• Could you elaborate on your answer? why is it easy when X is singular, and is the least singular value by definition the minimum of what you specified? I would really appreciate if you gave a full answer. @Exodd
– SAM
Jun 16, 2022 at 8:11

One easy way to prove it is to utilise Sylvester's secular theorem that $$ST$$ and $$TS$$ share the same multi-set of nonzero eigenvalues. In particular, $$ST$$ and $$TS$$ share the same spectra if both $$S$$ and $$T$$ are square matrices. Thus \begin{aligned} \lambda_n(X^TAX) &=\lambda_n\left((X^TA^{1/2})(A^{1/2}X)\right)\quad\text{(here we need A to be PSD)}\\ &=\lambda_n\left((A^{1/2}X)(X^TA^{1/2}))\right)\\ &\le\lambda_n\left(A^{1/2}\left(\|X\|_2^2I\right)A^{1/2})\right) \quad\text{(because XX^T\preceq \|X\|_2^2I)}\\ &=\|X\|_2^2\lambda_n(A)\\ &\le\|X\|_F^2\lambda_n(A).\\ \end{aligned} As we see in the above, $$\|X\|_2^2\lambda_n(A)$$ a sharper upper bound of $$\lambda_n(X^TAX)$$ than $$\|X\|_F^2\lambda_n(A)$$.