Matrix relations involving trace and eigenvalues

Question

Given the real symmetric positive definite matrix $A \succ 0$, consider the following inequality in $\alpha \in \mathbb{R}$.

$$ A \succ \alpha I $$

where $I$ is an identity matrix of appropriate dimension. Can I choose $ \lambda_{\min}(A) > \alpha $ to satisfy the inequality?

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Here is why I think I can. First, pre- and post-multiplying both sides by $x^\top$ and $x$, respectively,

$$ x^{\top} A x > x^{\text{T}} \alpha x $$

Then, using the identity $ \lambda_{\min}(A) \| x \|^2 \leq x^{\top} A x \leq \lambda_{\max}(A) \| x \|^2 $ on both sides yields the following

\begin{gather} \lambda_{\max}(A) \| x \|^2 \geq x^{\top} A x \geq \lambda_{\min}(A) \| x \|^2 > \lambda_{\max}(\alpha) \| x \|^2 \geq x^{\top} \alpha x \geq \lambda_{\min}(\alpha) \| x \|^2 \newline \implies \lambda_{\min}(A) \| x \|^2 > \lambda_{\max}(\alpha) \|\| x \|\|^2 \end{gather}

Dividing both sides by $\| x \|^2$ and recalling $\alpha$ is a scalar leads to

$$ \lambda_{\min}(A) > \alpha $$

Have I made any mistakes? Does this reasoning hold?

Answer 1

In general, it doesn't make sense to replace terms $A$, $B$ in inequalities (or equalities!) with inequalities for those terms. For instance, if $A = 1$ and $B = -1$, then $$ -100 \leq A \leq 100 \quad \text{and} \quad -10 \leq B \leq 10 \quad \text{and} \quad A > B $$ but it's certainly not true that $$ 100 \geq A \geq -100 > 10 \geq B > -10. $$ In your case, this means: Knowing $x^T A x \geq a$ and $b \geq x^T \alpha x$ and $x^T A x > x^T A x$ is not enough to conclude $a > b$ (which appears to be what you're trying to do).

(It is true that $A \succ \alpha I$ forces all the eigenvalues of $A$ to be larger than $\alpha$, but you'll need a different proof. Since $A$ is real and symmetric, it's diagonalizable, so what happens when you look at eigenvectors?)

Answer 2

Since the matrix $\bf A$ is symmetric, it has a spectral decomposition $\bf A = Q \Lambda Q^\top$. Hence,

$$ 0 < {\bf x}^\top \left( {\bf A} - t \, {\bf I}_n \right) {\bf x} = {\bf x}^\top {\bf Q} \left( {\bf \Lambda} - t \,{\bf I}_n \right) \underbrace{{\bf Q}^\top {\bf x}}_{=: {\bf y}} = {\bf y}^\top \left( {\bf \Lambda} - t \,{\bf I}_n \right) {\bf y} $$

and, thus, ${\bf \Lambda} - t \,{\bf I}_n \succ {\bf O}_n$. Therefore, $ \color{blue}{t < \lambda_{\min} ({\bf A})} $.

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_{Related: How to minimize $\| x {\bf I} - {\bf A} \|_2$?}