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Let $K$ be a positive integer and for each $j=1,\dots,K$ let $A_j\in\mathbb{R}^{p_j\times p_j}$ be symmetric matrices, where $p_j$ is a positive integer. Suppose that each $A_j$ has smallest eigenvalue greater than some universal constant $\eta>0$. Consider the matrix \begin{equation} \Sigma = \begin{pmatrix} A_1 & 0 &\cdots&0 \\ 0 & A_2 & \cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A_k \end{pmatrix}. \end{equation}

I am deducing from a past exam question that the smallest eigenvalue of $\Sigma$ will also be greater than the constant $\eta$, but I am not sure how to show this - advice would be greatly appreciated.

In a similar vein: I have also deduced this supposed property which I believe to be true but I'm not sure how to show. Say a symmetric matrix $X\in\mathbb{R}^{p\times p}$ has minimum eigenvalue $\lambda_0>0$. Is it true that $$\inf_{\beta\in\mathbb{R}^p: \beta\neq0}\frac{\beta^TX\beta}{\|\beta\|_2^2} = \lambda_0,$$ and if so how do I show this?

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    $\begingroup$ If $\Sigma (v_1,...,v_K) = (\lambda_1 v_1,...., \lambda_K v_K)$ you must have $\lambda_k \ge \eta$. $\endgroup$
    – copper.hat
    May 1, 2022 at 19:26
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    $\begingroup$ If the matric is symmetric, it is diagonalisable and you can conclude the latter accordingly. $\endgroup$
    – copper.hat
    May 1, 2022 at 19:31
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    $\begingroup$ Suppose $\Sigma (v_1,...,v_K) = \lambda (v_1,...,v_K) $. Then $A_k v_k = \lambda v_k$. So for each $k$, either $v_k = 0$ or $A v_k = \lambda v_k$ (and $\lambda$ is an eigenvalue of $A_k$). So, either $v=0$ or $\lambda$ is an eigenvalue of one of the blocks and so $\lambda \ge \eta$. $\endgroup$
    – copper.hat
    May 1, 2022 at 19:36
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    $\begingroup$ Great, thanks very much! $\endgroup$ May 1, 2022 at 19:42
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    $\begingroup$ Prove it for a diagonal $\Sigma$ first and then use a suitable basis of eigenvectors to express $\Sigma$ as a diagonal matrix. $\endgroup$
    – copper.hat
    May 1, 2022 at 20:03

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Regarding your first question, since the Matrix $\sum$ is in block diagonal form, the determinant and as such the determinant for calculating the characteristical polynomial, can be calculated by multiplying the determinants of the blocks: $$ det(\sum)=det(A_1) \cdot det(A_2) \cdot \ldots $$ This is analog for the calculation of the characteristical polynomial and implies that the roots of the characteristical polynomial of $\sum$ is equal to the combination of the roots of the characteristical polynomials of the Matrices $A_i$, $i \in \{1,2, \ldots , k\}$.

Note that the roots of the characteristical polynomial of a matrix correspond to it's Eigenvalues.

As a result we get $$Eigenvalues(\sum)=Eigenvalues(A_1) \cup \ldots \cup Eigenvalues(A_k)$$

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