# Eigenvalues of a block-diagonal matrix

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 $$$$\Sigma = \begin{pmatrix} A_1 & 0 &\cdots&0 \\ 0 & A_2 & \cdots&0 \\ \vdots&\vdots&\ddots&\vdots \\ 0&0&\cdots&A_k \end{pmatrix}.$$$$

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?

• If $\Sigma (v_1,...,v_K) = (\lambda_1 v_1,...., \lambda_K v_K)$ you must have $\lambda_k \ge \eta$. May 1, 2022 at 19:26
• If the matric is symmetric, it is diagonalisable and you can conclude the latter accordingly. May 1, 2022 at 19:31
• 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 $\lambda \ge \eta$. May 1, 2022 at 19:36
• Great, thanks very much! May 1, 2022 at 19:42
• Prove it for a diagonal $\Sigma$ first and then use a suitable basis of eigenvectors to express $\Sigma$ as a diagonal matrix. May 1, 2022 at 20:03

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 $$A_i$$, $$i \in \{1,2, \ldots , k\}$$.
As a result we get $$Eigenvalues(\sum)=Eigenvalues(A_1) \cup \ldots \cup Eigenvalues(A_k)$$