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Consider this square matrix $C = \begin{bmatrix} A& 0 \\ 0 &B \end{bmatrix}$, where $A$ and $B$ are also square matrices. Suppose $A$ is larger in the sense that is an $n \times n$ matrix, while $B$ is an $m \times m$ matrix, and $n > m$. Also suppose that both $A$ and $B$ are real symmetric matrices.

I know that the set of eigenvalues of $C$ is equal to the multi-set of eigenvalues of $A$ and $B$.

Question: Suppose the eigenvalues of $A$ are distinct from the eigenvalues of $B$. Must the largest eigenvalue of $C$ be an eigenvalue of $A$?

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Not necessarily. Take, for example, $$ A = \pmatrix{1&0&0\\0&1&0\\0&0&1},\quad B = \pmatrix{0&0\\0&2} $$

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  • $\begingroup$ Duh. I should have realized this. Thanks! +1 and accept. $\endgroup$ – Jean Valjean Jul 28 '14 at 19:06
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Denote by $E(T)$ the set of eigenvalues of any matrix $T$. I don't know what you mean by multi-set, but the set of eigenvalues of $C$ is given by $E(C)=E(A)\cup E(B)$. The largest eigenvalue is just the largest element in this union. It can be either in $E(A)$ or $E(B)$. It's very easy to find examples where this eigenvalue is not an eigenvalue of $A$. Just take any $A,B$ you like, such that $\max(E(B))>\max(E(A))$.

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  • $\begingroup$ "multi-set" means that you can repeat non-distinct elements, so that elements can have "multiplicities" within a multi-set. So, for example, $\{1,1,2\} \neq \{1,2\}$ for multi-sets. $\endgroup$ – Omnomnomnom Jul 28 '14 at 19:21

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