Let \(M_{n\times n}\) be the vector space of real \(n\times n\) matrices, identified with \(\mathbb R^{n^2}\), and let \(X\subset M_{n\times n}\) be compact. Let \(S\subset\mathbb C\) be the set of all numbers that occur as eigenvalues of at least one matrix in \(X\). Prove that \(S\) is compact.
Solution
If \(X\) is empty, then \(S\) is empty and hence compact. Assume henceforth that \(X\) is nonempty, and regard each real matrix in \(X\) as acting on \(\CC^n\).
1. There is a constant \(M\ge0\) such that\[ \norm{A}_{\operatorname{op}}\le M \]for every \(A\in X\).
The operator norm is a continuous function of the matrix entries. Since \(X\) is compact, it attains a maximum there. Let \[ M = \max_{A\in X}\norm{A}_{\operatorname{op}}. \]
2. Every \(\lambda\in S\) satisfies\[ \abs{\lambda}\le M. \]
Choose \(A\in X\) and a nonzero eigenvector \(v\in\CC^n\) such that \[ Av=\lambda v. \] After replacing \(v\) by \(v/\norm{v}\), assume \(\norm{v}=1\). Then step 1 gives \[ \abs{\lambda} = \norm{\lambda v} = \norm{Av} \le \norm{A}_{\operatorname{op}} \le M. \]
3. Define\[ K = \left\{ (A,\lambda,v): \begin{array}{l} A\in X,\quad \abs{\lambda}\le M,\quad \norm{v}=1,\\ Av=\lambda v \end{array} \right\}. \]Then \(K\) is compact.
The product \[ X \times \{\lambda\in\CC:\abs{\lambda}\le M\} \times \{v\in\CC^n:\norm{v}=1\} \] is compact. The map \[ (A,\lambda,v) \longmapsto Av-\lambda v \] is continuous, and \(K\) is its zero set inside that compact product. Thus \(K\) is closed in a compact space and therefore compact.
4. The projection of \(K\) onto its \(\lambda\)-coordinate is exactly\(S\).
If \((A,\lambda,v)\in K\), then \(\norm{v}=1\) and \(Av=\lambda v\), so \(v\neq0\) and \(\lambda\) is an eigenvalue of \(A\in X\). Hence \(\lambda\in S\).
Conversely, if \(\lambda\in S\), choose \(A\in X\) and a nonzero eigenvector \(v\) with \(Av=\lambda v\). Normalize \(v\) to have norm \(1\). Step 2 gives \(\abs{\lambda}\le M\), so \((A,\lambda,v)\in K\).
5. Therefore\[ \boxed{S\text{ is compact in }\CC}. \]
4. A continuous imageof a compact space is compact.
6. Q.E.D.
Step 5 is the required conclusion.