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The map $f:S_n(\mathbb R)\to \mathbb R$ such that $f(M)$ is the greatest eigenvalue of $M$ is continuous ($S_n(\mathbb R)$ is the set of symmetric matrices)

I need to prove this result in order to solve a broader problem.

I never feel comfortable when proving continuity of such maps.

I'd say that the map that takes $M$ to $\chi_M$ is continuous (why?), and the map that sends a real polynomial with only real roots $Q$ to its greatest root is continuous (why again?).

For the first question, I may argue that the coefficients of $\chi_M$ are polynomials in the coefficients of $M$ (doesn't sound very rigorous though)...

The original problem is the following

Let $\delta$ and $M$ be positive reals. Let $S_n(\mathbb R)^{++}$ denote the set of real positive definite matrices.

Prove the set $\{ A\in S_n(\mathbb R)^{++} | \det A \geq \delta \;\;\text{and}\; \; sp(A)\subset [0,M]\}$ is compact.

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Gabriel Romon
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2 Answers2

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On a normed space, the norm map $v \in V \mapsto \|v\| \in \mathbb R$ is continuous.

The spectral norm of a matrix $A$ is given by its largest singular value, which is also the square root of the largest eigenvalue of $AA^*$.

For real symmetric matrices, $AA^*=A^2$ and so the spectral norm of $A$ is largest eigenvalue of $A$.

Thus the map $A \mapsto \|A\|_2=\lambda_{max}(A)$ is continuous.

lhf
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Hint: You can use Rouché's Theorem in complex analysis to prove that the (complex) roots of a complex polynomial vary continuously with the coefficients of the polynomial.

Ted Shifrin
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