For a compact set $K\subset \mathbb M_n(\mathbb R)$, the eigenvalues of matrices in $K$ form a bounded set Let $K\subset \mathbb M_n(\mathbb R)$ be a compact subset. Then I have to show that :


All the eigen values of the elements of $K$ form a bounded set.


My work: consider the map $K \to det K$ which is continuous. Image set is compact in $\mathbb C$, hence closed bounded. If $\lambda_i (i=1\ldots,n)$ are the eigenvalues, then $\det K=\prod \lambda_i$ is bounded which in turn  gives $\lambda_i$ bounded.
Is my approach correct? Is there any better way to do it?

An easy approach for me:
Consider the map $f:K\to \Bbb C$ by $f(x)=\dfrac{x^TAx}{x^Tx}$
which is continuous and hence bounded on $K$.
$\text{Image} f=\{f(x):x\in K\}$ is bounded.
Now if $\lambda $ is an eigen value of $A$ corresponding to eigen vector $v$ then $Av=\lambda v\implies f(v)=\lambda$.
Hence $\lambda \in \text{Image }f$ which is bounded 
