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?
 A: You can consider a map $K\to \|K\|_2$, where $\|\cdot\|_2$ is a spectral norm of the matrix. This map is continuous (should be an easy exercise), and it has a great property - for any eigenvalue $\lambda$ of $K$ you have $|\lambda|\le \|K\|_2$.
A: Let $||*||$ be any norm on $ \mathbb R^n$ and let $||*||_O$ the matrix norm induced by $||*||$
Since $K$ is compact, $K$ is bounded. Thus, there is $c>0$ such that
$||A||_O \le c$ for all $A \in K$.
Now let  $A \in K$ and let $ \lambda$ be an eigenvalue of $A$. Then there is $x \in \mathbb R^n$ with $Ax= \lambda x$ and $||x||=1$.
We get
$$| \lambda|=|| \lambda x||= ||Ax|| \le ||A||_O*||x||=||A||_O \le c.$$
A: Your approach is not correct. In fact, $\prod_i \lambda_i$ can be bounded without $\max_i \lambda_i$ being bounded. (Why?)
A better way of going about it is this: for any $A \in \mathbb M_N(\mathbb R)$, any eigenvalue $\lambda$ of $A$ is bounded by
$$
\sum_{i = 1}^n \sum_{j = 1}^n |A_{i,j}|.
$$
Edit: It is even easier to do this when we take our bound $K$ equal to $n$ times the above sum. In that case, let $\lambda$ be an eigenvalue of eigenvector $x$, and let $x_i$ be the largest entry of $x$. Then in particular
$$
|\lambda| |x_i| = \left|\sum_{j=1}^n A_{i,j}x_j\right| \leq \sum_{j=1}^n |A_{i,j}||x_j| \leq \sum_{j=1}^n \frac Kn|x_i| = K|x_i|,
$$
hence $|\lambda| \leq K$.
A: 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 
