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?

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 $|\lambda| \leq K$.
