Let $A ∈M_{n×n}(\mathbb{R})$ be a matrix all entries in which are positive and let $c$ be a positive real number greater than the spectral radius of $A$. Show that $|cI −A| > 0$.
$c >\max\{\lambda_i|\lambda_i\in spec(A)\}$. Obviously $|cI-A|\neq 0$, since $c\notin spec(A)$, however, I cannot show this determinant positive.
First of all, as the matrix $A$ can have complex eigenvalues, as well, the assumption should be reformulated as $$ c>\max\{\lambda_k: \lambda_k\,\,\text{real eigenvalue of}\,\,A \}. $$
If $\lambda_k$, $k=1,\ldots,n$ are the eigenvalues of $a$, then $c-\lambda_k$, $k=1,\ldots,n$ are the eigenvalues of $cI-A$, and $$ \det(cI-A)=\prod_{k=1}^n (c-\lambda_k). $$ If $\lambda_k$ is real, then $c-\lambda_k>0$, due to the assumption.
If $\lambda_k$ is complex, then $\overline{\lambda}_k$ is also an eigenvalue, and $$ (c-\lambda_k)(c-\overline{\lambda}_k)=\lvert c-\lambda_k\rvert^2>0. $$ Then, as the complex eigenvalues of a real matrix come in conjugate pairs, then $$ \prod_{k=1}^n (c-\lambda_k)=\prod_{\text{$\lambda_k$ real}} (c-\lambda_k) \cdot \left(\prod_{\text{$\lambda_k$ complex}} |c-\lambda_k|^2\right)^{1/2}>0. $$
Edit. It can be proved in a more elegant way.
We clearly have that $$ \lim_{c\to\infty}\det(cI-A)=\infty, $$ and $\det(cI-A)=0$ if and only if $c$ is an eigenvalue of $A$. Therefore, because of the continuity of $f(c)=\det(cI-A)$, it follows that $$ f(c)=\det(cI-A)>0, \quad\text{for all $c>\max\{\lambda_k: \lambda_k\,\,\text{real eigenvalue of}\,\,A\}$.} $$