Is the characteristic polynomial positive for values greater than the spectral radius of the matrix?

Question

Because $$\det(tI-A)=t^n+...$$ So the characteristic polynomial is positive for big enough values of $t$.

If the characteristic polynomial is negative for some number greater than the spectral radius then it would be 0 for some point by the intermediate value theorem.

Is this correct?

Answer 1

Your argument is correct, but there is an elementary algebraic argument that does not rely on calculus. Factorise $\det(tI-A)$ as $$ \prod_{i=1}^{n_1}(t-p_i)\prod_{j=1}^{n_2}(t+q_j)\prod_{k=1}^{n_3}(t-\lambda_k)(t-\overline{\lambda}_k), $$ where the $p_i$s are nonnegative, the $q_j$s are positive and the $\lambda_k$s are non-real. Then

• for each $i$, since $\rho(A)\ge p_i$, we have $t-p_i>0$ whenever $t>\rho(A)$ (because $\rho(A)\ge p_i$);
• for each $j$, since $q_j$ is positive, $t+q_j>0$ whenever $t\ge0$;
• for each $k$, since $\lambda_k$ is not real, $(t-\lambda_k)(t-\overline{\lambda}_k)= |t-\lambda_k|^2>0$ whenever $t\in\mathbb R$.

It follows that $\det(tI-A)>0$ when $t>\rho(A)$.