Prove $p_{cA} = c^n \ p\left( \frac{t}{c}\right) $ is the characteristic polynomial of $\ cA $

Question

Suppose $p_A (t) $ is the characteristic polynomial of matrix $\ A \in \mathbb R^{n\times n} $. Prove that, for every $c \in \mathbb R$ such that $c \neq 0$, the characteristic polynomial $p_{cA}(t) $ of matrix $cA $ satisfies $$p_{cA}(t) = c^n p \left(\frac{t}{c}\right). $$

I don't really have any direction.

Answer 1

Hint: We have $\det(cA-tI)=c^n\,\det\left(A-\frac{t}{c}I\right)$.

Answer 2

The definition of the characteristic of $cA$ is \begin{align*} p_{cA}(t) = det(tI-cA) &= c^n det(\frac{t}{c} I - A) = c^n p_{A}\left(\frac{t}{c}\right)\\ \end{align*}

which proves your result. This equality is due to the fifth property of the determinant and can be checked easily