If two matrices have the same eigenvalues, will they have the same characteristic polynomial?

Question

This seems to hold, considering:

$$A=\left(\begin{matrix}2 &0\\0 &2\end{matrix}\right)$$ $$B=\left(\begin{matrix}2 &1\\0 &2\end{matrix}\right)$$

Is this always valid?

Answer 1

Two matrices have the same polynomial exactly when they have the same eigenvalues, each with the same algebraic multiplicity.

Note that the characteristic polynomial is always a polynomial with leading coefficient $1$ (or $(-1)^n$ if that's your definition), and that its zeros are precisely the eigenvalues of the matrix. In particular, suppose that $A$ has eigenvalues $\{\lambda_1,\dots,\lambda_n\}$. The only polynomial with exactly these zeros is $$ p(t) = (t-\lambda_1)(t-\lambda_2)\cdots(t-\lambda_n) $$ so, this must be the characteristic polynomial of $A$.

Note that the number of repetitions of an eigenvalue matters here. In particular, a matrix with eigenvalues $\{0,0,1\}$ will have characteristic polynomial $t^2(t-1)$, while a matrix with eigenvalues $\{0,1,1\}$ will have characteristic polynomial $t(t-1)^2$.