Find eigenvalues of A knowing its trace and the eigenvalues of $A^2+2A$.

Question

Let $A$ $\in$ $\mathbb{R}^{3x3}$ be a diagonalizable matrix with $\mathbf{trace}$($A$) = $-4$.
Let {$-1$, $3$, $8$} be the eigenvalues of $A^2+2A$.
Find the eigenvalues of $A$.

I tried stating that the sum of eigenvalues for $A$ has to be $-4$, which is also the coefficient for $\lambda^2$ in $\chi_{A}(\lambda)$ (minus a sign), but it has gotten me nowhere. Also I don't know how to work the part about the eigenvalues of $A^2+2A$.

Any help or hints are greatly appreciated, many thanks in advance.

Answer 1

The fact that $-1$ is an eigenvalue of $A^2+2A$ tells you that $\det(A^2+2A+I)=0$, hence that $\det(A+I)=0$, because $A^2+2A+I=(A+I)^2$.

In particular $-1$ is also an eigenvalue of $A$.

Next $\det(A^2+2A-3I)=0$, leading to $\det(A+3I)\det(A-I)=0$; also $\det(A^2+2A-8I)=0$, leading to $\det(A+4I)\det(A-2I)=0$.

Thus we have four possibilities: the remaining two eigenvalues, besides $-1$, can be $$ -3,-4\quad\text{or}\quad -3,2,\quad\text{or}\quad 1,-4,\quad\text{or}\quad 1,2 $$ Since we know that the trace of $A$ is $-4$…

The information that $A$ is diagonalizable is irrelevant.

Answer 2

This answer is less general than the one proposed by @egreg but, considering that the assumption that $A$ is diagonizable was included, it might be closer to the expected answer. If $\lambda(A) = \{\lambda_1, \lambda_2, \lambda_3\}$ is the set of eigenvalues of $A$, since $A$ is diagonizable, the set of eigenvalues of $P(A)=A^2+2A$ is $\lambda (P(A)) = \{\lambda_1^2 + 2 \lambda_1, \lambda_2^2+2 \lambda_2, \lambda_3^2+2 \lambda_3 \}$. Since you know that $\lambda(P(A))=\{-1,3,8\}$, you can conclude that $$ \lambda_1^2+2 \lambda_1 = -1, \quad \lambda_2^2+2\lambda_2 = 3, \quad \lambda_3^3+2 \lambda_3 = 8, $$

therefore you have that $\lambda_1=-1$, $\lambda_2=-3$ or $1$ and $\lambda_3 =-4$ or $2$. Since the trace is $-4$ (and the trace is the sum of the eigenvalues), the only way is if $\lambda_1=-1$, $\lambda_2=1$ and $\lambda_3=-4$.