Suppose that $A$ and $P$ are $3 \times 3$ matrices and $P$ is invertible matrix.
If
\[P^{-1}AP=\begin{bmatrix}
1 & 2 & 3 \\
0 &4 &5 \\
0 & 0 & 6
\end{bmatrix},\]
then find all the eigenvalues of the matrix $A^2$.

We give two proofs. The first version is a short proof and uses some facts without proving.
The second proof explains more details and give proofs of the facts which are not proved in the first proof.

Proof (short version).

Let $B=P^{-1}AP$. Since $B$ is an upper triangular matrix, its eigenvalues are diagonal entries $1, 4, 6$.

Since $A$ and $B=P^{-1}AP$ have the same eigenvalues, the eigenvalues of $A$ are $1, 4, 6$. Note that these are all the eigenvalues of $A$ since $A$ is a $3\times 3$ matrix.

It follows that all the eigenvalues of $A^2$ are $1, 4^2, 6^2$, that is, $1, 16, 36$.

Proof (long version.)

Let us put $B:=P^{-1}AP$.
The eigenvalues of $B$ are $1, 4, 6$ since $B$ is an upper triangular matrix and eigenvalues of an upper triangular matrix are diagonal entries.
We claim that the eigenvalues of $A$ and $B$ are the same.

To prove this claim, we show that their characteristic polynomials are equal.
Let $p_A(t), p_B(t)$ be the characteristic polynomials of $A, B$, respectively.
Then we have
\begin{align*}
p_B(t)&=\det(B-tI)=\det(P^{-1}AP-tI)\\
&= \det(P^{-1}(A-tI)P)\\
&=\det(P^{-1})\det(A-tI)\det(P)\\
&=\det(P)^{-1}p_A(t)\det(P)\\
&=p_A(t).
\end{align*}
Therefore, the characteristic polynomials of $A$ and $B$ are the same, and hence the matrices $A$ and $B$ have the same eigenvalues.
Thus $1, 4, 6$ are eigenvalues of the matrix $A$ and these are all the eigenvalues of $A$ since a $3\times 3$ matrix has at most three eigenvalues.
Now we claim that in general if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$. We prove this statement below. Assuming this claim for the moment, we finish the problem.
By the claim, the matrix $A^2$ has eigenvalues $1^2, 4^2, 6^2$. Thus all the eigenvalues of $A^2$ are
\[1, 16, 36.\]
To complete the solution, let us prove the last claim.
Since $\lambda$ is an eigenvalue of $A$, we have an eigenvector $\mathbf{x}$ such that
\[A\mathbf{x}=\lambda \mathbf{x}.\]

Multiplying by $A$ we obtain
\begin{align*}
A^2T\mathbf{x}&=\lambda (A\mathbf{x})\\
&=\lambda (\lambda \mathbf{x})\\
&=\lambda^2 \mathbf{x}
\end{align*}
and thus $\lambda^2$ is an eigenvalue of $A^2$.

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