# Eigenvalues of Squared Matrix and Upper Triangular Matrix

## Problem 184

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 $A^2$.

Let $A$ be an $n \times n$ matrix. Suppose that the matrix $A^2$ has a real eigenvalue $\lambda>0$. Then show...