# 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 $A^2$.

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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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