# Eigenvalues of Similarity Transformations

## Problem 452

Let $A$ be an $n\times n$ complex matrix.

Let $S$ be an invertible matrix.

**(a)** If $SAS^{-1}=\lambda A$ for some complex number $\lambda$, then prove that either $\lambda^n=1$ or $A$ is a singular matrix.

**(b)** If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.

**(c)** Suppose that all the eigenvalues of $A$ are integers and $\det(A) > 0$. If $n$ is odd and $SAS^{-1}=A^{-1}$, then prove that $1$ is an eigenvalue of $A$.

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Contents

## Proof.

### Basic Properties of Determinants

We use the following properties of determinants of matrices.

For $n\times n$ matrices $A, B$, we have

- $\det(AB)=\det(A)\det(B)$.
- $\det(A^{-1})=\det(A)^{-1}$ if $A$ is invertible.
- $\det(cA)=c^n\det(A)$ for any complex number $c$.

### (a) If $SAS^{-1}=\lambda A$, then prove that $\lambda^n=1$ or $A$ is a singular matrix.

Suppose that we have $SAS^{-1}=\lambda A$.

We consider the determinants of both sides and we have

\begin{align*}

\det(SAS^{-1})=\det(\lambda A).

\end{align*}

The left hand side becomes

\begin{align*}

\det(SAS^{-1})&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 1}\\

&=\det(S)\det(A)\det(S)^{-1} &&\text{by property 2}\\

&=\det(A).

\end{align*}

The right hand side is by property 3

\[\det(\lambda A)=\lambda^n \det(A).\]

Hence we obtain

\[\det(A)=\lambda^n \det(A).\]
It follows from this that either $\lambda^n=1$ or $\det(A)=0$.

In conclusion, either $\lambda^n=1$ or $A$ is a singular matrix.

### (b) If $n$ is odd and $SAS^{-1}=-A$, then prove that $0$ is an eigenvalue of $A$.

Suppose that we have $SAS^{-1}=-A$.

Then we have

\begin{align*}

&\det(A)=\det(S)\det(A)\det(S)^{-1} \\

&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 2}\\

&=\det(SAS^{-1}) &&\text{by property 1}\\

&=\det(-A) &&\text{by assumption}\\

&=(-1)^n\det(A) &&\text{by property 3}\\

&=-\det(A) &&\text{since $n$ is odd.}\\

\end{align*}

This yields that $\det(A)=0$.

Note that the product of all eigenvalues of $A$ is $\det(A)$.

(See the post “Determinant/Trace and Eigenvalues of a Matrix” for a proof.)

Thus, $0$ is an eigenvalue of $A$.

### (c) Note that as $\det(A) > 0$, the matrix $A$ is invertible.

Suppose that $SAS^{-1}=A^{-1}$.

Then we have

\begin{align*}

&\det(A)=\det(S)\det(A)\det(S)^{-1} \\

&=\det(S)\det(A)\det(S^{-1}) &&\text{by property 2}\\

&=\det(SAS^{-1}) &&\text{by property 1}\\

&=\det(A^{-1}) &&\text{by assumption}\\

&=\det(A)^{-1} &&\text{by property 2}.

\end{align*}

Thus we have $\det(A)^2=1$, hence $\det(A)=1$ as $\det(A) > 0$ by assumption.

Note again that the product of all eigenvalues of $A$ is $\det(A)$.

Since the eigenvalues of $A$ are integers by assumption, the eigenvalues of $A$ are either $1$ or $-1$.

If all of the $n$ eigenvalues of $A$ are $-1$, then the determinant is

\[\det(A)=(-1)^n=-1\]
since $n$ is odd, and this is a contradiction.

Thus, at least one of the eigenvalues must be $1$.

This completes the proof.

## Similar Transformation (conjugate)

A transformation $A$ to $SAS^{-1}$, for some invertible matrix $S$, is called a **similarity transformation** or **conjugation** of the matrix $A$.

We say that matrices $A$ and $B$ are **similar** if there exists an invertible matrix $S$ such that $B=SAS^{-1}$.

In other words, $A$ is similar to $B$ if there is a similarity transformation from $A$ to $B$.

Check out the following problems about similar matrices.

**Problem**. Is the matrix $A=\begin{bmatrix}

-1 & 6\\

-2& 6

\end{bmatrix}$ similar to the matrix $B=\begin{bmatrix}

1 & 2\\

-1& 4

\end{bmatrix}$?

For a solution together with similar problems, see the post “Determine whether given matrices are similar“.

**Problem**. Let $A, B$ be matrices. Show that if $A$ is diagonalizable and if $B$ is similar to $A$, then $B$ is diagonalizable.

For a proof, see the post “A matrix similar to a diagonalizable matrix is also diagonalizable“.

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