# Problems and Solutions About Similar Matrices

## Problem 319

Let $A, B$, and $C$ be $n \times n$ matrices and $I$ be the $n\times n$ identity matrix.

Prove the following statements.

**(a)** If $A$ is similar to $B$, then $B$ is similar to $A$.

**(b)** $A$ is similar to itself.

**(c)** If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

**(d)** If $A$ is similar to the identity matrix $I$, then $A=I$.

**(e)** If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.

**(f)** If $A$ is similar to $B$, then $A^k$ is similar to $B^k$ for any positive integer $k$.

Contents

- Problem 319
- Definition.
- Proof.
- (a) If $A$ is similar to $B$, then $B$ is similar to $A$.
- (b) $A$ is similar to iteself.
- (c) If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.
- (d) If $A$ is similar to the identity matrix $I$, then $A=I$.
- (e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.
- (f) If $A$ is similar to $B$, then $A^k$ is similar to $B^k$.

- Comment.

## Definition.

We say that a matrix $A$ is **similar** to a matrix $B$ is there exists a nonsingular (invertible) matrix $P$ such that

\[B=P^{-1}AP.\]

## Proof.

### (a) If $A$ is similar to $B$, then $B$ is similar to $A$.

If $A$ is similar to $B$, then there exists a nonsingular matrix $P$ such that $B=P^{-1}AP$.

Let $Q=P^{-1}$. Since $P$ is nonsingular, so is $Q$.

Then we have

\begin{align*}

Q^{-1}BQ&=(P^{-1})^{-1}BP^{-1}=PBP^{-1}\\

&=P(P^{-1}AP)P^{-1}=IAI=A.

\end{align*}

Hence $B$ is similar to $A$.

### (b) $A$ is similar to iteself.

Since the identity matrix $I$ is nonsingular and we have

\[A=I^{-1}AI,\]
the matrix $A$ is similar to $A$ itself.

### (c) If $A$ is similar to $B$ and $B$ is similar to $C$, then $A$ is similar to $C$.

If $A$ is similar to $B$, we have

\[B=P^{-1}AP,\]
for some nonsingular matrix $P$.

Also, if $B$ is similar to $C$, we have

\[C=Q^{-1}BQ,\]
for some nonsingular matrix $Q$.

Then we have

\begin{align*}

C&=Q^{-1}BQ\\

&=Q^{-1}(P^{-1}AP)Q\\

&=(PQ)^{-1}A(PQ).

\end{align*}

Let $R=PQ$. Since both $P$ and $Q$ are nonsingular, $R=PQ$ is also nonsingular.

The above computation yields that we have

\[C=R^{-1}AR,\]
hence $A$ is similar to $C$.

### (d) If $A$ is similar to the identity matrix $I$, then $A=I$.

Since $A$ is similar to $I$, there exists a nonsingular matrix $P$ such that

\[A=P^{-1}IP.\]
Since $P^{-1}IP=I$, we have $A=I$.

### (e) If $A$ or $B$ is nonsingular, then $AB$ is similar to $BA$.

Suppose first that $A$ is nonsingular. Then $A$ is invertible, hence the inverse matrix $A^{-1}$ exists.

Then we have

\begin{align*}

A^{-1}(AB)A=A^{-1}ABA=IBA=BA,

\end{align*}

hence $AB$ and $BA$ are similar.

Analogously, if $B$ is nonsingular, then the inverse matrix $B^{-1}$ exists.

We have

\begin{align*}

B^{-1}(BA)B=B^{-1}BAB=IAB=AB,

\end{align*}

hence $AB$ and $BA$ are similar.

### (f) If $A$ is similar to $B$, then $A^k$ is similar to $B^k$.

If $A$ is similar to $B$, then we have

\[B=P^{-1}AP\]
for some nonsingular matrix $P$.

Then we have for a positive integer $k$

\begin{align*}

B^{k}&=(P^{-1}AP)^k\\

&=\underbrace{(P^{-1}AP)(P^{-1}AP)\cdots (P^{-1}AP)}_{k \text{ times}} \\

&=P^{-1}A^kP

\end{align*}

since we can cancel $P$ and $P^{-1}$ in between.

Hence $A^k$ and $B^k$ are similar.

## Comment.

Part (a), (b), (c) show that similarity is an equivalence relation.

Add to solve later