# 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 $k$.

## 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.

Suppose that $A$ is a real $n\times n$ matrix. (a) Is it true that $A$ must commute with its transpose?...