Solution. To show that $\boldsymbol{A}$ is similar to $\boldsymbol{B}$, we must show that $\boldsymbol{A}=\boldsymbol{PBP}^{-1}$. The inverse of $\boldsymbol{P}$ is:

Now, $\boldsymbol{PBP}^{-1}$ is:

Therefore, $\boldsymbol{A}$ is similar to $\boldsymbol{B}$.

Proof. By definitionlink, if $\boldsymbol{A}$ is similar to $\boldsymbol{B}$, then there exists an invertible matrix $\boldsymbol{P}$ such that:

Let's perform some matrix manipulation:

This completes the proof.

Proof. If $\boldsymbol{A}\sim\boldsymbol{B}$, then:

By definition, we also have:

Now, let $\boldsymbol{R}=\boldsymbol{P}^{-1}$. This means that $\boldsymbol{R}^{-1}= \boldsymbol{P}$.

By definitionlink, we conclude that $\boldsymbol{B}\sim\boldsymbol{A}$.

The converse can be proven by switching $\boldsymbol{A}$ and $\boldsymbol{B}$ and performing exactly the same steps. This completes the proof.

Proof. To show that $\boldsymbol{A}$ is similar to itself, we must show that there exists an invertible matrix $\boldsymbol{P}$ such that:

Suppose $\boldsymbol{P}$ is the identity matrix, that is, $\boldsymbol{P}=\boldsymbol{I}$. This means that \eqref{eq:FrZhzta0E5oT9P3cMNS} becomes:

This completes the proof.

Proof. Since $\boldsymbol{A}\sim\boldsymbol{B}$ and $\boldsymbol{B}\sim\boldsymbol{C}$, there exists some invertible matrices $\boldsymbol{P}$ and $\boldsymbol{R}$ such that:

We substitute the second equation into the first equation to get:

Here, the last step comes from theoremlink. If we let matrix $\boldsymbol{W}=\boldsymbol{PR}$, then:

By definition, we conclude that $\boldsymbol{A}$ and $\boldsymbol{C}$ are similar.

Proof. Since $\boldsymbol{A}\sim\boldsymbol{B}$, we have that:

Taking the determinant of both sides gives:

By the multiplicative propertylink of determinants, we have that:

Recall the following propertylink of determinants:

Therefore, \eqref{eq:RM1WoFYIUlRiZMWJ4yA} is:

This completes the proof.

Proof. Since $\boldsymbol{A}\sim\boldsymbol{B}$, we have that:

Where $\boldsymbol{P}$ is an invertible matrix. Taking the rank of both sides gives:

By theoremlink, since $\boldsymbol{P}$ is invertible, we get:

Once again by theoremlink, since $\boldsymbol{P}^{-1}$ is invertible, we get:

This completes the proof.

Proof. Since $\boldsymbol{A}\sim\boldsymbol{B}$, we have that:

Where $\boldsymbol{P}$ is an invertible matrix. Taking the trace of both sides gives:

By theoremlink, we have that:

This completes the proof.

Proof. Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be square matrices of the same shape. Since $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar, we have that:

Where $\boldsymbol{P}$ is an invertible matrix. The characteristic polynomial of $\boldsymbol{B}$ is:

Here, we made use of the following theorems:

• the multiplicative propertylink of determinant.

• the theoremlink that states $\det(\boldsymbol{I})=1$.

This completes the proof.

Proof. By theoremlink, if $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar matrices, then they share the same characteristic polynomial. This immediately means that $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same eigenvalues because eigenvalues are the roots of the characteristic polynomial. This completes the proof.