Suppose $\boldsymbol{A}$ and $\boldsymbol{B}$ are $n\times{n}$ matrices. $\boldsymbol{A}$ is similar to $\boldsymbol{B}$ if and only if there exists an $n\times{n}$ invertible matrix $\boldsymbol{P}$ such that:

$$\boldsymbol{A}= \boldsymbol{P}\boldsymbol{B}\boldsymbol{P}^{-1}$$

To indicate that $\boldsymbol{A}$ is similar to $\boldsymbol{B}$, we write $\boldsymbol{A}\sim\boldsymbol{B}$.

Example.

Checking that two matrices are similar

Consider the following matrices:

$$\boldsymbol{A}=\begin{pmatrix} -1&13\\0&4 \end{pmatrix},\;\;\;\;\; \boldsymbol{B}=\begin{pmatrix} 0&2\\2&3 \end{pmatrix},\;\;\;\;\; \boldsymbol{P}=\begin{pmatrix} 3&5\\1&2 \end{pmatrix}$$

Show that $\boldsymbol{A}$ is similar to $\boldsymbol{B}$ using $\boldsymbol{P}$.

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:

$$\begin{align*} \boldsymbol{P}^{-1} &=\frac{1}{(3)(2)-(5)(1)}\begin{pmatrix} 2&-5\\-1&3 \end{pmatrix}\\ &=\begin{pmatrix} 2&-5\\-1&3 \end{pmatrix} \end{align*}$$

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

$$\begin{align*} \boldsymbol{PB}\boldsymbol{P}^{-1}&= \begin{pmatrix}3&5\\1&2\end{pmatrix} \begin{pmatrix}0&2\\2&3\end{pmatrix} \begin{pmatrix}2&-5\\-1&3\end{pmatrix}\\ &=\begin{pmatrix}-1&13\\0&4\end{pmatrix}\\ &=\boldsymbol{A} \end{align*}$$

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

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Theorem.

Equivalent definition of similar matrices

Matrix $\boldsymbol{A}$ is similar to matrix $\boldsymbol{B}$ if and only if there exists an invertible matrix $\boldsymbol{P}$ such that:

$$\boldsymbol{B}= \boldsymbol{P}^{-1}\boldsymbol{AP}$$

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

$$\boldsymbol{A} =\boldsymbol{P}\boldsymbol{B}\boldsymbol{P}^{-1}$$

Let's perform some matrix manipulation:

$$\begin{align*} \boldsymbol{A} &=\boldsymbol{P}\boldsymbol{B}\boldsymbol{P}^{-1}\\ \boldsymbol{AP}&=\boldsymbol{P}\boldsymbol{B}\\ \boldsymbol{P}^{-1}\boldsymbol{AP}&=\boldsymbol{B}\\ \end{align*}$$

This completes the proof.

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Theorem.

Commutative property of matrix similarity

Let $\boldsymbol{A}$ and $\boldsymbol{B}$ be $n\times{n}$ matrices. $\boldsymbol{A}$ is similar to $\boldsymbol{B}$ if and only if $\boldsymbol{B}$ is similar to $\boldsymbol{A}$, that is:

$$\boldsymbol{A}\sim \boldsymbol{B} \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \boldsymbol{B}\sim \boldsymbol{A}$$

Since matrix similarity is commutative, we typically say "$\boldsymbol{A}$ and $\boldsymbol{B}$ are similar" instead of "$\boldsymbol{A}$ is similar to $\boldsymbol{B}$".

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

$$\boldsymbol{A}= \boldsymbol{P}\boldsymbol{B}\boldsymbol{P}^{-1}$$

By definition, we also have:

$$\boldsymbol{B}= \boldsymbol{P}^{-1}\boldsymbol{AP}$$

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

$$\boldsymbol{B}= \boldsymbol{R}\boldsymbol{A}\boldsymbol{R}^{-1}$$

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.

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Theorem.

Reflexive property of similar matrices

Let $\boldsymbol{A}$ be a square matrix. $\boldsymbol{A}$ is similar to itself, that is $\boldsymbol{A}\sim\boldsymbol{A}$.

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

$$\begin{equation}\label{eq:FrZhzta0E5oT9P3cMNS} \boldsymbol{A}= \boldsymbol{P}\boldsymbol{A}\boldsymbol{P}^{-1} \end{equation}$$

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

$$\begin{align*} \boldsymbol{A}&= \boldsymbol{I}\boldsymbol{A}\boldsymbol{I}^{-1}\\ &=\boldsymbol{I}\boldsymbol{A}\boldsymbol{I}\\ &=\boldsymbol{A} \end{align*}$$

This completes the proof.

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Theorem.

Transitive property of similar matrices

If matrices $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar and matrices $\boldsymbol{B}$ and $\boldsymbol{C}$ are similar, then $\boldsymbol{A}$ and $\boldsymbol{C}$ are similar, that is:

$$(\boldsymbol{A}\sim\boldsymbol{B} \;\;\text{and}\;\; \boldsymbol{B}\sim\boldsymbol{C}) \;\;\;\;{\color{blue}\implies}\;\;\;\; \boldsymbol{A}\sim\boldsymbol{C}$$

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:

$$\begin{align*} \boldsymbol{A}&= \boldsymbol{P}\boldsymbol{B}\boldsymbol{P}^{-1}\\ \boldsymbol{B}&= \boldsymbol{R}\boldsymbol{C}\boldsymbol{R}^{-1}\\ \end{align*}$$

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

$$\begin{align*} \boldsymbol{A}&= \boldsymbol{P} (\boldsymbol{R}\boldsymbol{C}\boldsymbol{R}^{-1}) \boldsymbol{P}^{-1}\\ &=(\boldsymbol{P} \boldsymbol{R})\boldsymbol{C}(\boldsymbol{R}^{-1} \boldsymbol{P}^{-1})\\ &=(\boldsymbol{P} \boldsymbol{R})\boldsymbol{C}(\boldsymbol{P} \boldsymbol{R})^{-1} \end{align*}$$

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

$$\boldsymbol{A}= \boldsymbol{W}\boldsymbol{C}\boldsymbol{W}^{-1}$$

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

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Theorem.

Similar matrices have the same determinant

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar $n\times{n}$ matrices, then $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same determinant, that is:

$$\boldsymbol{A}\sim\boldsymbol{B} \;\;\;\;{\color{blue}\implies}\;\;\;\; \det(\boldsymbol{A})=\det(\boldsymbol{B})$$

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

$$\boldsymbol{B}= \boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}$$

Taking the determinant of both sides gives:

$$\det(\boldsymbol{B})= \det(\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P})$$

By the multiplicative propertylink of determinants, we have that:

$$\begin{equation}\label{eq:RM1WoFYIUlRiZMWJ4yA} \det(\boldsymbol{B})= \det(\boldsymbol{P}^{-1})\cdot \det(\boldsymbol{A})\cdot \det(\boldsymbol{P}) \end{equation}$$

Recall the following propertylink of determinants:

$$\det(\boldsymbol{P}^{-1})= \frac{1}{\det(\boldsymbol{P})}$$

Therefore, \eqref{eq:RM1WoFYIUlRiZMWJ4yA} is:

$$\begin{align*} \det(\boldsymbol{B})&= \frac{\det(\boldsymbol{A})\cdot \det(\boldsymbol{P})} {\det(\boldsymbol{P})}\\ &=\det(\boldsymbol{A}) \end{align*}$$

This completes the proof.

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Theorem.

Similar matrices have the same rank

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar $n\times{n}$ matrices, then $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same rank, that is:

$$\boldsymbol{A}\sim\boldsymbol{B} \;\;\;\;{\color{blue}\implies}\;\;\;\; \mathrm{rank}(\boldsymbol{A})= \mathrm{rank}(\boldsymbol{B})$$

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

$$\boldsymbol{B}= \boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}$$

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

$$\begin{align*} \mathrm{rank}(\boldsymbol{B})&= \mathrm{rank}(\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P})\\ &= \mathrm{rank}\big((\boldsymbol{P}^{-1}\boldsymbol{A}) \boldsymbol{P}\big)\\ \end{align*}$$

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

$$\mathrm{rank}(\boldsymbol{B})= \mathrm{rank}\big(\boldsymbol{P}^{-1}\boldsymbol{A}\big)$$

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

$$\mathrm{rank}(\boldsymbol{B}) = \mathrm{rank}\big(\boldsymbol{A}\big)$$

This completes the proof.

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Theorem.

Similar matrices have the same trace

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar $n\times{n}$ matrices, then $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same tracelink, that is:

$$\boldsymbol{A}\sim\boldsymbol{B} \;\;\;\;{\color{blue}\implies}\;\;\;\; \mathrm{tr}(\boldsymbol{A})= \mathrm{tr}(\boldsymbol{B})$$

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

$$\boldsymbol{B}= \boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P}$$

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

$$\begin{align*} \mathrm{tr} (\boldsymbol{B})&= \mathrm{tr} (\boldsymbol{P}^{-1} \boldsymbol{A} \boldsymbol{P})\\ &=\mathrm{tr} \big((\boldsymbol{P}^{-1} \boldsymbol{A}) \boldsymbol{P}\big)\\ \end{align*}$$

By theoremlink, we have that:

$$\begin{align*} \mathrm{tr} (\boldsymbol{B}) &=\mathrm{tr} \big((\boldsymbol{P}^{-1}\boldsymbol{A}) \boldsymbol{P}\big)\\ &=\mathrm{tr} \big((\boldsymbol{A}\boldsymbol{P}^{-1}) \boldsymbol{P}\big)\\ &=\mathrm{tr} (\boldsymbol{A}\boldsymbol{P}^{-1} \boldsymbol{P})\\ &=\mathrm{tr} (\boldsymbol{AI})\\ &=\mathrm{tr} (\boldsymbol{A})\\ \end{align*}$$

This completes the proof.

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Theorem.

Similar matrices have the same characteristic polynomial

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar $n\times{n}$ matrices, then $\boldsymbol{A}$ and $\boldsymbol{B}$ have the same characteristic polynomiallink.

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:

$$\boldsymbol{B}= \boldsymbol{P}^{-1}\boldsymbol{AP}$$

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

$$\begin{align*} \det\big(\boldsymbol{B}-\lambda\boldsymbol{I}\big) &= \det\big(\boldsymbol{P}^{-1}\boldsymbol{AP}-\lambda\boldsymbol{I}\big)\\ &= \det\big(\boldsymbol{P}^{-1}\boldsymbol{AP}- \lambda\boldsymbol{I}(\boldsymbol{P}^{-1}\boldsymbol{P})\big)\\ &= \det\big(\boldsymbol{P}^{-1}\boldsymbol{AP}- \lambda\boldsymbol{I}\boldsymbol{P}^{-1}\boldsymbol{P}\big)\\ &= \det\big(\boldsymbol{P}^{-1}\boldsymbol{AP}- \lambda\boldsymbol{P}^{-1}\boldsymbol{I}\boldsymbol{P}\big)\\ &=\det\big(\boldsymbol{P}^{-1}(\boldsymbol{AP}- \lambda\boldsymbol{I}\boldsymbol{P})\big)\\ &=\det\big(\boldsymbol{P}^{-1}(\boldsymbol{A}- \lambda\boldsymbol{I})\boldsymbol{P}\big)\\ &=\det\big(\boldsymbol{P}^{-1}\big)\cdot\det\big(\boldsymbol{A}- \lambda\boldsymbol{I}\big)\cdot\det\big(\boldsymbol{P}\big)\\ &=\det\big(\boldsymbol{P}^{-1}\big) \cdot\det\big(\boldsymbol{P}\big) \cdot\det\big(\boldsymbol{A}- \lambda\boldsymbol{I}\big)\\ &=\det\big(\boldsymbol{P}^{-1}\boldsymbol{P}\big) \cdot\det\big(\boldsymbol{A}- \lambda\boldsymbol{I}\big)\\ &=\det\big(\boldsymbol{I}\big) \cdot\det\big(\boldsymbol{A}- \lambda\boldsymbol{I}\big)\\ &=\det\big(\boldsymbol{A}- \lambda\boldsymbol{I}\big)\\ \end{align*}$$

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.

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Theorem.

Similar matrices have the same eigenvalues

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are similar matrices, then the eigenvalueslink of $\boldsymbol{A}$ and $\boldsymbol{B}$ are the same. In other words, if $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_k$ are the eigenvalues of $\boldsymbol{A}$, then $\boldsymbol{B}$ will also have eigenvalues $\lambda_1$, $\lambda_2$, $\cdots$, $\lambda_k$.

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.

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Published by Isshin Inada

Edited by 0 others

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