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# Invertibility, multiplicative and transpose properties of determinants

schedule Aug 12, 2023
Last updated
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Linear Algebra
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Theorem.

# Relationship between invertibility and determinant

A square matrix $\boldsymbol{A}$ is invertible if and only if $\det(\boldsymbol{A})\ne0$. In other words, $\boldsymbol{A}$ is singular if and only if $\det(\boldsymbol{A})=0$.

Proof. Let $\boldsymbol{A}$ be an invertible matrix. Let's first prove the following:

$$\boldsymbol{A}\text{ is invertible}\;\;\;\;\; {\color{blue}\implies}\;\;\;\;\;\det(\boldsymbol{A})\ne0$$

From theoremlink, we know that $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

$$\boldsymbol{A}=\boldsymbol{E}_k\cdots\boldsymbol{E}_2\boldsymbol{E}_1$$

Taking the determinant of both sides gives:

$$\det(\boldsymbol{A})= \det(\boldsymbol{E}_k\cdots\boldsymbol{E}_2\boldsymbol{E}_1)$$

Repeatedly applying theoremlink to the right-hand side gives:

$$\det(\boldsymbol{A})= \det(\boldsymbol{E}_k)\cdot\;\cdots\;\cdot\det(\boldsymbol{E}_2) \cdot\det(\boldsymbol{E}_1)$$

We know from theoremlink that the determinant of an elementary matrix cannot be $0$. Therefore, we have that:

$$\det(\boldsymbol{A})\ne0$$

Next, let's prove the converse, that is:

$$\det(\boldsymbol{A})\ne0 \;\;\;\;\; {\color{blue}\implies} \;\;\;\;\; \boldsymbol{A}\text{ is invertible}$$

Recall from our guide on proof by contraposition that we can equivalently prove the following contrapositive statement instead:

$$\boldsymbol{A}\text{ is not invertible}\;\;\;\;\; {\color{blue}\implies}\;\;\;\;\;\det(\boldsymbol{A})=0$$

We know from theoremlink that if $\boldsymbol{A}$ is not invertible, then the reduced row echelon form of $\boldsymbol{A}$ is not an identity matrix and instead has a row with all zeros. This reduced row echelon form is obtained by applying a series of elementary row operations on $\boldsymbol{A}$ like so:

$$\text{rref}(\boldsymbol{A})= \boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}$$

Taking the determinant of both sides gives:

$$\det\big(\text{rref}(\boldsymbol{A})\big)= \det\big(\boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}\big)$$

Because $\text{rref}(\boldsymbol{A})$ contains a row with all zeros, we know from theoremlink that its determinant is zero:

$$0= \det\big(\boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}\big)$$

After repeated application of theoremlink on the right-hand side, we get:

$$$$\label{eq:uh15d3QwVMp4dBC2l9I} 0= \det(\boldsymbol{E}_k)\cdot\;\cdots\;\cdot \det(\boldsymbol{E}_2)\cdot\det(\boldsymbol{E}_1) \cdot\det(\boldsymbol{A})$$$$

From theoremlink, the determinant of elementary matrices cannot be equal to zero. Therefore, for \eqref{eq:uh15d3QwVMp4dBC2l9I} to hold, $\det(\boldsymbol{A})=0$. This completes the proof.

Theorem.

# Multiplicative property of determinants

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are square matrices of the same shape, then:

$$\det(\boldsymbol{AB})= \det(\boldsymbol{A})\cdot\det(\boldsymbol{B})$$

Proof. First consider the case when $\boldsymbol{A}$ is invertible. By theoremlink, we know that $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

$$$$\label{eq:JH87eddXjoAVbCsnHhO} \boldsymbol{A}= \boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k$$$$

For reference later, let's take the determinant of both sides:

$$$$\label{eq:A3qN2BOhPSZhDTxp4FJ} \det(\boldsymbol{A})= \det(\boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k)$$$$

Now, let's multiply both sides of \eqref{eq:JH87eddXjoAVbCsnHhO} by some matrix $\boldsymbol{B}$ to get:

$$\boldsymbol{AB}= \boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k\boldsymbol{B}$$

Taking the determinant of both sides and repeatedly apply theoremlink to get:

\label{eq:EQDIFXv7fQvWqMq5jb7} \begin{aligned}[b] \det(\boldsymbol{AB})&= \det(\boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k\boldsymbol{B})\\ &= \det(\boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k)\cdot\det(\boldsymbol{B})\\ \end{aligned}

Substituting \eqref{eq:A3qN2BOhPSZhDTxp4FJ} into \eqref{eq:EQDIFXv7fQvWqMq5jb7} gives:

$$\det(\boldsymbol{AB})= \det(\boldsymbol{A})\cdot\det(\boldsymbol{B})$$

Now, suppose $\boldsymbol{A}$ is not invertible. By theoremlink, this means that $\det(\boldsymbol{A})=0$. Therefore, for any matrix $\boldsymbol{B}$, we have that:

$$\det(\boldsymbol{A})\cdot\det(\boldsymbol{B})=0$$

We must thus show that $\det(\boldsymbol{AB})=0$. The reduced row echelon form of $\boldsymbol{A}$ can be obtained by applying a series of elementary row operations:

$$\text{rref}(\boldsymbol{A})= \boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}$$

Now, let's multiply both sides by $\boldsymbol{B}$ to get:

$$\text{rref}(\boldsymbol{A})\boldsymbol{B}= \boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}\boldsymbol{B}$$

Taking the determinant of both sides gives:

$$$$\label{eq:DM0GgDYlP5mWd3hXtvz} \det\big(\text{rref}(\boldsymbol{A})\boldsymbol{B}\big)= \det\big(\boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}\boldsymbol{B}\big)$$$$

Because $\boldsymbol{A}$ is not invertible, the reduced row echelon form of $\boldsymbol{A}$ contains a row with all zeros. The matrix product $\mathrm{rref}(\boldsymbol{A})\boldsymbol{B}$ will therefore result in another matrix containing a row with all zeros. By theoremlink, this means that $\det\big(\text{rref}(\boldsymbol{A})\boldsymbol{B}\big)=0$. Therefore, \eqref{eq:DM0GgDYlP5mWd3hXtvz} becomes:

$$0= \det\big(\boldsymbol{E}_k\cdot\;\cdots\;\cdot \boldsymbol{E}_2\boldsymbol{E}_1 \boldsymbol{A}\boldsymbol{B}\big)$$

We now repeatedly apply theoremlink to get:

$$0= \det(\boldsymbol{E}_k)\cdot\;\cdots\;\cdot \det(\boldsymbol{E}_2)\cdot\det(\boldsymbol{E}_1)\cdot\det( \boldsymbol{A}\boldsymbol{B}\big)\\$$

By theoremlink, we know that the determinant of elementary matrices cannot be zero, and so $\det(\boldsymbol{AB})=0$. This completes the proof.

Theorem.

# Determinant of the inverse of a matrix is equal to the reciprocal of the determinant of the matrix

If $\boldsymbol{A}$ is an invertible matrix, then:

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

Proof. By the multiplicative propertylink of determinants, we have that:

\begin{align*} \det(\boldsymbol{A}^{-1}\boldsymbol{A})&= \det(\boldsymbol{A}^{-1})\cdot \det(\boldsymbol{A})\\ \det(\boldsymbol{I})&= \det(\boldsymbol{A}^{-1})\cdot \det(\boldsymbol{A})\\ \end{align*}

The determinant of the identity matrix is $1$ by theoremlink. Therefore, we get:

\begin{align*} 1&=\det(\boldsymbol{A}^{-1})\cdot \det(\boldsymbol{A})\\ \det(\boldsymbol{A}^{-1})&= \frac{1}{\det(\boldsymbol{A})} \end{align*}

This completes the proof.

Theorem.

# Determinant of kA where k is scalar

If $\boldsymbol{A}$ is an $n\times{n}$ matrix and $k$ is a scalar, then:

$$\det(k\boldsymbol{A})= k^n\cdot\det(\boldsymbol{A})$$

Proof. Consider an $n\times{n}$ matrix $\boldsymbol{A}$. The determinant of $k\boldsymbol{A}$ is:

\begin{align*} \det(k\boldsymbol{A})&= \det\Big(k(\boldsymbol{I}_n\boldsymbol{A})\Big)\\ &=\det\Big((k\boldsymbol{I}_n)\boldsymbol{A}\Big)\\ \end{align*}

Here, $\boldsymbol{I}_n$ is an $n\times{n}$ identity matrix. By the multiplicative property of determinantslink, we have that:

\begin{align*} \det(k\boldsymbol{A})&= \det(k\boldsymbol{I}_n)\cdot\det(\boldsymbol{A}) \end{align*}

Finally, by theoremlink, we have that $\det(k\boldsymbol{I}_n)=k^n$. Therefore, we conclude that:

\begin{align*} \det(k\boldsymbol{A})&= k^n\cdot\det(\boldsymbol{A}) \end{align*}

This completes the proof.

Theorem.

# Determinant of a transpose of a matrix

If $\boldsymbol{A}$ is a square matrix, then:

$$\det(\boldsymbol{A}^T)=\det(\boldsymbol{A})$$

Proof. Suppose $\boldsymbol{A}$ is not invertible. From theoremlink, we have that $\det(\boldsymbol{A})=0$. From theoremlink, if $\boldsymbol{A}$ is not invertible, then $\boldsymbol{A}^T$ is also not invertible, which means $\det(\boldsymbol{A}^T)=0$. Therefore, we have that:

$$\det(\boldsymbol{A}^T)=\det(\boldsymbol{A})$$

Now, suppose $\boldsymbol{A}$ is invertible. From theoremlink, $\boldsymbol{A}$ can be expressed as a product of elementary matrices:

$$$$\label{eq:VeE5NBxXWHXYDI9Q3Y5} \boldsymbol{A}= \boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k$$$$

For reference later, let's take the determinant of both sides and apply theoremlink repeatedly:

\label{eq:JUKQmcM56Yj39s7D7al} \begin{aligned}[b] \det(\boldsymbol{A})&= \det(\boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k) \\ &=\det(\boldsymbol{E}_1)\cdot\det(\boldsymbol{E}_2)\cdot\;\cdots\;\cdot \det(\boldsymbol{E}_k) \end{aligned}

Now, taking the transpose of both sides of \eqref{eq:VeE5NBxXWHXYDI9Q3Y5} and applying theoremlink gives:

\begin{align*} \boldsymbol{A}^T&= (\boldsymbol{E}_1\boldsymbol{E}_2\cdot\;\cdots\;\cdot \boldsymbol{E}_k)^T\\ &= \boldsymbol{E}_k^T\cdot\;\cdots\;\cdot \boldsymbol{E}_2^T\boldsymbol{E}_1^T \end{align*}

Taking the determinant of both sides and applying theoremlink repeatedly gives:

$$$$\label{eq:QvFCrdH1upnLdwJWLTK} \det(\boldsymbol{A}^T)= \det(\boldsymbol{E}_k^T) \cdot\;\cdots\;\cdot \det(\boldsymbol{E}_2^T) \cdot\det(\boldsymbol{E}_1^T)$$$$

From theoremlink, we know that the determinant of an elementary matrix is equal to the determinant of its transpose. Therefore, \eqref{eq:QvFCrdH1upnLdwJWLTK} becomes:

\begin{align*} \det(\boldsymbol{A}^T)&= \det(\boldsymbol{E}_k)\cdot\;\cdots\;\cdot \det(\boldsymbol{E}_2)\cdot \det(\boldsymbol{E}_1)\\ \end{align*}

This is the same as \eqref{eq:JUKQmcM56Yj39s7D7al}. Therefore, we conclude that:

$$\det(\boldsymbol{A}^T)=\det(\boldsymbol{A})$$

This completes the proof.

Published by Isshin Inada
Edited by 0 others
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