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# Comprehensive Guide on Transpose of Matrices

schedule Aug 10, 2023
Last updated
local_offer
Linear Algebra
Tags
Definition.

# Transpose of a matrix

The transpose of a matrix $\boldsymbol{A}$, denoted as $\boldsymbol{A}^T$, is a matrix obtained by swapping the rows and the columns of $\boldsymbol{A}$. This means that:

• the first row of $\boldsymbol{A}$ becomes the first column of $\boldsymbol{A}^T$.

• the second row of $\boldsymbol{A}$ becomes the second column of $\boldsymbol{A}^T$.

• and so on.

Mathematically, if $\boldsymbol{A}$ is an $m\times{n}$ matrix with entries $a_{ij}$, then the transpose of $\boldsymbol{A}^T$ is an $n\times{m}$ matrix with entries $a_{ji}$. In other words, the entry located at the $i$-th row $j$-th column in $\boldsymbol{A}$ will be located at the $j$-th row $i$-th column in $\boldsymbol{A}^T$.

Example.

# Finding the transpose of matrices

Find the transpose of the following matrices:

$$\boldsymbol{A}= \begin{pmatrix} 1&2\\3&4 \end{pmatrix},\;\;\;\;\; \boldsymbol{B}= \begin{pmatrix} 1&2&3\\4&5&6 \end{pmatrix},\;\;\;\;\; \boldsymbol{C}= \begin{pmatrix} 1&2&3\\4&5&6\\7&8&9 \end{pmatrix}$$

Solution. The transpose of these matrices is:

$$\boldsymbol{A}^T= \begin{pmatrix} 1&3\\2&4 \end{pmatrix},\;\;\;\;\; \boldsymbol{B}^T= \begin{pmatrix} 1&4\\2&5\\3&6 \end{pmatrix},\;\;\;\;\; \boldsymbol{C}^T= \begin{pmatrix} 1&4&7\\2&5&8\\3&6&9 \end{pmatrix}$$

Notice how taking the transpose affects the shape of the matrices:

• a $2\times2$ matrix remains a $2\times2$ matrix.

• a $3\times2$ matrix becomes a $2\times3$ matrix.

• a $3\times3$ matrix remains a $3\times3$ matrix.

In general, if $\boldsymbol{A}$ is an $m\times{n}$ matrix, then $\boldsymbol{A}^T$ will be an $n\times{m}$ matrix. Let's also understand the mathematical definition of the matrix transpose. Observe what happens to the entry $a_{21}=3$ in $\boldsymbol{A}$ after taking the transpose. Since the rows and columns are swapped, this entry is located at $a_{12}$ in $\boldsymbol{A}^T$.

Example.

# Shape of a matrix after taking its transpose

Suppose we take the transpose of a matrix $\boldsymbol{A}$ with $2$ rows and $3$ columns. What would the shape of $\boldsymbol{A}^T$ be?

Solution. Since we are swapping the rows and columns, the transpose of $\boldsymbol{A}$ would have $3$ rows and $2$ columns, that is, $\boldsymbol{A}^T\in\mathbb{R}^{3\times2}$.

# Properties of matrix transpose

Theorem.

## Transpose of a matrix transpose

Taking the transpose of a matrix transpose results in the matrix itself:

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

Proof. By definition, $\boldsymbol{A}^T$ is obtained by swapping the rows and columns of $\boldsymbol{A}$. Taking the transpose of $\boldsymbol{A}^T$ will swap back the rows and columns and so we end up with the original matrix $\boldsymbol{A}$.

To be more mathematically precise, suppose $\boldsymbol{A}$ is as follows:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{pmatrix}$$

The transpose of $\boldsymbol{A}$ is:

$$\boldsymbol{A}^T=\begin{pmatrix} a_{11}&a_{21}&\cdots&a_{m1}\\ a_{12}&a_{22}&\cdots&a_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}&a_{2n}&\cdots&a_{mn} \end{pmatrix}$$

The transpose of $\boldsymbol{A}^T$ is:

$$(\boldsymbol{A}^T)^T=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{pmatrix}= \boldsymbol{A}$$

This completes the proof.

Theorem.

## Transpose of A+B

If $\boldsymbol{A}$ and $\boldsymbol{B}$ are matrices, then:

$$(\boldsymbol{A}+\boldsymbol{B})^T= \boldsymbol{A}^T+\boldsymbol{B}^T$$

Proof. Consider the following matrices:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{pmatrix},\;\;\;\;\;\; \boldsymbol{B}=\begin{pmatrix} b_{11}&b_{12}&\cdots&b_{1n}\\ b_{21}&b_{22}&\cdots&b_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ b_{m1}&b_{m2}&\cdots&b_{mn} \end{pmatrix}$$

The transpose of $\boldsymbol{A}$ and $\boldsymbol{B}$ is:

$$\boldsymbol{A}^T=\begin{pmatrix} a_{11}&a_{21}&\cdots&a_{m1}\\ a_{12}&a_{22}&\cdots&a_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}&a_{2n}&\cdots&a_{mn} \end{pmatrix},\;\;\;\;\;\; \boldsymbol{B}^T=\begin{pmatrix} b_{11}&b_{21}&\cdots&b_{m1}\\ b_{12}&b_{22}&\cdots&b_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ b_{1n}&b_{2n}&\cdots&b_{mn} \end{pmatrix}$$

The sum $\boldsymbol{A}^T+\boldsymbol{B}^T$ is:

$$\begin{equation}\label{eq:yJ6VsJMu5sStPhtgG31} \boldsymbol{A}^T+\boldsymbol{B}^T=\begin{pmatrix} a_{11}+b_{11}&a_{21}+b_{21}&\cdots&a_{m1}+b_{m1}\\ a_{12}+b_{12}&a_{22}+b_{22}&\cdots&a_{m2}+b_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}+b_{1n}&a_{2n}+b_{2n}&\cdots&a_{mn}+b_{mn} \end{pmatrix} \end{equation}$$

Now, the sum $\boldsymbol{A}+\boldsymbol{B}$ is:

$$\boldsymbol{A}+\boldsymbol{B}=\begin{pmatrix} a_{11}+b_{11}&a_{12}+b_{12}&\cdots&a_{1n}+b_{1n}\\ a_{21}+b_{21}&a_{22}+b_{22}&\cdots&a_{2n}+b_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}+b_{m1}&a_{m2}+b_{m2}&\cdots&a_{mn}+b_{mn} \end{pmatrix}$$

Taking the transpose gives:

$$\begin{equation}\label{eq:CA2ZWVSz7LnrIMu6mZs} (\boldsymbol{A}+\boldsymbol{B})^T=\begin{pmatrix} a_{11}+b_{11}&a_{21}+b_{21}&\cdots&a_{m1}+b_{m1}\\ a_{12}+b_{12}&a_{22}+b_{22}&\cdots&a_{m2}+b_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}+b_{1n}&a_{2n}+b_{2n}&\cdots&a_{mn}+b_{mn} \end{pmatrix} \end{equation}$$

Notice that this matrix is equal to \eqref{eq:yJ6VsJMu5sStPhtgG31}. Therefore, we have that:

$$(\boldsymbol{A}+\boldsymbol{B})^T= \boldsymbol{A}^T+\boldsymbol{B}^T$$

This completes the proof.

Theorem.

# Diagonal entries of a square matrix do not change after taking its transpose

Let $\boldsymbol{A}$ be a square matrix. The diagonal entries of $\boldsymbol{A}$ and $\boldsymbol{A}^T$ are the same.

Proof. Suppose $\boldsymbol{A}$ is an $n\times{n}$ matrix. The diagonals of $\boldsymbol{A}$ is $a_{ii}$ for $i=1,2,\cdots,n$. Taking the transpose involves switching the two subscripts, but since they are identical, we also end up with $a_{ii}$ as the diagonal entries of $\boldsymbol{A}^T$. This completes the proof.

Theorem.

## Transpose of kA where k is a scalar

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

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

Proof. Consider the following matrix:

$$\boldsymbol{A}=\begin{pmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{pmatrix}$$

If $k$ is any scalar, then the product $k\boldsymbol{A}$ is:

$$k\boldsymbol{A}=\begin{pmatrix} ka_{11}&ka_{12}&\cdots&ka_{1n}\\ ka_{21}&ka_{22}&\cdots&ka_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ ka_{m1}&ka_{m2}&\cdots&ka_{mn} \end{pmatrix}$$

The transpose of $k\boldsymbol{A}$ is:

$$\begin{equation}\label{eq:VMQuvGftyd4iwd5MPJ0} (k\boldsymbol{A})^T=\begin{pmatrix} ka_{11}&ka_{21}&\cdots&ka_{m1}\\ ka_{12}&ka_{22}&\cdots&ka_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ ka_{1n}&ka_{2n}&\cdots&ka_{mn} \end{pmatrix} \end{equation}$$

The transpose of $\boldsymbol{A}$ is:

$$\boldsymbol{A}^T=\begin{pmatrix} a_{11}&a_{21}&\cdots&a_{m1}\\ a_{12}&a_{22}&\cdots&a_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}&a_{2n}&\cdots&a_{mn} \end{pmatrix}$$

The scalar-matrix product $k\boldsymbol{A}^T$ is:

$$\begin{equation}\label{eq:mEq1tU7quIWC01rlPIr} k\boldsymbol{A}^T=\begin{pmatrix} ka_{11}&ka_{21}&\cdots&ka_{m1}\\ ka_{12}&ka_{22}&\cdots&ka_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ ka_{1n}&ka_{2n}&\cdots&ka_{mn} \end{pmatrix} \end{equation}$$

This is equal to the matrix in \eqref{eq:VMQuvGftyd4iwd5MPJ0}. Therefore, we conclude that:

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

This completes the proof.

Theorem.

## Expressing dot product using matrix notation

If $\boldsymbol{v}$ and $\boldsymbol{w}$ are vectors, then:

$$\boldsymbol{v}\cdot{\boldsymbol{w}} \;\;\;{\color{green}=}\;\;\;\boldsymbol{v}^T\boldsymbol{w} \;\;\;{\color{green}=}\;\;\;\boldsymbol{w}^T\boldsymbol{v} \;\;\;{\color{green}=}\;\;\;\boldsymbol{v}\boldsymbol{w}^T$$

Proof. We will prove the case for $\mathbb{R}^3$ but this easily generalizes to $\mathbb{R}^n$. Let vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ be defined as follows:

$$\boldsymbol{v}= \begin{pmatrix} v_1\\ v_2\\ v_3\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{w}= \begin{pmatrix} w_1\\ w_2\\ w_3\\ \end{pmatrix}$$

Their dot product is:

\begin{align*} \boldsymbol{v}\cdot{\boldsymbol{w}}&= v_1w_1+v_2w_2+v_3w_3\\ &= \begin{pmatrix} v_1&v_2&v_3 \end{pmatrix} \begin{pmatrix} w_1\\w_2\\w_3 \end{pmatrix}\\ &=\boldsymbol{v}^T\boldsymbol{w} \end{align*}

Similarly, we have that:

\begin{align*} \boldsymbol{v}\cdot{\boldsymbol{w}}&= v_1w_1+v_2w_2+v_3w_3\\ &= \begin{pmatrix} w_1&w_2&w_3 \end{pmatrix} \begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}\\ &=\boldsymbol{w}^T\boldsymbol{v} \end{align*}

Similarly, we have that:

\begin{align*} \boldsymbol{v}\cdot{\boldsymbol{w}}&= v_1w_1+v_2w_2+v_3w_3\\ &= \begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix} \begin{pmatrix} w_1&w_2&w_3 \end{pmatrix}\\ &=\boldsymbol{v}\boldsymbol{w}^T \end{align*}

This completes the proof.

Theorem.

# Matrix product of a vector and its transpose is equal to the vector's squared magnitude

If $\boldsymbol{v}$ is a vector in $\mathbb{R}^n$, then:

$$\boldsymbol{v}^T\boldsymbol{v}= \Vert\boldsymbol{v}\Vert^2$$

Where $\Vert\boldsymbol{v}\Vert$ is the magnitudelink of $\boldsymbol{v}$.

Proof. By the previous propertylink and propertylink of dot product, we have that:

\begin{align*} \boldsymbol{v}^T\boldsymbol{v}&= \boldsymbol{v}\cdot{\boldsymbol{v}}\\ &=\Vert\boldsymbol{v}\Vert^2 \end{align*}

This completes the proof.

Theorem.

## Transpose of a product of two matrices

If $\boldsymbol{A}$ is an $m\times{n}$ matrix and $\boldsymbol{B}$ is an $n\times{p}$ matrix, then:

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

Proof. From the rules of matrix multiplication, we know that:

$$\begin{equation}\label{eq:hZKaAMSlx63jVMkEVY8} (\boldsymbol{AB})_{ij}= \sum^n_{k=1}a_{ik}\cdot{b_{kj}} \end{equation}$$

Here, the subscript $ij$ represents the value in the $i$-th row $j$-th column. \eqref{eq:hZKaAMSlx63jVMkEVY8} is true for all $i=1,2,\cdots,m$ and $j=1,2,\cdots,p$. For instance, $(\boldsymbol{AB})_{13}$ represents the entry at the $1$st row $3$rd column of matrix $\boldsymbol{AB}$.

We know from the definition of transpose that:

$$\begin{equation}\label{eq:aZZgDF30ZHVz6WidSNK} (\boldsymbol{AB})_{ij}= (\boldsymbol{AB})^T_{ji} \end{equation}$$

Equating \eqref{eq:aZZgDF30ZHVz6WidSNK} and \eqref{eq:hZKaAMSlx63jVMkEVY8} gives:

$$\begin{equation}\label{eq:LMNSG63uKXgNU6nkuDe} \sum^n_{k=1}a_{ik}\cdot{b_{kj}}= (\boldsymbol{AB})^T_{ji} \end{equation}$$

Now, consider the following:

\begin{equation}\label{eq:ZgdDoK2zrOpYmDYO8yN} \begin{aligned}[b] (\boldsymbol{B}^T\boldsymbol{A}^T)_{ji}&= \sum^n_{k=1}b^T_{jk}\cdot{a^T_{ki}}\\ &=\sum^n_{k=1}b_{kj}\cdot{a_{ik}} \end{aligned} \end{equation}

Here, we used the fact that the entry $b_{jk}$ in $\boldsymbol{B}^T$ is equal to the entry $b_{kj}$ in $\boldsymbol{B}$.

Equating \eqref{eq:LMNSG63uKXgNU6nkuDe} and \eqref{eq:ZgdDoK2zrOpYmDYO8yN} gives:

$$\begin{equation}\label{eq:Bd1Iy1tDn1Ilz2TpbMa} (\boldsymbol{B}^T\boldsymbol{A}^T)_{ji}= (\boldsymbol{AB})^T_{ji} \end{equation}$$

Equation \eqref{eq:Bd1Iy1tDn1Ilz2TpbMa} holds true for all $j=1,2,\cdots,p$, and $i=1,2,\cdots,m$. Since the values of two matrices $\boldsymbol{AB}^T$ and $\boldsymbol{B}^T\boldsymbol{A}^T$ are equivalent, these matrices are equivalent, that is:

$$(\boldsymbol{AB})^T= \boldsymbol{B}^T\boldsymbol{A}^T$$

This completes the proof.

Theorem.

## Transpose of a product of three matrices

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

$$(\boldsymbol{ABC})^T= \boldsymbol{C}^T\boldsymbol{B}^T\boldsymbol{A}^T$$

Proof. The proof makes use of our previous theorem $(\boldsymbol{AB})^T=\boldsymbol{B}^T\boldsymbol{A}^T$. We start from the left-hand side of our proposition:

\begin{align*} (\boldsymbol{ABC})^T &=\Big((\boldsymbol{AB})\boldsymbol{C}\Big)^T\\ &=\boldsymbol{C}^T(\boldsymbol{AB})^T\\ &=\boldsymbol{C}^T\boldsymbol{B}^T\boldsymbol{A}^T \end{align*}

This completes the proof.

We will now generalize this theorem for any number of matrices. The proof uses induction and follows a similar logic.

Theorem.

## Transpose of a product of n matrices

If $\boldsymbol{A}_i$ for $i=1,2,\cdots,n$ are matrices, then:

$$(\boldsymbol{A}_1 \boldsymbol{A}_2 \cdots \boldsymbol{A}_n)^T= \boldsymbol{A}_n^T \cdots \boldsymbol{A}_2^T \boldsymbol{A}_1^T$$

Proof. We will prove the theorem by induction. Consider the base case when we have $2$ matrices. We have already shown in theoremlink that:

$$(\boldsymbol{A}_1\boldsymbol{A}_2)^T= \boldsymbol{A}_2^T\boldsymbol{A}_1^T$$

Therefore, the base case holds. We now assume that the theorem holds for $n-1$ matrices:

$$\begin{equation}\label{eq:IHQoYjMe8rq0BGQ1T2f} (\boldsymbol{A}_1 \boldsymbol{A}_2\cdots \boldsymbol{A}_{n-1} )^T= \boldsymbol{A}_{n-1}^T\cdots\boldsymbol{A}_2^T\boldsymbol{A}_1^T \end{equation}$$

Our goal is to show that the theorem holds for $n$ matrices:

\begin{align*} (\boldsymbol{A}_1 \boldsymbol{A}_2\cdots \boldsymbol{A}_{n-1} \boldsymbol{A}_{n} )^T&= \Big((\boldsymbol{A}_1 \boldsymbol{A}_2\cdots \boldsymbol{A}_{n-1}) \boldsymbol{A}_{n}\Big)^T\\ &=\boldsymbol{A}_{n}^T(\boldsymbol{A}_1 \boldsymbol{A}_2\cdots \boldsymbol{A}_{n-1})^T \end{align*}

We now use the inductive assumption \eqref{eq:IHQoYjMe8rq0BGQ1T2f} to get:

$$(\boldsymbol{A}_1 \boldsymbol{A}_2\cdots \boldsymbol{A}_{n-1} \boldsymbol{A}_{n} )^T= \boldsymbol{A}_{n}^T \boldsymbol{A}_{n-1}^T\cdots\boldsymbol{A}_2^T \boldsymbol{A}_1^T$$

By the principle of mathematical induction, the theorem holds for the general case of $n$ matrices. This completes the proof.

The next theorem is useful for certain proofs.

Theorem.

## Another expression for the dot product of Av and w

If $\boldsymbol{A}$ is an $m\times{n}$ matrix and $\boldsymbol{v}\in\mathbb{R}^n$ and $\boldsymbol{w}\in\mathbb{R}^{m}$ are vectors, then:

$$\boldsymbol{A}\boldsymbol{v}\cdot\boldsymbol{w}= \boldsymbol{v}\cdot\boldsymbol{A}^T\boldsymbol{w}$$

Proof. The matrix-vector product $\boldsymbol{A}\boldsymbol{v}$ results in a vector. We can use theoremlink to convert the dot product of $\boldsymbol{Av}$ and $\boldsymbol{w}$ into a matrix product:

\begin{align*} \boldsymbol{A}\boldsymbol{v}\cdot\boldsymbol{w}&= \boldsymbol{w}^T(\boldsymbol{A}\boldsymbol{v})\\ &=(\boldsymbol{w}^T\boldsymbol{A})\boldsymbol{v}\\ &=(\boldsymbol{A}^T\boldsymbol{w})^T\boldsymbol{v}\\ &=\boldsymbol{v}\cdot\boldsymbol{A}^T\boldsymbol{w} \\ \end{align*}

Note the following:

• the third step uses theoremlink, that is, $(\boldsymbol{A}\boldsymbol{B})^T= \boldsymbol{B}^T\boldsymbol{A}^T$.

• the final step uses theoremlink to convert the matrix product into a dot product.

This completes the proof.

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