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

Example.

Finding the transpose of matrices

Find the transpose of the following matrices:

A = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}), B = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{matrix}), C = (\begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{matrix})

Solution. The transpose of these matrices is:

A^{T} = (\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}), B^{T} = (\begin{matrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{matrix}), C^{T} = (\begin{matrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{matrix})

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

a $2 \times 2$ matrix remains a $2 \times 2$ matrix.
a $3 \times 2$ matrix becomes a $2 \times 3$ matrix.
a $3 \times 3$ matrix remains a $3 \times 3$ matrix.

In general, if $A$ is an $m \times n$ matrix, then $A^{T}$ 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 $A$ after taking the transpose. Since the rows and columns are swapped, this entry is located at $a_{12}$ in $A^{T}$ .

Example.

Shape of a matrix after taking its transpose

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

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

Properties of matrix transpose

Theorem.

Transpose of a matrix transpose

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

(A^{T})^{T} = A

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

To be more mathematically precise, suppose $A$ is as follows:

A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{matrix})

The transpose of $A$ is:

A^{T} = (\begin{matrix} a_{11} & a_{21} & \dots & a_{m 1} \\ a_{12} & a_{22} & \dots & a_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{1 n} & a_{2 n} & \dots & a_{m n} \end{matrix})

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

(A^{T})^{T} = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{matrix}) = A

This completes the proof.

Theorem.

Transpose of A+B

If $A$ and $B$ are matrices, then:

(A + B)^{T} = A^{T} + B^{T}

Proof. Consider the following matrices:

A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{matrix}), B = (\begin{matrix} b_{11} & b_{12} & \dots & b_{1 n} \\ b_{21} & b_{22} & \dots & b_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ b_{m 1} & b_{m 2} & \dots & b_{m n} \end{matrix})

The transpose of $A$ and $B$ is:

A^{T} = (\begin{matrix} a_{11} & a_{21} & \dots & a_{m 1} \\ a_{12} & a_{22} & \dots & a_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{1 n} & a_{2 n} & \dots & a_{m n} \end{matrix}), B^{T} = (\begin{matrix} b_{11} & b_{21} & \dots & b_{m 1} \\ b_{12} & b_{22} & \dots & b_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ b_{1 n} & b_{2 n} & \dots & b_{m n} \end{matrix})

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

\begin{matrix} (1) & A^{T} + B^{T} = (\begin{matrix} a_{11} + b_{11} & a_{21} + b_{21} & \dots & a_{m 1} + b_{m 1} \\ a_{12} + b_{12} & a_{22} + b_{22} & \dots & a_{m 2} + b_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{1 n} + b_{1 n} & a_{2 n} + b_{2 n} & \dots & a_{m n} + b_{m n} \end{matrix}) \end{matrix}

Now, the sum $A + B$ is:

A + B = (\begin{matrix} a_{11} + b_{11} & a_{12} + b_{12} & \dots & a_{1 n} + b_{1 n} \\ a_{21} + b_{21} & a_{22} + b_{22} & \dots & a_{2 n} + b_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} + b_{m 1} & a_{m 2} + b_{m 2} & \dots & a_{m n} + b_{m n} \end{matrix})

Taking the transpose gives:

\begin{matrix} (2) & (A + B)^{T} = (\begin{matrix} a_{11} + b_{11} & a_{21} + b_{21} & \dots & a_{m 1} + b_{m 1} \\ a_{12} + b_{12} & a_{22} + b_{22} & \dots & a_{m 2} + b_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{1 n} + b_{1 n} & a_{2 n} + b_{2 n} & \dots & a_{m n} + b_{m n} \end{matrix}) \end{matrix}

Notice that this matrix is equal to $(1)$ . Therefore, we have that:

(A + B)^{T} = A^{T} + B^{T}

This completes the proof.

Theorem.

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

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

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

Theorem.

Transpose of kA where k is a scalar

If $A$ is a matrix and $k$ is any scalar, then:

(k A)^{T} = k A^{T}

Proof. Consider the following matrix:

A = (\begin{matrix} a_{11} & a_{12} & \dots & a_{1 n} \\ a_{21} & a_{22} & \dots & a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} \end{matrix})

If $k$ is any scalar, then the product $k A$ is:

k A = (\begin{matrix} k a_{11} & k a_{12} & \dots & k a_{1 n} \\ k a_{21} & k a_{22} & \dots & k a_{2 n} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ k a_{m 1} & k a_{m 2} & \dots & k a_{m n} \end{matrix})

The transpose of $k A$ is:

\begin{matrix} (3) & (k A)^{T} = (\begin{matrix} k a_{11} & k a_{21} & \dots & k a_{m 1} \\ k a_{12} & k a_{22} & \dots & k a_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ k a_{1 n} & k a_{2 n} & \dots & k a_{m n} \end{matrix}) \end{matrix}

The transpose of $A$ is:

A^{T} = (\begin{matrix} a_{11} & a_{21} & \dots & a_{m 1} \\ a_{12} & a_{22} & \dots & a_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{1 n} & a_{2 n} & \dots & a_{m n} \end{matrix})

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

\begin{matrix} (4) & k A^{T} = (\begin{matrix} k a_{11} & k a_{21} & \dots & k a_{m 1} \\ k a_{12} & k a_{22} & \dots & k a_{m 2} \\ ⋮ & ⋮ & ⋱ & ⋮ \\ k a_{1 n} & k a_{2 n} & \dots & k a_{m n} \end{matrix}) \end{matrix}

This is equal to the matrix in $(3)$ . Therefore, we conclude that:

(k A)^{T} = k A^{T}

This completes the proof.

Theorem.

Expressing dot product using matrix notation

If $v$ and $w$ are vectors, then:

v \cdot w = v^{T} w = w^{T} v = v w^{T}

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

v = (\begin{matrix} v_{1} \\ v_{2} \\ v_{3} \end{matrix}), w = (\begin{matrix} w_{1} \\ w_{2} \\ w_{3} \end{matrix})

Their dot product is:

\begin{aligned} v \cdot w & = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} \\ = (\begin{array}{c} v_{1} & v_{2} & v_{3} \end{array}) (\begin{array}{c} w_{1} \\ w_{2} \\ w_{3} \end{array}) \\ = v^{T} w \end{aligned}

Similarly, we have that:

\begin{aligned} v \cdot w & = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} \\ = (\begin{array}{c} w_{1} & w_{2} & w_{3} \end{array}) (\begin{array}{c} v_{1} \\ v_{2} \\ v_{3} \end{array}) \\ = w^{T} v \end{aligned}

Similarly, we have that:

\begin{aligned} v \cdot w & = v_{1} w_{1} + v_{2} w_{2} + v_{3} w_{3} \\ = (\begin{array}{c} v_{1} \\ v_{2} \\ v_{3} \end{array}) (\begin{array}{c} w_{1} & w_{2} & w_{3} \end{array}) \\ = v w^{T} \end{aligned}

This completes the proof.

Theorem.

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

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

v^{T} v = ‖ v ‖^{2}

Where $‖ v ‖$ is the magnitudelink of $v$ .

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

\begin{aligned} v^{T} v & = v \cdot v \\ = ‖ v ‖^{2} \end{aligned}

This completes the proof.

Theorem.

Transpose of a product of two matrices

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

(A B)^{T} = B^{T} A^{T}

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

\begin{matrix} (5) & (A B)_{i j} = \sum_{k = 1}^{n} a_{i k} \cdot b_{k j} \end{matrix}

Here, the subscript $i j$ represents the value in the $i$ -th row $j$ -th column. $(5)$ is true for all $i = 1, 2, \dots, m$ and $j = 1, 2, \dots, p$ . For instance, $(A B)_{13}$ represents the entry at the $1$ st row $3$ rd column of matrix $A B$ .

We know from the definition of transpose that:

\begin{matrix} (6) & (A B)_{i j} = (A B)_{j i}^{T} \end{matrix}

Equating $(6)$ and $(5)$ gives:

\begin{matrix} (7) & \sum_{k = 1}^{n} a_{i k} \cdot b_{k j} = (A B)_{j i}^{T} \end{matrix}

Now, consider the following:

\begin{matrix} (8) & \begin{aligned} (B^{T} A^{T})_{j i} & = \sum_{k = 1}^{n} b_{j k}^{T} \cdot a_{k i}^{T} \\ = \sum_{k = 1}^{n} b_{k j} \cdot a_{i k} \end{aligned} \end{matrix}

Here, we used the fact that the entry $b_{j k}$ in $B^{T}$ is equal to the entry $b_{k j}$ in $B$ .

Equating $(7)$ and $(8)$ gives:

\begin{matrix} (9) & (B^{T} A^{T})_{j i} = (A B)_{j i}^{T} \end{matrix}

Equation $(9)$ holds true for all $j = 1, 2, \dots, p$ , and $i = 1, 2, \dots, m$ . Since the values of two matrices $A B^{T}$ and $B^{T} A^{T}$ are equivalent, these matrices are equivalent, that is:

(A B)^{T} = B^{T} A^{T}

This completes the proof.

Theorem.

Transpose of a product of three matrices

If $A$ , $B$ and $C$ are matrices, then:

(A B C)^{T} = C^{T} B^{T} A^{T}

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

\begin{aligned} (A B C)^{T} & = ((A B) C)^{T} \\ = C^{T} (A B)^{T} \\ = C^{T} B^{T} A^{T} \end{aligned}

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 $A_{i}$ for $i = 1, 2, \dots, n$ are matrices, then:

(A_{1} A_{2} \dots A_{n})^{T} = A_{n}^{T} \dots A_{2}^{T} 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:

(A_{1} A_{2})^{T} = A_{2}^{T} A_{1}^{T}

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

\begin{matrix} (10) & (A_{1} A_{2} \dots A_{n - 1})^{T} = A_{n - 1}^{T} \dots A_{2}^{T} A_{1}^{T} \end{matrix}

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

\begin{aligned} (A_{1} A_{2} \dots A_{n - 1} A_{n})^{T} & = ((A_{1} A_{2} \dots A_{n - 1}) A_{n})^{T} \\ = A_{n}^{T} (A_{1} A_{2} \dots A_{n - 1})^{T} \end{aligned}

We now use the inductive assumption $(10)$ to get:

(A_{1} A_{2} \dots A_{n - 1} A_{n})^{T} = A_{n}^{T} A_{n - 1}^{T} \dots A_{2}^{T} 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 $A$ is an $m \times n$ matrix and $v \in R^{n}$ and $w \in R^{m}$ are vectors, then:

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

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

\begin{aligned} A v \cdot w & = w^{T} (A v) \\ = (w^{T} A) v \\ = (A^{T} w)^{T} v \\ = v \cdot A^{T} w \end{aligned}

Note the following:

the third step uses theoremlink, that is, $(A B)^{T} = B^{T} A^{T}$ .
the final step uses theoremlink to convert the matrix product into a dot product.

This completes the proof.

Practice problems

Consider a $3 \times 4$ matrix $A$ . What would the shape of $A^{T}$ be?

3 \times 3

4 \times 4

4 \times 3

3 \times 4

Since $A^{T}$ has the columns of $A$ as its rows while the rows of $A$ as its columns, we have that $A^{T}$ has shape $4 \times 3$ .

Consider the following matrix:

A = (\begin{matrix} 3 & 4 & 1 \\ 5 & 6 & 8 \end{matrix})

Find $A^{T}$ .

Your answer

Submit answer

Show solution

We flip the rows and columns to get:

A^{T} = (\begin{matrix} 3 & 5 \\ 4 & 6 \\ 1 & 8 \end{matrix})

Published by Isshin Inada

Edited by 0 others

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