Matrix Transpose

The transpose of a matrix is the first matrix operation introduced in this course not defined in terms of scalar algebra.

The transpose of some matrix $A$ is denoted by the symbol $A^T$ . If $A$ is an $m \times n$ matrix, $A^T$ will be an $n \times m$ matrix.

The transpose of a matrix is obtained by swapping the rows with the columns of $A$ . More formally, for each row $i$ and column $j$ of $A$ ,

[A^T]_{ij} = [A]_{ji}.

In other words, if

A = \begin{bmatrix} 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{bmatrix},

then

A^T = \begin{bmatrix} a_{11} & a_{21} & \cdots & a_{n1} \\ a_{12} & a_{22} & \cdots & a_{n2} \\ \vdots & \vdots & \ddots & \vdots \\ a_{1m} & a_{2m} & \cdots & a_{nm} \end{bmatrix}.

By transposing a matrix twice, you return to the starting matrix. That is

(A^T)^T = A.

Example

Here is a simple numerical example to ease understanding. Let

A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}.

Then,

A^T = \begin{bmatrix} 1 & 3 & 5 \\ 2 & 4 & 6 \end{bmatrix}.

One helpful trick to compute the transpose of a matrix manually is to think of it in the following way:

The first column of $A$ - which had the entries $1$ , $3$ , and $5$ - became the first row of $A^T$ . The second column of $A$ - with the entries $2$ , $4$ , and $6$ - became the second row of $A^T$ .

By proceeding in this fashion, the process becomes rather trivial.

Properties of the Transpose of a Matrix

The following properties hold for the transposition of matrices:

Distributivity Over Matrix Addition

Let $A$ and $B$ be two matrices of the same shape. Then

(A + B)^T = A^T + B^T.

Since the proof is very short, if you’re interested, here it is:

For each $i$ and $j$ ,

\begin{align*} \left[ (A + B)^T \right]_{ij} &= [A + B]_{ji} \\ &= [A]_{ji} + [B]_{ji} \\ &= \left[ A^T \right]_{ij} + \left[ B^T \right]_{ij} \\ &= \left[ A^T + B^T \right]_{ij}. \end{align*}

We proved that all entries of $(A + B)^T$ and $A^T + B^T$ are the same, therefore

(A + B)^T = A^T + B^T.

Distributivity Over Scalar Multiplication

Let $A$ be a matrix and let $c$ be a scalar. Then,

(cA)^T = cA^T.

This proof is also very simple, so if you’re interested, read along:

For each $i$ and $j$ ,

\begin{align*} \left[ (cA)^T \right]_{ij} &= [cA]_{ji} \\ &= c[A]_{ji} \\ &= c \left[ A^T \right]_{ij}. \end{align*}

Again, we proved that the entries of $(cA)^T$ and $cA^T$ are all the same, therefore

(cA)^T = cA^T.