Transpose of a Matrix

This lesson explains the transpose of a matrix and uses it to prove three things: that a matrix plus its transpose is symmetric, that a matrix minus its transpose is skew symmetric, and that the transpose of a product equals the product of the transposes taken in reverse order.

What the transpose is

The transpose of a matrix is obtained by interchanging its rows and columns. For a matrix $A$, the transpose is written $A^{\mathsf T}$.

$A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \implies A^{\mathsf T} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$

The first row $\begin{bmatrix} 1 & 2 \end{bmatrix}$ becomes the first column, and the second row becomes the second column.

Sum with the transpose is symmetric

Given the matrix

$A = \begin{bmatrix} 0 & 1 \\ 2 & 4 \end{bmatrix},$

we show that $A + A^{\mathsf T}$ is symmetric.

First write the transpose by swapping rows and columns:

$A^{\mathsf T} = \begin{bmatrix} 0 & 2 \\ 1 & 4 \end{bmatrix}.$

Then add:

$A + A^{\mathsf T} = \begin{bmatrix} 0 & 3 \\ 3 & 8 \end{bmatrix}.$

A matrix is symmetric when it equals its own transpose. Taking the transpose of this result swaps the rows and columns, but the off diagonal entries are both $3$, so nothing changes:

$\left(A + A^{\mathsf T}\right)^{\mathsf T} = \begin{bmatrix} 0 & 3 \\ 3 & 8 \end{bmatrix} = A + A^{\mathsf T}.$

Since $\left(A + A^{\mathsf T}\right)^{\mathsf T} = A + A^{\mathsf T}$, the sum is symmetric.

Difference is skew symmetric

Now subtract the transpose from $A$:

$A - A^{\mathsf T} = \begin{bmatrix} 0 & 1 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$

A matrix is skew symmetric when its transpose equals its negative. Taking the transpose swaps the off diagonal entries:

$\left(A - A^{\mathsf T}\right)^{\mathsf T} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = -\left(A - A^{\mathsf T}\right).$

Since the transpose equals the negative, $A - A^{\mathsf T}$ is skew symmetric.

Transpose of a product reverses the order

Given

$A = \begin{bmatrix} 0 & 4 \\ 2 & 3 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 2 \\ 4 & 1 \end{bmatrix},$

we prove that $\left(AB\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$.

The transposes are

$A^{\mathsf T} = \begin{bmatrix} 0 & 2 \\ 4 & 3 \end{bmatrix}, \qquad B^{\mathsf T} = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix}.$

Left side

Multiply $A$ by $B$, taking each row of $A$ against each column of $B$:

$AB = \begin{bmatrix} 0\cdot 1 + 4\cdot 4 & 0\cdot 2 + 4\cdot 1 \\ 2\cdot 1 + 3\cdot 4 & 2\cdot 2 + 3\cdot 1 \end{bmatrix} = \begin{bmatrix} 16 & 4 \\ 14 & 7 \end{bmatrix}.$

Now take the transpose by swapping rows and columns:

$\left(AB\right)^{\mathsf T} = \begin{bmatrix} 16 & 14 \\ 4 & 7 \end{bmatrix}.$

Right side

Multiply $B^{\mathsf T}$ by $A^{\mathsf T}$:

$B^{\mathsf T} A^{\mathsf T} = \begin{bmatrix} 1\cdot 0 + 4\cdot 4 & 1\cdot 2 + 4\cdot 3 \\ 2\cdot 0 + 1\cdot 4 & 2\cdot 2 + 1\cdot 3 \end{bmatrix} = \begin{bmatrix} 16 & 14 \\ 4 & 7 \end{bmatrix}.$

Both sides give the same matrix, so $\left(AB\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$.

Properties of the transpose

For a square matrix $A$, the main properties are:

$\left(A^{\mathsf T}\right)^{\mathsf T} = A$

$\left(A + B\right)^{\mathsf T} = A^{\mathsf T} + B^{\mathsf T}$

$\left(AB\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$

If $A$ is symmetric then $A^{\mathsf T} = A$, and if $A$ is skew symmetric then $A^{\mathsf T} = -A$.

Key takeaways

• The transpose swaps the rows and columns of a matrix, and transposing twice returns the original matrix.
• For any square matrix, $A + A^{\mathsf T}$ is symmetric and $A - A^{\mathsf T}$ is skew symmetric.
• The transpose of a product reverses the order of the factors: $\left(AB\right)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$.