Transpose of Matrices: Exercise 3.3 Solutions

This lesson works through the "sure questions" of Exercise 3.3 on the transpose of a matrix: finding transposes, verifying the standard transpose identities, and splitting a matrix into symmetric and skew-symmetric parts.

Finding the transpose

The transpose is found by interchanging rows and columns: the first row becomes the first column, the second row becomes the second column, and so on.

A column matrix. Given the column matrix, its transpose is the row matrix: $\begin{bmatrix} 5 \\ 1 \\ 2 \\ -1 \end{bmatrix}^{\mathsf T} = \begin{bmatrix} 5 & 1 & 2 & -1 \end{bmatrix}$

A square matrix. $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}^{\mathsf T} = \begin{bmatrix} 1 & 2 \\ -1 & 3 \end{bmatrix}$

A 3 by 3 matrix. $\begin{bmatrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \end{bmatrix}^{\mathsf T} = \begin{bmatrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \end{bmatrix}$

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

We are given $A^{\mathsf T}$ and $B$, so first recover $A$ and $B^{\mathsf T}$.

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

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

Left side. Add $A$ and $B$, then transpose: $A + B = \begin{bmatrix} 2 & 1 & 3 \\ 5 & 4 & 4 \end{bmatrix}, \qquad (A+B)^{\mathsf T} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 3 & 4 \end{bmatrix}$

Right side. Add the two transposes: $A^{\mathsf T} + B^{\mathsf T} = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 4 \\ 3 & 4 \end{bmatrix}$

Both sides agree, so $(A+B)^{\mathsf T} = A^{\mathsf T} + B^{\mathsf T}$. (The companion result $(A-B)^{\mathsf T} = A^{\mathsf T} - B^{\mathsf T}$ is left as practice.)

Finding $(A + 2B)^{\mathsf T}$

Given $A^{\mathsf T} = \begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$.

Recover $A$ and scale $B$: $A = \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix}, \qquad 2B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$

Then add and transpose: $A + 2B = \begin{bmatrix} -4 & 1 \\ 5 & 6 \end{bmatrix}, \qquad (A + 2B)^{\mathsf T} = \begin{bmatrix} -4 & 5 \\ 1 & 6 \end{bmatrix}$

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

Here $A$ is a column matrix and $B$ is a row matrix: $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}, \qquad B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

Left side. The product $AB$ is a $3 \times 3$ matrix (each entry is one element of $A$ times one element of $B$): $AB = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix}, \qquad (AB)^{\mathsf T} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

Right side. With $B^{\mathsf T} = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$ and $A^{\mathsf T} = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$: $B^{\mathsf T} A^{\mathsf T} = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$

The two results are equal, so $(AB)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$.

Rotation matrix: verifying $A^{\mathsf T} A = I$

Let $A = \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}$, so $A^{\mathsf T} = \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}$.

Multiply $A^{\mathsf T}$ by $A$ and use $\sin^2\alpha + \cos^2\alpha = 1$: $A^{\mathsf T} A = \begin{bmatrix} \cos^2\alpha + \sin^2\alpha & \cos\alpha\sin\alpha - \sin\alpha\cos\alpha \\ \sin\alpha\cos\alpha - \cos\alpha\sin\alpha & \sin^2\alpha + \cos^2\alpha \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I$

Symmetric and skew-symmetric matrices

For $A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$ we have $A^{\mathsf T} = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$.

$A + A^{\mathsf T}$ is symmetric (it equals its own transpose): $A + A^{\mathsf T} = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}$

$A - A^{\mathsf T}$ is skew-symmetric (its transpose is its negative): $A - A^{\mathsf T} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \qquad (A - A^{\mathsf T})^{\mathsf T} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = -(A - A^{\mathsf T})$

Expressing a matrix as a sum

Any square matrix splits as $A = P + Q$, where $P = \tfrac{1}{2}(A + A^{\mathsf T})$ is symmetric and $Q = \tfrac{1}{2}(A - A^{\mathsf T})$ is skew-symmetric. For $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$:

$P = \tfrac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}, \qquad Q = \tfrac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

Indeed $P + Q = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = A$.

Key takeaways

• The transpose interchanges rows and columns, and taking the transpose twice returns the original matrix.
• The transpose distributes over sums, $(A+B)^{\mathsf T} = A^{\mathsf T} + B^{\mathsf T}$, but reverses the order in products, $(AB)^{\mathsf T} = B^{\mathsf T} A^{\mathsf T}$.
• A matrix is symmetric when $A^{\mathsf T} = A$ and skew-symmetric when $A^{\mathsf T} = -A$; every square matrix is the sum of a symmetric part $\tfrac{1}{2}(A + A^{\mathsf T})$ and a skew-symmetric part $\tfrac{1}{2}(A - A^{\mathsf T})$.