Introduction

The transpose of a matrix is one of the simplest and most useful operations in linear algebra.
If you already know the basics of matrices—entries, rows, columns, and simple operations—then you’re ready for this article.

In short:

• The transpose flips a matrix across its main diagonal.
• Rows become columns.
• Columns become rows.

This operation appears everywhere: solving systems, defining inner products, working with transformations, and more.

What Is the Transpose?

The transpose of a matrix $A$ is written as $A^T$.

To compute it:

• Take each row of $A$.
• Turn it into a column in the same order.

Example: $$A = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} \quad\Rightarrow\quad A^T = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$$ Key ideas:

• $A$ is $3 \times 2$
• $A^T$ is $2 \times 3$
• Shape always flips.

Why the Transpose Matters

The transpose is important because it:

• Converts between row and column viewpoints.
• Helps define symmetric matrices ($A = A^T$).
• Appears in formulas for projections and least squares.
• Lets us rewrite expressions in cleaner ways.

Some simple but powerful facts:

• $(A^T)^T = A$
• $(A + B)^T = A^T + B^T$
• $(cA)^T = cA^T$ for any number $c$
• $(AB)^T = B^T A^T$ (note the reversed order!)

These properties make the transpose behave nicely with other matrix operations.

Working Through Examples

Example 1: Transposing a Small Matrix

Let $$A = \begin{bmatrix} 2 & -1 \\ 7 & 4 \end{bmatrix}.$$ Then $$A^T = \begin{bmatrix} 2 & 7 \\ -1 & 4 \end{bmatrix}.$$

Example 2: Rectangular Matrix

Let $$B = \begin{bmatrix} 3 & 0 & 5 \\ -2 & 1 & 4 \end{bmatrix}.$$ Then $$B^T = \begin{bmatrix} 3 & -2 \\ 0 & 1 \\ 5 & 4 \end{bmatrix}.$$

Example 3: Using Properties

If $$C = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, \quad D = \begin{bmatrix} 0 & 5 \\ -1 & 2 \end{bmatrix},$$ then $$(C + D)^T = C^T + D^T.$$ You can verify this by computing both sides.

Calculator

Exercises

1. Compute the transpose of $$A = \begin{bmatrix} 4 & 2 \\ -1 & 3 \end{bmatrix}.$$

   Solution

   $$A^T = \begin{bmatrix} 4 & -1 \\ 2 & 3 \end{bmatrix}.$$
2. Let $$B = \begin{bmatrix} 1 & 0 & -2 \\ 5 & 3 & 4 \end{bmatrix}.$$ Compute $B^T$.

   Solution

   $$B^T = \begin{bmatrix} 1 & 5 \\ 0 & 3 \\ -2 & 4 \end{bmatrix}.$$
3. True or false: If $A$ is $2 \times 3$, then $A^T$ is $3 \times 2$.

   Solution

   True.
   A $2 \times 3$ matrix always becomes $3 \times 2$ when transposed.
4. Compute $(A + C)^T$ for $$A = \begin{bmatrix} 2 & 1 \\ 0 & -3 \end{bmatrix}, \quad C = \begin{bmatrix} -1 & 4 \\ 5 & 2 \end{bmatrix}.$$

   Solution

   
   First compute $$A + C = \begin{bmatrix} 2 + (-1) & 1 + 4 \\ 0 + 5 & -3 + 2 \end{bmatrix} = \begin{bmatrix} 1 & 5 \\ 5 & -1 \end{bmatrix}.$$ Then $$(A + C)^T = \begin{bmatrix} 1 & 5 \\ 5 & -1 \end{bmatrix}.$$ (This one happens to be symmetric.)
5. Compute $(AB)^T$ for $$A = \begin{bmatrix} 1 & 2 \\ 3 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & -1 \\ 2 & 5 \end{bmatrix}.$$ Then verify that $(AB)^T = B^T A^T$.

   Solution

   
   Compute $$AB = \begin{bmatrix} 1\cdot4 + 2\cdot2 & 1\cdot(-1) + 2\cdot5 \\ 3\cdot4 + 0\cdot2 & 3\cdot(-1) + 0\cdot5 \end{bmatrix} = \begin{bmatrix} 8 & 9 \\ 12 & -3 \end{bmatrix}.$$ So $$(AB)^T = \begin{bmatrix} 8 & 12 \\ 9 & -3 \end{bmatrix}.$$ Now compute $$B^T = \begin{bmatrix} 4 & 2 \\ -1 & 5 \end{bmatrix}, \quad A^T = \begin{bmatrix} 1 & 3 \\ 2 & 0 \end{bmatrix}.$$ Then $$B^T A^T = \begin{bmatrix} 4\cdot1 + 2\cdot2 & 4\cdot3 + 2\cdot0 \\ -1\cdot1 + 5\cdot2 & -1\cdot3 + 5\cdot0 \end{bmatrix} = \begin{bmatrix} 8 & 12 \\ 9 & -3 \end{bmatrix}.$$ Verified: $(AB)^T = B^T A^T$.

6. Describe in words what the transpose does to the rows and columns of a matrix.

   Solution

   
   The transpose turns rows into columns and columns into rows.
   It flips the matrix across its main diagonal.
7. Compute the transpose of $$D = \begin{bmatrix} 7 \\ -2 \\ 5 \end{bmatrix}.$$

   Solution

   $$D^T = \begin{bmatrix} 7 & -2 & 5 \end{bmatrix}.$$