Transpose of a Matrix

The transpose of a matrix is obtained by interchanging its rows and columns. In this page we present the definition, key properties, worked examples, and practice questions with detailed solutions.

Definition of the Transpose of a Matrix

If \( M \) is an \( m \times n \) matrix, then the transpose of \( M \), denoted \( M^T \), is the \( n \times m \) matrix obtained by interchanging the rows and columns of \( M \).

The rows of \( M^T \) are the columns of \( M \), and the columns of \( M^T \) are the rows of \( M \).

Key idea. Transposing twice returns the original matrix: \((A^T)^T = A\). The transpose operation is an involution — it is its own inverse.

(a) Row Matrix to Column Matrix

\( A = \begin{bmatrix} 1 & -1 & 3 \end{bmatrix} \) has size \( 1 \times 3 \). Its transpose is a \( 3 \times 1 \) column matrix:

\[ A^T = \begin{bmatrix} 1 \\ -1 \\ 3 \end{bmatrix}. \]

(b) Column Matrix to Row Matrix

\( B = \begin{bmatrix} -2 \\ 0 \\ 1 \\ 4 \end{bmatrix} \) has size \( 4 \times 1 \). Its transpose is a \( 1 \times 4 \) row matrix:

\[ B^T = \begin{bmatrix} -2 & 0 & 1 & 4 \end{bmatrix}. \]

(c) \( 2 \times 3 \) Matrix to \( 3 \times 2 \) Matrix

\[ C = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \quad\Longrightarrow\quad C^T = \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix}. \]

(d) \( 3 \times 3 \) Square Matrix

\[ D = \begin{bmatrix} 2 & -5 & 9 \\ -7 & 0 & 9 \\ 1 & -2 & 11 \end{bmatrix} \quad\Longrightarrow\quad D^T = \begin{bmatrix} 2 & -7 & 1 \\ -5 & 0 & -2 \\ 9 & 9 & 11 \end{bmatrix}. \]

Properties of the Transpose of Matrices

  1. \( (A^T)^T = A \)
  2. \( (AB)^T = B^T A^T \)
  3. \( (A + B)^T = A^T + B^T \)
  4. \( (kA)^T = k A^T \), where \( k \) is a scalar.
  5. \( (A^T)^{-1} = (A^{-1})^T \) (when \( A \) is invertible).
  6. \( \det(A^T) = \det(A) \)
  7. \( A^T = A \) if and only if \( A \) is a symmetric matrix.
  8. \( A^{-1} = A^T \) if and only if \( A \) is an orthogonal (square) matrix.
  9. For any matrix \( A \), the products \( A^T A \) and \( A A^T \) are symmetric.
  10. \( \operatorname{tr}(A^T) = \operatorname{tr}(A) \), where \( \operatorname{tr} \) denotes the trace (sum of diagonal entries).
Why the order reverses in \((AB)^T = B^T A^T\). For \( AB \) to be defined, the columns of \( A \) must match the rows of \( B \). After transposing, the columns of \( B^T \) must match the rows of \( A^T \), so the order reverses.
Complex matrices. For complex matrices, the relevant operation is the conjugate transpose \( A^H = \overline{A^T} \), which is essential for defining Hermitian and unitary matrices.

Examples with Solutions

Example 1 — Find the Transpose of a Matrix

Find the transpose of each matrix:

a) \( A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \end{bmatrix} \)

b) \( B = \begin{bmatrix} -2 & 0 & 1 \end{bmatrix} \)

c) \( C = \begin{bmatrix} 5 & -7 \\ 1 & -4 \\ 0 & -1 \\ 7 & -4 \end{bmatrix} \)

Solution

a) \( A^T = \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ 1 & 0 \end{bmatrix} \)

b) \( B^T = \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} \)

c) \( C^T = \begin{bmatrix} 5 & 1 & 0 & 7 \\ -7 & -4 & -1 & -4 \end{bmatrix} \)

Example 2 — Show That \((AB)^T = B^T A^T\)

Given \(A = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 2 & 0 \end{bmatrix}\) and \(B = \begin{bmatrix} -1 \\ 1 \\ 2 \end{bmatrix}\), show that \((AB)^T = B^T A^T\).

Solution

\[ AB = \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \qquad (AB)^T = \begin{bmatrix} 3 & 1 \end{bmatrix}. \]

\[ A^T = \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ 1 & 0 \end{bmatrix}, \qquad B^T = \begin{bmatrix} -1 & 1 & 2 \end{bmatrix}. \]

\[ B^T A^T = \begin{bmatrix} -1 & 1 & 2 \end{bmatrix} \begin{bmatrix} -1 & 1 \\ 0 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \end{bmatrix}. \]

Hence \((AB)^T = B^T A^T\).

Example 3 — Show That \((A^T)^{-1} = (A^{-1})^T\)

Given \(A = \begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}\), verify that \((A^T)^{-1} = (A^{-1})^T\).

Solution

\[ A^T = \begin{bmatrix} -1 & 1 \\ 0 & 2 \end{bmatrix}. \]

\[ (A^T)^{-1} = \frac{1}{(-1)(2) - (1)(0)} \begin{bmatrix} 2 & -1 \\ 0 & -1 \end{bmatrix} = -\frac{1}{2}\begin{bmatrix} 2 & -1 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & \tfrac{1}{2} \\ 0 & \tfrac{1}{2} \end{bmatrix}. \]

\[ A^{-1} = \frac{1}{(-1)(2) - (0)(1)} \begin{bmatrix} 2 & 0 \\ -1 & -1 \end{bmatrix} = -\frac{1}{2}\begin{bmatrix} 2 & 0 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ \tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}. \]

\[ (A^{-1})^T = \begin{bmatrix} -1 & \tfrac{1}{2} \\ 0 & \tfrac{1}{2} \end{bmatrix}. \]

Hence \((A^T)^{-1} = (A^{-1})^T\).

Example 4 — Show That \(\det(A^T) = \det(A)\)

Given \(A = \begin{bmatrix} -1 & 0 & 1 \\ 1 & 2 & 2 \\ -2 & 0 & 1 \end{bmatrix}\), verify that \(\det(A^T) = \det(A)\).

Solution

Expand along row 1:

\[ \det(A) = (-1)\det\begin{bmatrix} 2 & 2 \\ 0 & 1 \end{bmatrix} - 0 + 1 \cdot \det\begin{bmatrix} 1 & 2 \\ -2 & 0 \end{bmatrix} = (-1)(2) + (0 + 4) = -2 + 4 = 2. \]

\[ A^T = \begin{bmatrix} -1 & 1 & -2 \\ 0 & 2 & 0 \\ 1 & 2 & 1 \end{bmatrix}. \]

Expand along row 2 (easiest, two zeros):

\[ \det(A^T) = 2 \cdot \det\begin{bmatrix} -1 & -2 \\ 1 & 1 \end{bmatrix} = 2 \cdot (-1 + 2) = 2. \]

Hence \(\det(A^T) = \det(A) = 2\).

Example 5 — Use \(A^T \neq A\) to Show a Matrix Is Not Symmetric

Show that the following matrix is not symmetric:

\[ A = \begin{bmatrix} 0 & -1 & 3 \\ -1 & 2 & 5 \\ 3 & -5 & 1 \end{bmatrix}. \]

Solution

\[ A^T = \begin{bmatrix} 0 & -1 & 3 \\ -1 & 2 & -5 \\ 3 & 5 & 1 \end{bmatrix}. \]

For instance, \(A_{2,3} = 5\) but \((A^T)_{2,3} = -5\). Hence \(A^T \neq A\), so \(A\) is not symmetric.

Questions with Solutions

Question 1 — Symmetry of \(A^T A\) and \(A A^T\)

Given \(A = \begin{bmatrix} 1 & 1 & -3 \\ -1 & 0 & 0 \end{bmatrix}\), show that \(A^T A\) and \(A A^T\) are both symmetric.

Solution

\[ A^T = \begin{bmatrix} 1 & -1 \\ 1 & 0 \\ -3 & 0 \end{bmatrix}. \]

\[ A^T A = \begin{bmatrix} 2 & 1 & -3 \\ 1 & 1 & -3 \\ -3 & -3 & 9 \end{bmatrix}, \qquad A A^T = \begin{bmatrix} 11 & -1 \\ -1 & 1 \end{bmatrix}. \]

Both results are symmetric.

Question 2 — Transpose of a Matrix Sum

Given \(A = \begin{bmatrix} -5 & -2 \\ -1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 0 & 3 \\ 2 & -3 \end{bmatrix}\), find \((A+B)^T\).

Solution

\[ A+B = \begin{bmatrix} -5 & 1 \\ 1 & -1 \end{bmatrix}. \]

Since \(A+B\) is symmetric, \((A+B)^T = A+B\).

Question 3 — Orthogonal Matrix

Let \(A = \begin{bmatrix} 0 & \tfrac{5}{3\sqrt{5}} & \tfrac{2}{3} \\[4pt] \tfrac{1}{\sqrt{5}} & -\tfrac{4}{3\sqrt{5}} & \tfrac{2}{3} \\[4pt] \tfrac{2}{\sqrt{5}} & \tfrac{2}{3\sqrt{5}} & -\tfrac{1}{3} \end{bmatrix}\). Show that \(A A^T = A^T A = I_3\) and therefore \(A^{-1} = A^T\).

Solution

The rows of \(A\) form an orthonormal set: each row has norm 1, and any two distinct rows are orthogonal. Hence

\[ A A^T = I_3. \]

The columns of \(A\) also form an orthonormal set, giving

\[ A^T A = I_3. \]

Since \(A A^T = A^T A = I_3\), we conclude \(A^{-1} = A^T\) — the defining property of an orthogonal matrix.

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