Transpose of a matrix

Take any matrix, fold it along its main diagonal like a sheet of paper, and read off where every entry lands — that folded copy is the transpose. Rows become columns and columns become rows, nothing more mysterious than that. It sounds like a bookkeeping trick, but it is the hinge on which the whole next section of the chapter turns: symmetric and skew-symmetric matrices are defined entirely in terms of how a matrix compares with its own transpose. Get comfortable with the row-to-column swap here, notice which entries never move, and the definitions that follow will feel almost obvious rather than freshly memorised.

The core insight

The transpose $A^T$ of an $m \times n$ matrix $A = [a_{ij}]$ is the $n \times m$ matrix $[a_{ji}]$ obtained by interchanging its rows and columns, leaving every diagonal entry fixed.

The core idea

Definition. If $A = [a_{ij}]_{m \times n}$ , its transpose is written $A^T$ (sometimes $A'$ ) and is defined entry-by-entry by swapping the two indices:

A^T = [a_{ji}]_{n \times m}

where: $a_{ij}$ is the entry in row $i$ , column $j$ of $A$ ; the entry in row $i$ , column $j$ of $A^T$ is $a_{ji}$ , the entry that sat in row $j$ , column $i$ of $A$ . If $A$ has order $m \times n$ , then $A^T$ has order $n \times m$ — for a non-square matrix the dimensions genuinely swap.

Rows become columns. The first row of $A$ becomes the first column of $A^T$ , the second row becomes the second column, and so on — nothing is reversed or rotated, each whole row is simply re-laid as a column.

The diagonal never moves. For a square matrix, picture folding along the main diagonal $a_{11}, a_{22}, a_{33}, \dots$ ; every off-diagonal entry lands on the position it swaps with, while diagonal entries have $i = j$ , so swapping the indices changes nothing. This single fact is exactly why symmetric matrices — matrices for which the fold leaves everything unchanged — are defined using the diagonal.

Special shapes. A row matrix's transpose is a column matrix, and vice versa; a $1 \times n$ matrix becomes $n \times 1$ . This case is easy to miss because it feels too simple to be "real" transposition, yet it follows from the same rule.

Figure

A 2 by 3 grid folding along its diagonal so each row becomes a column in a 3 by 2 grid

Derivation AThe double transpose returns the original matrix

The goal is to prove, using nothing beyond the entry-swap definition itself, that transposing a matrix twice always gives back the original matrix, so that $(A^T)^T = A$ holds for every matrix $A$ of every order, not merely as a pattern noticed in one worked example. This property is used repeatedly later in the chapter to simplify proofs about symmetric matrices, so it is worth seeing why it is forced by the definition rather than just believed.

Step 1. Start from the definition. Let $A = [a_{ij}]_{m \times n}$ . By the rule $A^T = [a_{ji}]_{n \times m}$ , the entry sitting in row $p$ , column $q$ of $A^T$ is $a_{qp}$ — the roles of the two indices are exchanged once.

Step 2. Transpose again. Apply the same rule to the matrix $A^T$ itself, treating $A^T$ as the matrix being transposed. The entry in row $p$ , column $q$ of $(A^T)^T$ equals the entry in row $q$ , column $p$ of $A^T$ .

Step 3. Substitute. From Step 1, the entry in row $q$ , column $p$ of $A^T$ is $a_{pq}$ (swap $p$ and $q$ in the formula $a_{qp}$ ). So the entry in row $p$ , column $q$ of $(A^T)^T$ is $a_{pq}$ — exactly the entry that sits in row $p$ , column $q$ of the original matrix $A$ .

Step 4. Conclude. Since every entry of $(A^T)^T$ matches the corresponding entry of $A$ , and the order also returns to $m \times n$ (swapped twice), the two matrices are identical: $(A^T)^T = A$ . Two index swaps undo each other, exactly as folding a sheet along a crease twice returns every point to where it started.

Worked example

Find the transpose of $A = \begin{bmatrix} 2 & 5 & -1 \\ 0 & 3 & 4 \end{bmatrix}$ , and then verify $(A^T)^T = A$ .

Step 1. Note the order of $A$ : it is $2 \times 3$ , so $A^T$ will have order $3 \times 2$ .

Step 2. The first row of $A$ , which is $2, 5, -1$ , becomes the first column of $A^T$ ; the second row, $0, 3, 4$ , becomes the second column. Assembling gives

A^T = \begin{bmatrix} 2 & 0 \\ 5 & 3 \\ -1 & 4 \end{bmatrix}

Step 3. Transpose $A^T$ itself to check the double-transpose rule: the first row of $A^T$ , which is $2, 0$ , becomes the first column; the second row, $5, 3$ , becomes the second column; the third row, $-1, 4$ , becomes the third column.

Step 4. Assembling, $(A^T)^T = \begin{bmatrix} 2 & 5 & -1 \\ 0 & 3 & 4 \end{bmatrix}$ , which is exactly the original $A$ , confirming $(A^T)^T = A$ .

Answer: $A^T = \begin{bmatrix} 2 & 0 \\ 5 & 3 \\ -1 & 4 \end{bmatrix}$ , and transposing it again returns $A$ unchanged.

Competitive stretch

For $B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ , quickly check whether $B^T = B$ . Swapping rows and columns gives $B^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$ , which is not equal to $B$ since the off-diagonal entries $2$ and $3$ have swapped places. $B$ is not symmetric, because $B^T \ne B$ — a one-line check that previews the exact test used to define symmetric matrices next.

Common traps

Confusing transpose with reversing the order of entries — it is a precise row-column swap, not a reversal or a rotation of the matrix.
Forgetting the order changes for a non-square matrix — an $m \times n$ matrix becomes $n \times m$ , so $A$ and $A^T$ have different shapes unless $A$ is square.
Transposing only the diagonal entries and leaving the rest alone — every entry relocates, not just some of them.
Assuming $A^T = A$ always holds — this is only true for the special case of symmetric matrices, not for a general matrix.
Thinking transpose changes the value of an entry — it only relocates entries; no number is ever altered.
Overlooking that a row matrix transposes to a column matrix (and vice versa), a case students often expect only "genuinely rectangular" matrices to show.

Ask the tutor: Want more transpose drills, including checking symmetric and skew-symmetric matrices, or larger $3\times3$ examples? Ask the AI tutor for extra practice problems.

Is transpose the same as rotating the matrix 90 degrees?

No, and this mix-up is common. Rotating a grid of numbers by 90 degrees reorders entries along both directions in a way that does not match transpose at all — try it on a $2\times2$ matrix and you will get a different result from $A^T$ . Transpose is specifically the row-to-column, column-to-row swap defined by exchanging the two subscripts, $a_{ij} \to a_{ji}$ . Stick to the definition rather than a mental picture of spinning the grid.

Why does the order have to change for a non-square matrix?

Because every row of the original becomes a column of the transpose, and rows and columns can have different lengths when $m \neq n$ . A $2 \times 3$ matrix has rows of length $3$ ; once each row becomes a column, those columns have length $3$ too, giving a $3 \times 2$ result. The number of rows and the number of columns simply trade places, so $m \times n$ becomes $n \times m$ — and only when $m = n$ , a square matrix, does the order stay the same.

Do I need to memorise the algebraic properties of transpose separately?

The properties like $(A^T)^T = A$ , $(A+B)^T = A^T + B^T$ and $(kA)^T = kA^T$ are covered in the companion "Properties of transpose" article and are separately examinable, but they all flow from the same entry-swap definition used here. Once you trust that $A^T$ is defined by $a_{ij} \to a_{ji}$ , most of those properties can be re-derived quickly in an exam rather than recalled from memory, exactly as the double-transpose proof above shows.

Transpose of a matrix

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads