Transpose of a Matrix

3.5 Transpose of a Matrix

A natural operation on a matrix is to flip it — turning rows into columns and columns into rows. This is called transposition, and the result is the transpose of the original. It leads to two important families of matrices — symmetric and skew-symmetric — that appear throughout mathematics and physics.

What is the Transpose?

Take a matrix $A$ of size $m \times n$ with entry $a_{ij}$ in row $i$, column $j$:

Swapping the roles of rows and columns moves the entry at $(i, j)$ to $(j, i)$. The result is the transpose of $A$, denoted $A'$ or $A^T$ (NCERT uses $A'$), of size $n \times m$.

If $A = [a_{ij}]_{m \times n}$, then $A' = [a_{ji}]_{n \times m}$.

Example.

Let

Writing each row of $A$ as a column gives

The entry $a_{12} = 3$ in $A$ becomes the $(2,1)$ entry of $A'$; $a_{31} = -3$ becomes the $(1,3)$ entry, and so on.

Properties of the Transpose

Each property below follows directly from the definition; full proofs are given.

Property (I): Transpose of a Transpose

$(A')' = A$ — transposing twice returns the original matrix.

Proof.

Let $A = [a_{ij}]_{m \times n}$, so $A' = [a_{ji}]_{n \times m}$. The $(i, j)$ entry of $(A')'$ is the $(j, i)$ entry of $A'$, which is $a_{ij}$. Hence $(A')' = [a_{ij}]_{m \times n} = A$. ∎

Property (II): Transpose of a Sum

$(A + B)' = A' + B'$, provided $A$ and $B$ are of the same order.

Proof.

Let $A = [a_{ij}]_{m \times n}$ and $B = [b_{ij}]_{m \times n}$, so $A + B = [a_{ij} + b_{ij}]$. The $(i, j)$ entry of $(A+B)'$ is the $(j, i)$ entry of $A+B$, namely $a_{ji} + b_{ji}$ — which is the $(i,j)$ entry of $A' + B'$. Hence $(A+B)' = A' + B'$. ∎

Property (III): Transpose of a Scalar Multiple

$(kA)' = kA'$, where $k$ is any scalar.

Proof.

Let $A = [a_{ij}]_{m \times n}$, so $kA = [k a_{ij}]$. The $(i, j)$ entry of $(kA)'$ is the $(j, i)$ entry of $kA$, i.e. $k a_{ji}$ — exactly $k$ times the $(i,j)$ entry of $A'$. Hence $(kA)' = k A'$. ∎

Property (IV): Transpose of a Product

$(AB)' = B' A'$, provided $A$ and $B$ are conformable for multiplication.

This is the most important property — note the reversal of order. It is not $(AB)' = A'B'$.

Proof.

Let $A = [a_{ij}]_{m \times n}$ and $B = [b_{jk}]_{n \times p}$, so $AB$ is $m \times p$ with

Then $(AB)'$ is $p \times m$, and its $(k, i)$ entry is the $(i, k)$ entry of $AB$:

Now $B'$ is $p \times n$ with $(B')_{kj} = b_{jk}$, and $A'$ is $n \times m$ with $(A')_{ji} = a_{ij}$, so …