# Transpose of a matrix - properties and formulas -

Definition

A matrix in which the row and column elements of a matrix are exchanged is called the

transposed matrix of that matrix.
The transpose of a matrix $A$ is represented as $A^ {T}$.
The $i$-th row, j-th column element of $A^T$ is the $j$-th row, $i$-th column element of $A:$

**Explanation**
The $i$-th row, $j$-th column element of $A^T$ is the $j$-th row, $i$-th column element of $A$.
For example, if $i=1$ and $j=2$,

And if $i=2$ and $j=1$,

By transposition,
each element is exchanged corssing diagonal elements.

Since $(A^{T})_{ii} = A_{ii}$,
each diagonal element is unchanged.

In general,
the size of a traspose matrix is different from its original one.
If $A$ is an $m \times n$ matrix,
then $A^{T}$ is an $n \times m$ matrix.

If $A$ is a square matrix,
the size of $A^{T}$ is the same as that of $A$.

A column matrix has only one column but any number of rows.
A row matrix has only one row but any number of columns.
Transpose transforms a column matrix to a row matrix, and vice versa.

Various matrices are defined by transpose. For example,

- If $A^{T}=A$, $A$ is called a symmetric matrix.
- If $A^{T}A=AA^{T}=I$, $A$ is called an orthogonal matrix.

Examples

The transposed matrix of matrix

is

. The transposed matrix of matrix

is

.

Product

The transposed matrix of the product of two matrices
is equal to the product of the transposed matrices in which the order of the products is reversed:

.

**Proof**
Let $A$ be a $l \times m$ matrix, and
$B$ be a $m \times n$ matrix, written as

Using

the definition of transposed matrix and the definition of product of matrix,
for the $i$-th row, $j$-th column element of $(AB)^T$,
we see that

. Since this holds for any element, we obtain

.

Inverse

If a matrix $A$ is invertible,
the transpose of $A$ is also invertible,
and the inverse matrix of $A^{T}$ is $(A^{-1})^{T}$, that is,

.

**Proof**
Let $A$ be an invertible matrix.
There exists $A^{-1}$ for which the following holds

$$
\tag{4.1}
$$
, where $I$ is the identity matrix.
By

the propery of transpose of product of two matrices,
and the first equation of $(4.1)$, we see that

$$
\tag{4.2}
$$
Simimary, by the second equation of $(4.2)$, we see that

$$
\tag{4.3}
$$
Eq. $(4.2)$ and $(4.3)$ give

.
This expression shows that $ (A ^ {-1}) ^ {T} $ is the inverse matrix of $ A ^ {T} $, that is,

.

Determinant

Let $A$ be a square matrix,
and $|A^{T}|$
be the determinant of

transpose of $A$.
$|A^{T}|$ is equal to the determinant of $A$, that is,

**Proof**
By

definition,
the determinant of transpose of an arbitray $n \times n$ matrix $A$
is written as

,
where $\sigma$ denotes a function,
called permutation,
that reorders the set of integers,
$
\{
1,2,\cdots,n
\}
$
$S_{n}$ denotes
the set of all such permutations,
$\mathrm{sgn}(\sigma) $ is
a sign depending on the permutation,
and
a sum $\sum_{\sigma \in S_{n}}$ involves all permutations.
(For details, see

"definition of determinant".)

Since $ A_ {ij} ^ {T} = A_ {ji} $
from

the definition of transpose,

.
The permutation is bijective.
Therefore,
for an arbitray permutation $ \sigma $,
there exists an inverse map $ \sigma^{-1} $
such that

, where $i=1,2,\cdots,n$.
Using this, we have

$$
\tag{5.1}
$$
.
Since $\sigma$
is a map that maps
$ \{1,2, \cdots, n \}$
to the same set
$ \{1,2, \cdots, n \}$,
there exists $j$ $(j=1,\cdots,n)$
such that $\sigma(j) = 1$.
Therefore $(5.1)$ can be written as

, where
,in the last line,
the order of multiplication is just changed.
Similary,
there exists $k$ $(k=1,\cdots,n, k\neq j)$
such that $\sigma(k) = 2$.
We see that

By repeating the same procedure to the end,
we obtain

$$
\tag{5.2}
$$
Since $\sigma^{-1}$ of an even permutation $\sigma$
is also an even permutation,
and $\sigma^{-1}$ of an odd permutation $\sigma$ is also an odd permutation,
$\mathrm{sgn}(\sigma^{-1}) = \mathrm{sgn}(\sigma)$ holds.
Therefore, we can write

$\sigma$
is a map that maps
$ \{1,2, \cdots, n \}$
to the same set
$ \{1,2, \cdots, n \}$,
and similary,
$\sigma^{-1}$
is a map that maps
$ \{1,2, \cdots, n \}$
to the same set
$ \{1,2, \cdots, n \}$.
The set of
all $ \sigma^{-1}$
is the same as the set of
all $\sigma$,
Therefore, in Eq.$(5.2)$,
$ \sum_{\sigma \in S_{n}}$
can be replaced with
$\sum_{\sigma^{-1} \in S_{n}}$

Since $\sigma^{-1}$ is one of permutation,
we can rewrite $\sigma^{-1}$ as $\xi \hspace{1mm} (\in S_{n})$,

In this way,
the determinant of transpose of a square matrix is equal to the determinant of the square matrix.