Transpose of a matrix   - properties and formulas -

- Definition
- Examples

- Product
- Inverse
- Determinant
- Trace
- Linearity
  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 $A:$
definition of transpose matrix

  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.

  The transposed matrix of matrix
transpose matrix example
  . The transposed matrix of matrix
  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:
transposed matrix of product
  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
transposed matrix of product

  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,
  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,

  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,
determinant of transpose of matrix

  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.

  Let $A$ a squared matrix and $ A ^ {T} $ be the transpose of $A$. The trace of $A^{T}$ is equal to the trace of $ A $:
trace of a matrix

  Let $A$ an $n \times n$ matrix, and $A_{ij}$ be the $i$-th row, $j$-th column element of $A$. By the definition of transpose, we have
trace of a matrix
, whrere ($i,j=1,2 \cdots , n$). This and the defintion of the trace give
trace of a matrix

  Let $A$ and $B$ be $m \times n$ matrices and alpha be a scalar. The transpse is a linear transformation:
matrix transpose Linearyty

  Suppose $A$ and $B$ are represented as
We see that