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1.5. The transpose of a matrix

In this subsection, we consider arbitrary rectangular matrices. A concept that turns out to be useful in practice is the transpose of a matrix. Roughly, starting with a matrix $A$ , we take its flip about the diagonal terms $a_{ii}$ of $A$ , so that the rows of $A$ form the columns of $A^{t}$ and vice-versa.

Definition 1.5.1.

Let $A=(a_{ij})\in\operatorname{M}_{{n\times m}}({{\mathbb{R}}})$ , for some integers $n,m\geq 1$ . The transpose of $A$ is the matrix $A^{t}=(a^{\prime}_{ij})\in\operatorname{M}_{{m\times n}}({{\mathbb{R}}})$ with coefficients

a^{\prime}_{ij}=a_{ji}.

Example 1.5.2.

$\begin{pmatrix}1&2\\ 3&4\end{pmatrix}^{t}=\begin{pmatrix}1&3\\ 2&4\end{pmatrix}\quad\hbox{,}\quad\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}^{t}=\begin{pmatrix}1&2&3\end{pmatrix}\quad\hbox{and}\quad\begin% {pmatrix}1&2&3\\ 4&5&6\end{pmatrix}^{t}=\begin{pmatrix}1&4\\ 2&5\\ 3&6\end{pmatrix}.$

Remark 1.5.3.

Some authors write the transpose of a matrix as $A^{T}$ , or even $A^{\prime}$ .

The next result outlines the main properties of the transpose. See Section 3 for the definition of the inverse $A^{-1}$ of a matrix $A\in\operatorname{M}_{{n\times n}}({{\mathbb{R}}})$ .

Theorem 1.5.4.

Let $A\in\operatorname{M}_{{n\times m}}({{\mathbb{R}}})$ . The following properties hold.

(i)

$(A^{t})^{t}=A$ .
(ii)

If $B\in\operatorname{M}_{{m\times p}}({{\mathbb{R}}})$ , then $A B$ and $B^{t}A^{t}$ are defined, and $(AB)^{t}=B^{t}A^{t}$ .

Proof.

For the proof, we write $X_{ij}$ for the $(i,j)$ coefficient of a matrix $X$ .

(i)

By definition of the transpose, we have $(A^{t})_{ij}=A_{ji}$ . By iterating the transpose, we obtain

$\bigl{(}(A^{t})^{t}\bigr{)}_{ij}=(A^{t})_{ji}=A_{ij}\quad\hbox{for all indices% $i,j$,}\quad$

and so, $(A^{t})^{t}=A$ .
(ii)

We have $B^{t}\in\operatorname{M}_{{p\times m}}({{\mathbb{R}}})$ and $A^{t}\in\operatorname{M}_{{m\times n}}({{\mathbb{R}}})$ , so $B^{t}A^{t}$ is defined and both $(AB)^{t}$ and $B^{t}A^{t}$ belong to $\operatorname{M}_{{p\times n}}({{\mathbb{R}}})$ . We need to check that $\big{(}(AB)^{t}\big{)}_{ij}=(B^{t}A^{t})_{ij}$ for all indices $i, j$ . We have by definition of the transpose and matrix multiplication

$\big{(}(AB)^{t}\big{)}_{ij}=(AB)_{ji}=\sum_{1\leq k\leq m}A_{jk}B_{ki}$

which we compare with

$(B^{t}A^{t})_{ij}=\sum_{1\leq k\leq m}(B^{t})_{ik}(A^{t})_{kj}=\sum_{1\leq k% \leq m}B_{ki}A_{jk}.$

Since $A_{jk}B_{ki}=B_{ki}A_{jk}$ for all indices $i, j, k$ , the products are equal, saying that $(AB)^{t}=B^{t}A^{t}$ .

∎

Many matrices that arise naturally, such as the correlation matrix in statistics, have a special property: they are symmetric.

Definition 1.5.5.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ . We say that $A$ is symmetric if $A^{t}=A$ . We say that $A$ is skew-symmetric if $A^{t}=-A$ .

Remark 1.5.6.

(i)

The terms symmetric and skew-symmetric are defined for square matrices only.
(ii)

If $A$ is skew-symmetric then the elements on the diagonal are all zero.
(iii)

Most matrices are neither symmetric, nor skew-symmetric.
(iv)

Compare the notions of (skew-)symmetry of matrices with the concept of parity of functions in analysis.

Example 1.5.7.

1. (a)
  
  The matrix $\begin{pmatrix}1&-1&4\\ -1&2&5\\ 4&5&3\end{pmatrix}$ is symmetric.
  
  The matrix $\begin{pmatrix}0&1&-2\\ -1&0&4\\ 2&-4&0\end{pmatrix}$ is skew-symmetric.
2. (b)
  
  Let $A$ be any square matrix. Then $AA^{t}$ is symmetric. Indeed, by Theorem 1.5.4, we have
  
  $(AA^{t})^{t}=(A^{t})^{t}A^{t}=AA^{t}.$
  
  Thus, $AA^{t}$ is a symmetric matrix.
3. (c)
  
  Let $A$ be a symmetric matrix and $B$ a skew-symmetric matrix. Then $A B A$ is skew-symmetric. Indeed, by Theorem 1.5.4, we have
  
  $(ABA)^{t}=A^{t}B^{t}A^{t}=A(-B)A=-(ABA).$
  
  So, $A B A$ is skew-symmetric.