Question

For the matrix A = \left( {\begin{array}{*{20}{c}} 1&5 \\\ 6&7 \end{array}} \right),verify that
(i) $\left( {A + A'} \right)$is a symmetric matrix.
(ii) $\left( {A - A'} \right)$is a skew symmetric matrix.

Answer 1

Hint: Use the property of symmetric and skew symmetric matrices directly on the given matrix expression.

Given the matrix A = \left( {\begin{array}{*{20}{c}} 1&5 \\\ 6&7 \end{array}} \right)

We know that the transpose of a matrix is obtained by switching the rows with its columns
\Rightarrow A' = \left( {\begin{array}{*{20}{c}} 1&6 \\\ 5&7 \end{array}} \right)

A Symmetric matrix is the one in which the matrix is equal to the transpose of itself.
$\Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }}A = A'$

We need to prove $\left( {A + A'} \right)$is a symmetric matrix.
A + A' = \left( {\begin{array}{*{20}{c}} 1&5 \\\ 6&7 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&6 \\\ 5&7 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 2&{11} \\\ {11}&{14} \end{array}} \right){\text{ (1)}}
Also,{\left( {A + A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}} 2&{11} \\\ {11}&{14} \end{array}} \right){\text{ (2)}}

From equations $(1)$and$(2)$ , we get ${\left( {A + A'} \right)^\prime } = \left( {A + A'} \right)$, which satisfies the above-mentioned condition of symmetric matrices.
Hence $\left( {A + A'} \right)$is a symmetric matrix.

A Skew symmetric matrix is the one in which the negative of the matrix is equal to the transpose of itself.

$\Rightarrow if{\text{ }}A = {\left[ {{a_{ij}}} \right]_{n \times m}}{\text{ }}and{\text{ }}A' = {\left[ {{a_{ij}}} \right]_{m \times n}}{\text{ ,}}then{\text{ }} - A = A'$

We need to prove $\left( {A - A'} \right)$is a skew symmetric matrix.
A - A' = \left( {\begin{array}{*{20}{c}} 1&5 \\\ 6&7 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 1&6 \\\ 5&7 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0&{ - 1} \\\ 1&0 \end{array}} \right){\text{ (3)}}
Also,{\left( {A - A'} \right)^\prime } = \left( {\begin{array}{*{20}{c}} 0&1 \\\ { - 1}&0 \end{array}} \right){\text{ (4)}}

From equations $(3)$and$(4)$ , we get ${\left( {A - A'} \right)^\prime } = - \left( {A - A'} \right)$, which satisfies the above-mentioned condition of skew symmetric matrices.
Hence $\left( {A - A'} \right)$is a skew symmetric matrix verified.

Note: The above-mentioned results are true for all square matrices. Similarly using above results, it can be proved that a square matrix can be expressed as the sum of a symmetric and skew symmetric matrix.