Question

Let A be a square matrix. Which of the following is/are not skew-symmetric matrix/matrices?
(a)$A - {A^T}$ (b)${A^T} - A$
(c)$A{A^T} - {A^T}A$ (d)$A + {A^T}$, when$A$is skew-symmetric

Answer 1

Hint: For a matrix $A$ to be skew-symmetric ${A^T} = - A$ must be true.

Let $A$ be any square matrix $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$
The condition for a matrix $A$ to be skew-symmetric is
If ${A^T} = - A$, then $A$ is skew-symmetric. …(1)
Let us consider all the given options one by one and find out if they are skew-symmetric or not.
Option (a) is $A - {A^T}$. For this to be skew-symmetric, from (1) the condition is
${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$ …(2)
Let us find ${A^T}$. The transpose of a matrix is written by interchanging the rows and columns into columns and rows.
${A^T} = \left( {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right)$
$
  A - {A^T} = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right) \\
  A - {A^T} = \left( {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right) \\
$
${\left( {A - {A^T}} \right)^T} = \left( {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right)$ …(3)
$ - \left( {A - {A^T}} \right) = - \left( {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right)$ …(4)
From (2), (3), and (4), we find that ${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$
 holds true. So, $A - {A^T}$ is skew-symmetric.
Option (b) is ${A^T} - A$. For this to be skew-symmetric, from (1) the condition is
${\left( {{A^T} - A} \right)^T} = - \left( {{A^T} - A} \right)$ …(5)
$
  {A^T} - A = \left( {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) \\
  {A^T} - A = \left( {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right) \\
$
${\left( {{A^T} - A} \right)^T} = \left( {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right)$ …(6)
$ - \left( {{A^T} - A} \right) = - \left( {\begin{array}{*{20}{c}}
  0&1 \\
  { - 1}&0
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&0
\end{array}} \right)$ …(7)
From (5), (6), and (7), we find that ${\left( {{A^T} - A} \right)^T} = - \left( {{A^T} - A} \right)$
 holds true. So, ${A^T} - A$ is skew-symmetric.
Option (c) is $A{A^T} - {A^T}A$. For this to be skew-symmetric, from (1) the condition is
${\left( {A{A^T} - {A^T}A} \right)^T} = - \left( {A{A^T} - {A^T}A} \right)$ …(8)
$
  A{A^T} = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right) \\
  A{A^T} = \left( {\begin{array}{*{20}{c}}
  {1\left( 1 \right) + 2\left( 2 \right)}&{1\left( 3 \right) + 2\left( 4 \right)} \\
  {3\left( 1 \right) + 4\left( 2 \right)}&{3\left( 3 \right) + 4\left( 4 \right)}
\end{array}} \right) \\
  A{A^T} = \left( {\begin{array}{*{20}{c}}
  5&{11} \\
  {11}&{25}
\end{array}} \right) \\
$
$
  {A^T}A = \left( {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right)\left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right) \\
  {A^T}A = \left( {\begin{array}{*{20}{c}}
  {1\left( 1 \right) + 3\left( 3 \right)}&{1\left( 2 \right) + 3\left( 4 \right)} \\
  {2\left( 1 \right) + 4\left( 3 \right)}&{2\left( 2 \right) + 4\left( 4 \right)}
\end{array}} \right) \\
  {A^T}A = \left( {\begin{array}{*{20}{c}}
  {10}&{14} \\
  {14}&{20}
\end{array}} \right) \\
$
$
  A{A^T} - {A^T}A = \left( {\begin{array}{*{20}{c}}
  5&{11} \\
  {11}&{25}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
  {10}&{14} \\
  {14}&{20}
\end{array}} \right) \\
  A{A^T} - {A^T}A = \left( {\begin{array}{*{20}{c}}
  { - 5}&{ - 3} \\
  { - 3}&5
\end{array}} \right) \\
$
${\left( {A{A^T} - {A^T}A} \right)^T} = \left( {\begin{array}{*{20}{c}}
  { - 5}&{ - 3} \\
  { - 3}&5
\end{array}} \right)$ …(9)
$ - \left( {A{A^T} - {A^T}A} \right) = - \left( {\begin{array}{*{20}{c}}
  { - 5}&{ - 3} \\
  { - 3}&5
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  5&3 \\
  3&{ - 5}
\end{array}} \right)$ …(10)
From (8), (9), and (10), we find that ${\left( {A{A^T} - {A^T}A} \right)^T} \ne - \left( {A{A^T} -
{A^T}A} \right)$. So, $A{A^T} - {A^T}A$ is not skew-symmetric.
Option (d) is $A + {A^T}$, when $A$ is skew-symmetric. It is given that $A$ is skew-symmetric. So, from (1),
${A^T} = - A$ …(11)
For $A + {A^T}$ to be skew-symmetric, we need to prove that ${\left( {A + {A^T}} \right)^T} =
  - \left( {A + {A^T}} \right)$
Using (11) here, we get
$
  LHS = {(A + {A^T})^T} = {(A + {( - A)^T})^T} = \left( {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right) \\
  RHS = - (A + {A^T}) = - (A + ( - A)) = \left( {\begin{array}{*{20}{c}}
  0&0 \\
  0&0
\end{array}} \right) \\
$
LHS=RHS
So, ${\left( {A + {A^T}} \right)^T} = - \left( {A + {A^T}} \right)$ holds true. Hence, $A + {A^T}$ is skew-symmetric, when $A$ is skew-symmetric.
Hence, the only option (c) is not skew-symmetric.

Note: This problem can be alternatively solved without using the sample matrix and just by
 using the formula ${\left( {A + B} \right)^T} = {A^T} + {B^T}$ for each of the statements given
 to prove the condition for a skew-symmetric matrix. For example, for this $A - {A^T}$ option,
 the condition is ${\left( {A - {A^T}} \right)^T} = - \left( {A - {A^T}} \right)$. $LHS = {\left( {A -
 {A^T}} \right)^T} = {A^T} - {\left( {{A^T}} \right)^T} = {A^T} - A = - \left( {A - {A^T}} \right) =
RHS$. Similarly, this can be done for all options to prove it.
.