Question

: Let A be a square matrix, then prove that $A - {A^T}$is a skew symmetric matrix.

Answer 1

Hint: In this question, here we came across the term skew symmetric. So, a symmetric and skew symmetric matrix both are square matrices. But the difference between them is, the symmetric matrix is equal to its transpose whereas skew symmetric matrix is a matrix whose transpose is equal to its negative. If A is a symmetric matrix, then $A = {A^T}$and if A is a skew symmetric matrix then ${A^T} = - A$.

Complete step-by-step answer:
 Here we will discuss some properties of skew-symmetric matrix:
When we add two skew symmetric matrices then the resultant matrix is also skew-symmetric.
The determinant of skew symmetric matrix is non-negative.
When the identity matrix is added to the skew symmetric matrix then the resultant matrix is invertible.
The diagonal of skew symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero.
Scalar product of skew symmetric matrix is also a skew symmetric matrix
Given, A be a square matrix.
Let C = $A - {A^T}$
Now, ${C^T} = {A^T} - {({A^T})^T}$(Transposing both the side of LHS and RHS)
$ \Rightarrow {C^T} = {A^T} - {({A^T})^T}$
$ \Rightarrow {C^T} = - (A - {A^T})$ (We know that ${({A^T})^T} = A$)
$\therefore {C^T} = - C$
As we can see that ${C^T} = - C$, Hence it is proved that C that is $A - {A^T}$is a skew-symmetric.

Note: Only basic difference between a symmetric and skew symmetric matrix is that the transpose of the symmetric matrix is equal to the original matrix. The transpose of the skew symmetric matrix is equal to negative of the original matrix.
Here we see some examples of skew symmetric matrices are:
A = \[\left( {\begin{array}{*{20}{c}}
  0&{ - 6} \\
  6&0
\end{array}} \right)\]
B = $\left( {\begin{array}{*{20}{c}}
  0&3&{ - 8} \\
  { - 3}&0&4 \\
  8&{ - 4}&0
\end{array}} \right)$