Question

If A is a square matrix then $A-\overline{A^T}$ is a skew hermitian matrix.

• A. True
• B. False

Answer 1

The correct option is A

True

Lets take general element from matrix A and verify if $A-\overline{A^T}$ is a skew hermitian matrix.

Assume

aij = p + iq

aji = r + is

The expression $A-\overline{A^T}$ will translate to $aij-\overline{aji}$ at the elementary level in the resultant matrix.

If mij is the ijth element in resultant matrix

$mij=aij-\overline{aji}=(p+iq)-(r-is)$

=(p – r)+ i (q + s) ….(1)

$mji=aij-\overline{aij}=(r+is)-(p-iq)$

= r – p + i (q+s) ….(2)

From (1) & (2) we can see that

$mij=-\overline{mji}$

This implies that the resultant matrix or $A-\overline{A^T}$ is skew hermitian.