Answer to Question #102657 in Linear Algebra for BIVEK SAH

Question

Show that matrix A given below is skew-hermitian:


A=[0 -5-i -2+3i]

[ 5-i -3i -6]

[2+3i 6 6i ]

Answer 1

The matrix $A$ is skew-hermitian if it satisfies the relation

$A$ skew-hermitian ⟺ $A^H = -A$

where $A^{H}$denotes the conjugate transpose of the matrix $A$. In component form, this means that

$A$ skew-hermitian ⟺ $a_{ij} = -\overline{a}_{ji}$

for all indices $i$  and $j$ , where $a_{ij}$ is the element in the $j$th row and $i$th column of $A$ , and the overline denotes complex conjugation.

Therefore, to show that given matrix $A$ is skew-hermitian, you should show that $A^H = -A$.

$A = \begin{bmatrix} 0 & -5-i & -2+3i\\ 5-i & -3i & -6 \\ 2+3i & 6 & 6i \end{bmatrix}$

Determine $-A$.

$-A = \begin{bmatrix} 0 & 5+i & 2-3i\\ -5+i & 3i & 6 \\ -2-3i & -6 & -6i \end{bmatrix}$

This equals complex conjugation:

$= \begin{bmatrix} \overline{0} &\overline{5-i} & \overline{2+3i}\\ \overline{-5-i} & \overline{-3i} & \overline{6} \\ \overline{-2+3i} & \overline{-6} & \overline{6i} \end{bmatrix}$

It equals the transposed matrix:

$= \begin{bmatrix} \overline{0} &\overline{-5-i} & \overline{-2+3i}\\ \overline{5-i} & \overline{-3i} & \overline{-6} \\ \overline{2+3i} & \overline{6} & \overline{6i} \end{bmatrix}^{T} = A^H$

Thus, the matrix $A$ is skew-hermitian.

Use information from Wikipedia: https://en.wikipedia.org/wiki/Skew-Hermitian_matrix