Question

For $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],a,b,c,d\in C$, put $A^{\theta }=\left[\begin{array}{cc}\overline{a}& \overline{c}\\ \overline{b}& \overline{d}\end{array}\right]$. We say matrix $A$ is a Hermitian matrix, if $A=A^{\theta }$. Suppose $A$ is Hermitian matrix such that $A^2=0$, then

• A. $A=A^{\prime }$
• B. $A=-A^{\prime }$
• C. $A=0$
• D. $A=I$

Answer 1

Answer: (C)

$A=0$

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],a,b,c,d\in C$.

given $A$ is a Hermitian matrix and $A^2=0$

$\Rightarrow A=A^{\theta }$ where $A^{\theta }=\left[\begin{array}{cc}\overline{a}& \overline{c}\\ \overline{b}& \overline{d}\end{array}\right]$

$A^2=0$

$\Rightarrow A{A}^{\theta }=0$

$\Rightarrow \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}\overline{a}& \overline{c}\\ \overline{b}& \overline{d}\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{a}^2+b^2& a\overline{c}+b\overline{d}\\ c\overline{a}+d\overline{b}& c^2+d^2\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow a^2+b^2=0$ and $c^2+d^2=0$

$\Rightarrow a=0;b=0;c=0$ and $d=0$

$\therefore A$ is a null matrix.

Hence, option C.