Question

Let two matrices $A=\text{diag}(-1,\omega ,i)$ and $B=\text{diag}(1,{\omega }^2,-i),$ then which of the following statement(s) is(are) correct ?
$(\text{where}{i}^2=-1\text{and}\omega \text{is a cube root of unity.})$

• A. $Tr\left(AB\right)=\omega$
• B. $Tr\left(A^2\right)={\omega }^2$
• C. $Tr\left(B^2\right)=\omega$
• D. $Tr\left(A^2{B}^2\right)=3$

Answer 1

Answer: (B), (C), (D)

$Tr\left(A^2{B}^2\right)=3$

Given : $A=\text{diag}(-1,\omega ,i)$
Using property of diagonal matrix,
$\Rightarrow A^2=\text{diag}\left(\right(-1{)}^2,{\omega }^2,i^2)=\text{diag}(1,{\omega }^2,-1)$
So, $Tr\left(A^2\right)=1+{\omega }^2-1={\omega }^2$
and $B=\text{diag}(1,{\omega }^2,-i)$
$\Rightarrow B^2=\text{diag}(1^2,{\omega }^4,(-i{)}^2)=\text{diag}(1,\omega ,-1)$
So, $Tr\left(B^2\right)=1+\omega -1=\omega$
Now,
$A^2{B}^2=\text{diag}\left(\right(1\cdot 1),({\omega }^2\cdot \omega ),(-1\cdot -1\left)\right)=\text{diag}(1,{\omega }^3,1)=\text{diag}(1,1,1)$
$(\because {\omega }^3=1)$
So, $Tr\left(A^2{B}^2\right)=3$
and $AB=\text{diag}\left(\right(-1\cdot 1),(\omega \cdot {\omega }^2),(i\cdot -i\left)\right)$
$\Rightarrow AB=\text{diag}(-1,{\omega }^3,1)$
So, $Tr\left(AB\right)=-1+1+1=1$