If $\omega$ is a complex cube root of unity, show that
$$ \left( \begin{bmatrix}
1 & \omega & \omega^2 \\
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega \\
\end{bmatrix} + \begin{bmatrix}
\omega & \omega^2 & 1 \\
\omega^2 & 1 & \omega \\
\omega & \omega^2 & 1 \\
\end{bmatrix} \right)
\begin{bmatrix}
1 \\
\omega \\
\omega^2 \\
\end{bmatrix}=\begin{bmatrix}
0 \\
0 \\
0 \\
\end{bmatrix}$$
I tried to solve this and I reduced the L.H.S. to
$$\begin{bmatrix}
2 +2\omega+2\omega^2 \\
2 +2\omega+2\omega^2 \\
2 +2\omega+2\omega^2 \\
\end{bmatrix}
$$ (since $\omega^3=1$)
but couldn't equate it to R.H.S.
Please provide your assistance.
Thank you
Best Answer
$$\omega^3=1 \Rightarrow ω^3−1=0 \Rightarrow (ω−1)(ω^2+ω+1)=0 \Rightarrow ω−1=0 \lor ω^2+ω+1= 0$$
Since, $\omega \neq 1 $ as it is complex , $ω^2+ω+1=0$
Hence, LHS = RHS
- Answer originally posted as comment by Git Gud