If $\omega$ is a complex cube root of unity, show that the following equals null matrix.

Question

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

@EwanDelanoy That's good enough for an answer, in my opinion. — Git Gud, May 19 '13 at 10:50
@chndn Yes. Now what do you know about $\omega ^2+\omega +1$? What is $\omega$? — Git Gud, May 19 '13 at 10:52
@GitGud okay take common 2 outside. Well what is w? I know $\omega^3=1$ — chndn, May 19 '13 at 11:02
@chndn I just realised what you're trying to prove is false. It's not enough $\omega$ is a complex cubic root of $1$. You need $\omega \in \Bbb C\setminus \Bbb R$. Assuming this, what more can you conclude from the fact that $\omega ^3=1$? — Git Gud, May 19 '13 at 11:08
@chndn $\omega ^3=1\implies \omega ^3-1=0\implies(\omega-1)(\omega ^2+\omega +1)=0 \implies \omega -1=0 \lor \omega^2+\omega +1=0$. Can you conclude? — Git Gud, May 19 '13 at 11:20
@GitGud So, if the first one is zero, how will it be equal to $0$? ... +1 — chndn, May 19 '13 at 11:24
@chndn The first one can't be $0$. Why? The answer is somewhere in my comments. — Git Gud, May 19 '13 at 11:26
@chndn Now that you've figured out the answer,I suggest you write it yourself and accept it so this question doesn't show as unanswered. — Git Gud, May 19 '13 at 11:45

score 0 · Accepted Answer · answered May 20 '13 at 14:38

0

$$\omega^3=1 \Rightarrow ω^3−1=0 \Rightarrow (ω−1)(ω^2+ω+1)=0 \Rightarrow ω−1=0 \lor ω^2+ω+1= 0$$

Since, $\omega \neq 1 $ as it is complex , $ω^2+ω+1=0$

Hence, LHS = RHS

- Answer originally posted as comment by Git Gud

answered May 20 '13 at 14:38

chndn

2,863

I have presented the answer for the stackexchange community as per Git Gud's comment. Hope you understood. Thank you – chndn May 22 '13 at 12:36

If $\omega$ is a complex cube root of unity, show that the following equals null matrix.

1 Answers1