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Let $\omega$ be a complex number such that $\omega^3=1$ , but $\omega\ne1$. If: $$A= \begin{pmatrix}1& \omega & \omega^2 \\ \omega& \omega^2&1\\\omega^2&\omega&1\end{pmatrix}$$ then which of the following are true?

  1. $A$ is invertible.
  2. $\operatorname{rank}(A)=2$.
  3. $0$ is an eigenvalue of $A$.
  4. There exist linearly independent vectors $v,w\in\mathbb{C}^3$ such that $Av=Aw=0$ .

Statement 1 is false as rank is $2$. So 2 and 3 are true, but I have no idea about 4. Please help me.

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Do you know the rank-nullity theorem? – Matthew Pressland Dec 13 '12 at 16:11
here nullity is 1 but if 4 gives nullity is 4 is my approach is correct? – dekchi Dec 13 '12 at 16:19
Yes is correct. – P.. Dec 13 '12 at 16:40

$$A= \begin{pmatrix}1& \omega & \omega^2 \\ \omega& \omega^2&1\\\omega^2&\omega&1\end{pmatrix}\stackrel{R_2-\omega R_1\,,\,R_3-\omega^2 R_1}\longrightarrow \begin{pmatrix}1& \omega & \omega^2 \\ 0&0&0\\0&\omega-1&1-\omega\end{pmatrix}$$

This proves $\,\dim\ker(A)=1\,$ and thus $\,(4)\,$ is false.

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