Let $\omega$ be a complex number such that $\omega^3=1$, but $\omega\neq1$. If $$ A=\begin{pmatrix} 1&\omega&\omega^2\\ \omega&\omega^2&1 \\ \omega^2&\omega&1\end{pmatrix}\quad $$ then show that there exist linearly independent vectors $v, w \in \mathbb C^3$ such that $Av=Aw=0.$ I tried to show $0$ is an at least $2$-fold eigenvalue of $A$ but couldn't.
Show that $Av=Aw=0.$
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linear-algebra
eigenvalues-eigenvectors
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1I think you mean to say $v,w \in \mathbb{C}^3.$ – 2012-12-06
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0Sorry ... corrected my question. – 2012-12-06
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1I attempted it and it seems like the eigenspace for eigenvalue 0 is only of dimension 1... – 2012-12-06
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0Did you mean the last row to be $(\omega^2, 1, \omega)$ instead? – 2012-12-06
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0@copper.hat: Nope. That would make it trivial. – 2012-12-06
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2It would make it possible. As the question stands, it is impossible. See my answer below. – 2012-12-06
1 Answers
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It is impossible. It is easy to see that $A (1,1,1)^T = 0$, hence $\dim \ker A \geq 1$. However, the vectors $(1, \omega, \omega^2)^T$ and $(\omega, \omega^2, \omega)^T$ are linearly independent, hence $\dim {\cal R} (A) \geq 2$. It follows from the rank-nullity theorem that $\dim \ker A = 1$.