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I want to determine the vectors $v$ and $w$, given the following product: \begin{align*} P_x &= \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, \ P_y = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix},\\ P_x v &= -\frac v2 + \frac{\sqrt{3}w}{2},\\ P_x w &= -\frac v2 - \frac{\sqrt{3}w}{2},\\ P_y v &= v,\\ P_y w &= -w, \end{align*} where $v$ and $w$ are nonzero vectors.

For this problem, I let $v$ and $w$ be arbitrary nonzero vectors and then, worked out the computation by substitution and taking product of matrices. However, I get zero vectors as the results.

Any advices or comments?

user1551
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NasuSama
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  • Have you figured out the eigenspaces of $P_y$? – Jyrki Lahtonen May 21 '13 at 04:58
  • You are correct. There are no nontrivial solutions. $v$ and $w$ must be zero. – user1551 May 21 '13 at 07:37
  • Should the second equation be $P_x w=\dfrac{-\sqrt{3}}{2}v-\dfrac12w$? – Daryl May 21 '13 at 07:49
  • Ah hah! There are zero vectors, so I need to switch either of the equation with a half and $\sqrt{3}$'s around. – NasuSama May 21 '13 at 14:14
  • @Daryl Then, there will be infinite number of solutions. Actually, yes there will be nonzero solutions. – NasuSama May 21 '13 at 15:30
  • @NasuSama To get non-zero solutions there will be an infinite number of solutions. You are essentially solving a homogeneous linear system of size $12\times6$. The only way to get non-zero solutions is for an infinite number of them. – Daryl May 21 '13 at 19:32

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