# Find an invertible matrix $B$ such that all eigenvectors of $B$ are scalar multiples of a given vector.

Let $u = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ 0 \\ \end{array}} \right)$. Find an invertible matrix $B$ such that all eigenvectors of $B$ are scalar multiples of $u$.

My attempt: Let $\left( {\begin{array}{*{20}{c}} \alpha \\ 0 \\ 0 \\ \end{array}} \right)$ be an eigenvector of B corresponding to eigenvalue $\lambda$. Then $B\left( {\begin{array}{*{20}{c}} \alpha \\ 0 \\ 0 \\ \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{b_{11}}} & {{b_{12}}} & {{b_{13}}} \\ {{b_{21}}} & {{b_{22}}} & {{b_{23}}} \\ {{b_{31}}} & {{b_{32}}} & {{b_{33}}} \\ \end{array}} \right)\left( {\begin{array}{*{20}{c}} \alpha \\ 0 \\ 0 \\ \end{array}} \right) = \lambda \left( {\begin{array}{*{20}{c}} \alpha \\ 0 \\ 0 \\ \end{array}} \right) \Rightarrow \left( {\begin{array}{*{20}{c}} {\alpha {b_{11}}} \\ {\alpha {b_{21}}} \\ {\alpha {b_{31}}} \\ \end{array}} \right) = \lambda \left( {\begin{array}{*{20}{c}} \alpha \\ 0 \\ 0 \\ \end{array}} \right)$. Therefore $b_{11}=\lambda$ and $b_{21}=b_{31}=0$. I don't know how to find the other entries of $B$...

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If $\lambda\ne \mu$ eigenvales, then the corresponding eigenvectors must be indepent.
1. In case, we work over $\Bbb R$, we can set 2 other eigenvalues to be not reals, but complex conjugated to each other, for example $B$ restricted to the $y,z$ plane is a rotation.