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Got the following problem:

Let $F:R^3 \rightarrow R^3$ a linear transformation so that $F(0,0,1) = (2,-1,\alpha)$

$(1,0,-1)$ and $(0,1,0)$ are eigenvectors of F with associated eigenvalues $3$ and $-9$ respectively. Find $\alpha$ knowing that $-2$ is eigenvalue.

What I guess is that I have to find the eigenvector's basis first, then find $(0,0,1)$ and $(2,-1,\alpha)$ coordinates in that basis, but I can't figure out how to find the third eigenvector that I need to have that basis.

Any hint will be appreciated. Thanks in advance for your time!

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up vote 3 down vote accepted

Here is a plan how you can solve this: find the matrix of $F$ in the basis $\left((1,0,-1),(0,1,0),(0,0,1)\right)$. The matrix will be triangular, and on the diagonal it will have numbers $3,\,-9$, something. This "something" will depend on $\alpha$.

On the other hand, you know that numbers on the diagonal are exactly the eigenvalues, so that "something" must be equal to the last eigenvalue, $-2$. So you will be able to find $\alpha$ from the equation "something"$=-2$.

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Thanks @danshved for your answer. One last question, this plan is possible to do because if, suppose, B and B' are bases of a vector space V, and F: V -> V is a linear transformation, then F matrices have the same eigenvalues, right? I just wanna have a better understanding. Thanks! – Lucas Nov 19 '12 at 17:16
More or less. But I would put it in different words: if $B$ is a basis in $V$ and $F\colon V \to V$ is a linear transformation, then eigenvalues of $F$ are the same as the eigenvalues of its matrix in basis $B$. – Dan Shved Nov 19 '12 at 17:49

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