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If $T: R^3 \rightarrow R^3$ is diagonalizable and has distinct eigenvalues, find $0 \neq v \in R^3$ such that
{$v,T(v), T^2(v)$} is a basis for $R^3$.

I think that statement: {$v,T(v), T^2(v)$} is a basis for $R^3$, is obvious by theorem, but don't know what the question asks to find. It means that I really have to find exact value of $v$, $v=(1,0,0)$ something like that?

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I don't think this is true. Take $T$ to be the identity matrix. You likely want it to have distinct entries on the diagonal. – Calvin Lin May 14 '13 at 13:49
@CalvinLin Sorry, I omitted one information. $T$ has distinct eigenvalues so it is not an identity matrix. – noname May 14 '13 at 13:52

Since $T$ is diagonalizable, let $T=diag (t_1, t_2, t_3)$. Suppose $v=(v_1, v_2, v_3)\in R^3$ is a desired vector, then $$ \det \pmatrix{ v_1& v_2& v_3\\ t_1v_1& t_2v_2& t_3v_3\\ t_1^2v_1& t_2^2v_2& t_3^2v_3 }=v_1v_2v_3\text{Vandermonde matrix} $$ so $v_1v_2v_3\ne 0$ and $t_1, t_2, t_3$ are distinct.

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I know the concept of Vandermonde matrix, but not familiar to it. Can you explain it more precisely? I mean, why you put $(v,T(v),T^2(v))$ as a column vector? Also, I know the determinant of Vandermonde matrix is the product of $(c_j-c_i)$ where $1\le i<j \le n$ but where $v_1 v_2 v_3$ come from? – noname May 14 '13 at 14:17
v can be belong to only one eigen vectore space but you consider it's coordinate seperated why? – Somaye May 14 '13 at 16:42
in other word if v_1,v_2,v_3 \in R^3 then v =(v_1,v_2,v_3) will not be in R^3 – Somaye May 14 '13 at 16:57
@somaye $v=(v_1, v_2, v_3)\in R^3$ means $v_1, v_2, v_3 \in R^1$. – Ma Ming May 15 '13 at 8:33

Since you have distinct eigenvalues, you have a complete eigenbasis, which I will denote as $\{ v_1, v_2, v_3 \}$

Claim: $ v = v_1 + v_2 + v_3 $ works.

Complete this. Show that $av + bTv + cT^2 v = 0 \Leftrightarrow a, b, c = 0 $.

Hint: You have a quadratic equation that is satisfied by 3 distinct values.

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