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Let $f: \mathbb{C^3}\to \mathbb{C^3}$ has matrix $A=\begin{bmatrix} -1&-2&-3\\0&2&3\\0&-3&-4\end{bmatrix} $ determine if there exist basis where matrix of $f$ has form: $\begin{bmatrix} -1&1&1\\0&-1&0\\0&0&-1\end{bmatrix}$

I found Jordan form of $f$ and it's $\begin{bmatrix} -1&1&0\\0&-1&1\\0&0&-1\end{bmatrix}$ but don't know how to find solution to my problem

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Hint: In order to show that the second matrix does not have the same Jordan form as the first, it suffices to note that $$ f+I = \pmatrix{0&1&1\\0&0&0\\0&0&0} $$ has a kernel of dimension $2$.

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I did not check, but the characteristic polynomial is $(x+1)^{3}$ by looking at the Jordan form. Now, it means that you have an eigenvector $v_{1}$ with eigenvalue $-1$. And then we find $v_{1}$. Now, we look for $v_{2}$ such that $Av_{2}=v_{1}-v_{2}$ so $(A+I)v_{2}=v_{1}$, and then for $v_{3}$ such that $Av_{3}=v_{2}-v_{3}$. Could you guess what the change of basis mtrix would be?

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    $\begingroup$ isn't sufficient to check if both matrices has the same Jordan form ? $\endgroup$ – Jess Jun 12 '15 at 15:45
  • $\begingroup$ Ohhh yes! I accidently read find a basis! $\endgroup$ – mich95 Jun 12 '15 at 15:48
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If $(e_1,e_2,e_3)$ is a Jordan basis, we have $f(e_1)=-e_1$, $\,f(e_2)=-e_2+e_1$, $\,f(e_3)=-e_3+e_2$, hence $f(e_3+e_2)=-e_3+e_1$, and $(e_1, e_2, e_3+e_2)$ is clearly another basis. In this basis the matrix of $f$ has the required form.

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