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Let $\ T: \mathbf R^n \rightarrow \mathbf R^n $ be a linear transformation such that there is a vector $\ u \in \mathbf R^n $, $\ T^2(u) \not = 0 $ and $\ \dim\ker T = n-1 $

Prove that $\operatorname{Im} T = \operatorname{span}\{T(u)\} $ and that $ T $ can be diagonalised.

My solution so far is that $T^2(u) \not =0 \rightarrow T(T(u)) \neq 0 \rightarrow T(u) \notin \ker T \rightarrow T(u) \in \operatorname{Im} T $

and because

$$\dim \ker T = n-1 \Rightarrow \dim\operatorname{Im} T = 1 = \dim(\operatorname{span}\{T(u)\} $$

and so $\dim(\operatorname{span}\{T(u)\}) \in \operatorname{Im} T $ and therefore $\operatorname{Im} T = \operatorname{span}\{T(u)\} $

But how do I prove that $T$ is diagonalizable?

My first guess is that because $\dim \ker T = n-1 $ then $ 0 $ is an eigenvalue with $n-1$ eigenvectors but what can I conclude about the last vector?

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    $\begingroup$ $T^2u\in\operatorname{image}T=\operatorname{span}\{Tu\}$ and is nonzero, so $Tu$ is an eigenvector with nonzero eigenvalue. $\endgroup$ – user10354138 Nov 5 '18 at 10:21
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If we denote by $\;V_\lambda\;$ the eigenspace corresponding to eigenvector $\;\lambda\;$ , then we simply have that $\;\ker T= T_0\;$ so $\;\dim T_0=n-1\;$, and taking $\;n-1\;$ lin. independent vectors here together with any non-zero vector mapping onto Im$\;T\;$ automatically gives us a basis of $\;\Bbb R^n\;$ (why?), and this basis' elements are all of them eigenvectors of $\;T\iff T\;$ is diagonalizable. Fill in details now.

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Take any basis of ker($T$) and include $Tu$ . This will give you a basis of $\mathbf{R}^n$. Check that with respect to this basis matrix of $T$ is diagonal.

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Since $\dim \ker T=n-1$ consider a basis $e_1,...,e_n$, where $e_1,...,e_{n-1}\in \ker T$, write the matrix of $T$ in this basis, you have $T(e_n)=ae_n+a_1e_1+..+a_{n-1}e_{n-1}$ and the polynomial characteristic of $T$ is $(x-a)x^{n-1}$, $a\neq 0$ otherwise $Im T\subset \ker T$ and $T^2=0$, let $u$ be an eigenvector associated to $a$, $T$ is diagonalizable in $(e_1,..,_{n-1},u)$.

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