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Let $V$ be a finite-dimensional vector space and let $T: V\rightarrow V$ be such that $\dim \text{Im} T=1$.

Show that there exists some $w\in V$ such that $V=\ker T\oplus \text{Sp}\left \{ w \right \}$

According to inuition, I'm pretty sure $0\neq w\in \text{Im} T$, so $\text{Sp}\left \{ w \right \}=\text{Im} T$. We're supposed to take a basis of $\ker T$- $\left \{ v_1,\cdots ,v_{n-1} \right \}$ (where $\dim V=n$). However I don't know where to make progress....

Thanks in advance.

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  • $\begingroup$ Since the dimension of the image is one, there exists some $w$ such that Im$T$ =$\text{span}(w)$. Then all you need to do is show that any vector can be expressed as the sum of two vectors, one from the kernel and one from the image. Indeed $Tw \neq 0$ as you suggest. $\endgroup$ – John Martin Sep 2 '16 at 14:09
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    $\begingroup$ How do you know $T(w)\neq 0$? $\endgroup$ – Theorem Sep 2 '16 at 14:14
  • $\begingroup$ Second tentative.... Let $u\not =0$ in $Im(T)$, and $w$ such that $T(w)=u$ (hence $w\not =0$). For $x\in V$, we have $T(x)=bu$. Write $x=x-bw+bw$ $\endgroup$ – Kelenner Sep 2 '16 at 14:33
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Let $w$ be any vector such that $T(w)\neq 0$. If $v\in \ker T\cap span\{w\}$ then $v=cw$ and $T(v)=0$ so $cT(w)=0$ thus $c=0$.

On the other hand if $v\in V$ is any vector then there is a $c$ such that $T(v)=cT(w)$ because the image is one dimensional. Now it is easy to see $v-cw\in \ker T$.

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