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Let $f$ be endomorphism on $\mathbb{R}^3$ such that $ \ker f =\bigl\{(x,y,z) \in \mathbb{R}^3/x+y=0 \bigl\}$ ,There exists an $k \in \mathbb{R} $ such that $f(0,1,0)=k(0,1,0)$ and $f \circ f =2f$ find matrix of $f$ in the canonical basis.
My attempts:
we have that $ \ker f =\bigl\{(x,y,z) \in \mathbb{R}^3/x+y=0\bigl\}$ then $\ker f= \text{span} \bigl\{[-1,1,0]^\dagger\bigl\}$ and we have $f(0,1,0)=k(0,1,0)$ so $f(e_2)=ke_2$. what we can conclude from $f \circ f =2f$.

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    $\begingroup$ The kernel is two dimensional. You're missing a vector in it! $\endgroup$
    – Pedro
    Commented Sep 18, 2023 at 19:13
  • $\begingroup$ @Pedro yes the vector$ (0,0,1)$ $\endgroup$
    – DARK Orn
    Commented Sep 18, 2023 at 19:59
  • $\begingroup$ The answer to your question is that you can conclude what the value of $k$ is. Very obviously, too. You do know what $f\circ f$ means, don't you? $\endgroup$ Commented Sep 19, 2023 at 19:16
  • $\begingroup$ @PaulSinclair $k=0$ or $k=2$ $\endgroup$
    – DARK Orn
    Commented Sep 19, 2023 at 19:32
  • $\begingroup$ Since you know the kernel of $f$, one of those can be eliminated. Now that you know what $k$ is, note that you know the value of $f$ for $(0,1,0), (0,0,1)$ and $(-1,1,0)$. What can you figure out from that? $\endgroup$ Commented Sep 19, 2023 at 19:40

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