# Find matrix of $f$ in the canonical basis such that $\ker f =\bigl\{(x,y,z) \in \mathbb{R}^3/x+y=0 \bigl\}$ and $f \circ f =2f$

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 $$f \circ f =2f$$.

• The kernel is two dimensional. You're missing a vector in it!
– Pedro
Commented Sep 18, 2023 at 19:13
• @Pedro yes the vector$(0,0,1)$ Commented Sep 18, 2023 at 19:59
• 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? Commented Sep 19, 2023 at 19:16
• @PaulSinclair $k=0$ or $k=2$ Commented Sep 19, 2023 at 19:32
• 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? Commented Sep 19, 2023 at 19:40