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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1$\begingroup$ The kernel is two dimensional. You're missing a vector in it! $\endgroup$– Pedro ♦Sep 18 at 19:13
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$\begingroup$ @Pedro yes the vector$ (0,0,1)$ $\endgroup$– Khalid MardiSep 18 at 19:59
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$\begingroup$ 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? $\endgroup$– Paul SinclairSep 19 at 19:16
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$\begingroup$ @PaulSinclair $k=0$ or $k=2$ $\endgroup$– Khalid MardiSep 19 at 19:32
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$\begingroup$ 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? $\endgroup$– Paul SinclairSep 19 at 19:40
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