I have a problem:
Let's find an endomorphism of $\mathbb{R}^3$ such that $\operatorname{Im}(f) \subset \operatorname{Ker}(f)$.
How would you do it?
The endomorphism must be not null.
Tell me more
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Well, you could always take $f$ to be the null function... Not the only solution, but certainly the simplest :) If you want $f$ to be non-null, then you just need to make sure that $f^2=0$. Either try to find a $3 \times 3$ matrix for which this holds (look up nilpotent matrices) or look at this endomorphism and see how it can be adapted : $$f : (x,y,z) \mapsto (y,z,0)$$ |
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Observe that such operator has the property that $f(f(\vec x))=0$, since $f(\vec x)\in\ker f$. Any matrix $M$ in $M_3(\mathbb R)$ such that $M^2=0$ will give you such operator. For example, then, $f(x_1,x_2,x_3)=(x_3,0,0)$, for the canonical basis the matrix looks like this:$$\begin{pmatrix} 0&0&1\\0&0&0\\0&0&0\end{pmatrix}$$ See now that $f^2=0$, but $f\neq 0$. |
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