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My example: Find some non zero endomoprhism $f$ such that $Kerf = Imf$.

  • Maybe this question has been already answered somewhere on this page, but I have found it, then sorry.

I just struggle to imagine this transformation. The vectors from the kernel always give me the zero vectors after a linear transformation, then how can I satisfy this equation $Kerf = Imf$, if $Imf$ is nonzero?

I tried to use the matrix mapping: $f_A: V \to V$, $\;$ $f_A(x) = Ax$,

$Ker(f_A) = \{x \in \Bbb F^n, Ax = 0 \}$

$Im(f_A) = \{y \in \Bbb F^m, Ax = y \}$

No idea what should I do next, please help me.

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    As a side observation, $\ker 0\ne\operatorname{im}0$ to begin with (unless $\dim V=0$). – 2017-02-27
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    Ok. And If I look at the image of my transformation such that $Ax = y$, is $Ay = 0$ true? If I map the nonzero vector $x$ into $V$, I get vector $y$ that lies in that Kernel, then $Imf = Kerf$? Or am I totally wrong? – 2017-02-27

2 Answers 2

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For example: $f: \mathbb R^2 \to \mathbb R^2$, $(a, b) \mapsto (b, 0)$. $\ker f = \mathrm{Im}\, f = \{(a, 0) | a \in \mathbb R\}$.

Or did I misunderstand the question?

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Since we have $n=\dim\mathop{Ker}f+\dim\mathop{Im}f=2\dim\mathop{Ker} f$, we need $n$ to be even. Let's try $n=2$. We want the kernel to be 1-dimensional. Let's try to get $\mathop{Ker}f = \mathop{Im} f=\mathbb Fe_1$. What would $Ae_1$ have to be? What would $Ae_2$ have to be? Does that work?