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Let $V$ be an $n$-dimensional vector space.

To prove: $\DeclareMathOperator{im}{im}$ There is a linear map $\varphi:V\to V$ with $\ker \varphi= \im \varphi $ if and only if $n$ is even.

Solution:

"$\Rightarrow$" If $\ker \varphi=\im \varphi $ then $n = \dim \ker \varphi+ \dim \im \varphi = 2\cdot \dim \ker \varphi = 2\cdot \dim \im \varphi$

So $n$ is even.

How do I show the other direction "$\Leftarrow$"?

Can I say if $n$ is even, we know from "$\Rightarrow$" that there is a linear map?

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Select a basis $\{e_i\}_{i=1}^{2N}$. Map $\{e_i\}_{i=1}^{N}$ to 0, and $\{e_i\}_{i=N+1}^{2N}$ to $\{e_i\}_{i=1}^{N}$ (by taking $e_i$ to $e_{-N+i} $). This is a map with the desired property.

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  • $\begingroup$ That would be $\varphi = (0_n,\dotsc, 0_n, e_1, \dotsc e_N)$. The first half mapped to zero the second half to a set of $N$ linear independent vectors. $\endgroup$ – mvw Jul 26 '18 at 21:11
  • $\begingroup$ Yes, and the matrix in the answer below represents this transformation. $\endgroup$ – ertl Jul 26 '18 at 21:13
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Can I say if $n$ is even, we know from "$\Rightarrow$" that there is a linear map?$\DeclareMathOperator{im}{im}$

Note: You want to infer from the even dimension, and $V$ being a vector space, that there exists a linear map $\varphi$ from $V$ to $V$ with $\ker \phi = \im \phi$.

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