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My idea:
When $\ker f_{n}=\operatorname{image}(f_{n})$ then $\dim(ker(f_{n}))=\dim(\operatorname{image}(f_{n}))$. Moreover I know that for $g: V \rightarrow W$ I have $\dim(V)=\dim (\ker g) + \dim(\operatorname{image} g)$, so in this case: $n=2\dim(\ker f)$, so this task is true for even $n$.

Unfortunately I am afraid that it is incorrect solution and please rate it.

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The assertion is correct. It has been shown already that $n$ must be even. For the converse part consider $V=\mathbb R^{2m}$. Define $f: V \to V$ by $f(x_1,x_2,\cdots,x_{2m})=(x_{m+1},x_{m+2},\cdots, x_{2m},0,0, \cdots,0)$. Then $f$ is linear and its range coincides with the kernel.

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  • $\begingroup$ Why I have to consider the converse part? $\endgroup$ – MP3129 Jan 13 at 0:05
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    $\begingroup$ 'For which numbers $n$' means you have to precisely identify all $n$ with this property. It is not enough to say that if the result is true then $n$ must be even. You also have prove that such a linear functional exists for every even $n$. $\endgroup$ – Kavi Rama Murthy Jan 13 at 0:10

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