# For which numbers $n \in \mathbb N$ there is a linear map $f_{n}: \mathbb R^{n} \rightarrow \mathbb R^{n}$ such that $ker(f_{n})=im(f_{n})$

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.

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.
• '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 $n$. – Kavi Rama Murthy Jan 13 at 0:10