Dimensions with Rank-Nullity question $f:V \rightarrow V$, $\dim(V) = n$
a) If $n$ is odd, prove that there does not exist a linear transformation with $\operatorname{im}(f) = \ker(f)$
b) Is it false to affirm the previous statement if $n$ is even?
My answer:
$$
i = \dim(\operatorname{im}(f))\\
k = \dim(\ker(f))
\begin{cases}
n=2i                           & \text {if $i=k$} \\
n=i+k   & \text{if $n\neq k$} \\
\end{cases} 
$$
b) Yes.
Is that a valid proof?
 A: These are correct, but for part $(a)$ I would recommend (and require in my classes) writing your solution using full sentences, and mention somewhere "by the rank-nullity theorem, we have $\ldots$"
For part $(b)$, you should produce a counterexample to show the claim is invalid.  This is not too dificult, just look for examples in the case that $\dim V=2 $.
A: I wouldn't accept this as a valid answer. It is hard to understand what you are trying to say, and your introduction of the variables $i,k$ is more confusing than useful. Moreover, you need to provide a counterexample for (b).
Example of answers I'd be looking for:
By the rank-nullity theorem, any linear transformation $f:V \to V$ satisfies $\dim V = \dim \ker f + \dim \operatorname{im} f$. Therefore, if $\ker f = \operatorname{im} f$, then $\dim V = 2 \dim \ker f$ must be even.
Conversely, if $\dim V = 2n$ is even, then such a linear transformation exists. Indeed, let $(e_i)_{i=1}^{2n}$ be a basis of $V$, and define $f$ by on its action on the basis as follows: $f(e_i) = 0$ for $i=1,\ldots,n$, and $f(e_i) = e_{i-n}$ for $i = n+1,\ldots,2n$. Then $$\ker f = \operatorname{im} f =  \operatorname{span}(e_i)_{i=1}^n.$$
