Let $V$ be an $n$-dimensional vector space and $T: V \to V$ a linear transformation such that the image and kernel of $T$ are identical.

• a) Prove that $n$ is even:

I figure that since $n$ is the dimension of the vector space,

By the rank-nullity theorem: $n = \dim(\mathrm{im}(T)) + \dim(\ker(T))$

But since image and kernel are identical, by calling $x = dim(ker(T))$ we have:

$n = 2x$ so n can only be even.

Am I right with this reasoning?

• b) Give an example of such linear transformation

What example comes to your mind? I don't know if I should think at a certain example or generalize the answer.

Thank you

b) A general example (I mean, for all $n$) would be$$f(x_1,x_2,\dots,x_{2n-1},x_{2n})=(x_2,0,x_4,0,\ldots,x_{2n},0).$$Then$$\operatorname{Im}f=\ker f=\{(x_1,x_2,\ldots,x_{2n})\,|\;x_2=x_4=\cdots=x_{2n}=0\}.$$
• What's even cooler is that Jose's example is in fact the only possibility! By which I mean that given any $f$ where $Im(f) = Ker(f)$, there is a basis $\{v_i\}$ for $V$ where $f(v_1) = 0, f(v_2) = v_1, ...$ and so on pairwise. It's not obvious but it's not too hard to prove either. – JonathanZ Jun 17 '17 at 23:55
Define $T:\mathbb{R}^{2} \to \mathbb{R}^{2}$ as follows: $T((x,y))=(y,0)$. Observe that both the image and kernel is the x-axis.