What Happens if Ker(T)=Im(T) for a linear transformation from V to W?

Question

I know that this is impossible for dim(V) odd and i think that i proved that the only possibility is for V=0 but i just want to make sure

Answer 1

No, this is possible for any even- or infinite-dimensional space. For instance, consider T:R²→R² given by T(x) = Mx, where M=

[0, 1]
[0, 0]

Im (T) = span ([1, 0]) and Ker (T) = span ([1, 0]), so Ker (T) = Im (T). In general, wherever V⊆W, you may select any basis whatsoever of V, partition that basis into two equinumerous sets A and B, set up a bijection f:B → A (this is possible because A and B are specified to be equinumerous), and then define T to send every vector in A to 0 and every vector x in B to f(x) . Then Ker (T) = span (A), and Im (T) = span ({T(x): x∈B}) = span (A), so Ker (T) = Im (T).

Note that the reason this doesn't work for V being odd-dimensional is that it's impossible to partition the basis into equinumerous sets in that case.

If Im (T) = Ker (T), then T will be a nilpotent transformation. In particular, T²(x) = 0 for all x.

Edit: fixed some typos.

Answer 2

I believe it means you have the zero map. Ker(T) = set of everything that goes to zero. Im(T) = set of all possible outcomes.