Demonstrating that (im τ) \cap (ker τ) is the Zero Space

[alexfloo]

Consider a linear operator τ:V→V (where V is finite-dimensional) such that rk τ²=rk τ. Show that (im τ) \cap (ker τ) is the zero space.

Here's where I am:

Its easy to see that I am τ=im τ², since it is a subspace with the same dimension. I also know that if (im τ) \cap (ker τ) contains a nonzero vector, then it has positive dimension.

My intuition is that I can use that positive dimension and the rank plus nullity theorem to show a contradiction, but I just can't seem to figure out how. Rank-plus-nullity gives me null τ = null τ², and I know that dim ((im τ) \cap (ker τ)) ≤ null τ. Any idea where to go next?

 

[alexfloo]

I think I got it:

Let σ denote the restriction of τ to its own image, and consider it as a function I am τ→im τ².

It's easy to see that this function is surjective. Since it is a surjective map between finite-dimensional spaces of equal dimension, it is also injective, so ker σ = {0}.

However, ker σ = ker τ \cap I am τ, so this completes the proof.