Proving that $0$ is an eigenvalue of this transformation

Question

I'm currently self-studying linear algebra and am interested in proving the following:

Let $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$, where $ \operatorname{rank}(T)<n$, prove that $0$ is an eigenvalue of the transformation.

What I'd like to do is use the inveritable matrix theorem, where since $ \operatorname{rank}(T)$ doesn't equal $n$, I can use every part of theorem negated: http://mathworld.wolfram.com/InvertibleMatrixTheorem.html

One such part is: "18. 0 fails to be an eigenvalue of A."

Negated: $0$ is an eigenvalue of A, in this case, T.

Thus proving $0$ is an eigenvalue of T.

What do you think of this proof? If I had the wrong idea, could you go over one way to prove it?

Answer 1

An eigenvector for the value $0$ is just a non-trivial member of the null space or kernel of $T$: a vector $x \neq 0$ such that $Tx=0$.

And the nullity rank theorem says that $\dim(\operatorname{ker}(T)) + \operatorname{rank}(T) = n (= \dim(\mathbb{R}^n))$, so if the rank is $< n$ we know $\dim(\operatorname{ker}(T)) > 0$ and so take any non-zero vector in it.

Answer 2

We have the Kernel of T is non empty, since the Rank of T is less than the dimension $n$.

That is $Tv=0$ for some $v\ne 0$

$Tv=0=0v$ implies that $0$ is an eigenvalue with eigenvector $v$