Let $T$ be a linear operator on a finite dimensional vector space $V$ such that $\text{rank}(T) = \text{rank}(T^2)$. I want to show that the kernel and image of $T$ are disjoint, i.e., that they have only the zero vector in their intersection.
I am not able to use the information that the rank of $T$ and $T^2$ is same. I know that would imply that the nullity of $T$ and $T^2$ is also same.
I started as follows: Let $v$ be a vector in the intersection of kernel and image or $T$. I wish to show that $v$ is the zero vector. I'm not able to do it.
Any help is appreciated but hints are more appreciated than the answer itself.
Best Answer
A first thing to realise is that the image of $T^2$ is always contained in the image of $T$, since any vector of the form $T(T(v))$ is in particular of the form $T(w)$. Then the given fact that $T$ and $T^2$ have the same rank means that they actually have the same image; call this common image$~W$.
Now the restriction of $T$ to its own image $W$ defines a linear map $\tilde T:W\to W$ (one might call it $T|_W$, but note that both domain and codomain have changed). And the image of $\tilde T$ is by definition the image of $T^2$, which is all of$~W$, so $\tilde T$ is surjective. Then the kernel of $\tilde T$ is $\{0\}$, and by definition this kernel is the intersection of the kernel of$~T$ with its image$~W$.