Prove or give a counterexample: $V=\text{null}T+\text{range}T$

Question

If $V$ is a finite-dimensional vector space and $T:V\rightarrow V$ is linear then

$$V= \text{null}(T) + \text{range} (T).$$

I know a counterexample if the sum is replaced with a direct sum. However I can't see why the statement above would be false. Both $\text{null}(T)$ and $\text{range} (T)$ are subspaces of $V$ and $\text{null}(T) \cup\text{range} (T)=V$, so why wouldn't this statement be true?

I just want to know if there is something I am missing here...

Answer 1

Let $V=\mathbb{R}^2$ and $$T=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$ then the range and nullspace of $T$ are both equal to $$\Big\{\begin{bmatrix}t\\0\end{bmatrix}:t\in\mathbb{R}\Big\}$$ so their sum is a proper subspace of $V$. This example also shows that contrary to your claim, it's not true in general that $V=\mathrm{range}(T)\cup \mathrm{null}(T)$.

Answer 2

What is true is that the sum of the dimensions of $\text{range}(T)$ and $\text{null}(T)$ is the dimension of $V$. But $\text{range}(T)$ and $\text{null}(T)$ can be any linear subspaces with dimensions that add up to the dimension of $V$.