# Let $T: V \to V$ be a linear map on an $m$-dimensional vector space $V$. Prove the following:

Let $$T:V \to V$$ be a linear map on an $$m$$-dimensional vector space $$V$$. Prove the following:

(a): Suppose $$V=imT + kerT$$, then $$V=imT \oplus kerT$$.

(b): Suppose $$imT \cap kerT = \{0\}$$, then $$V= imT \oplus kerT$$.

I'm stuck on this problem and don't know where to start. My initial instinct with statement (a) is:

$$T:V \to V\Rightarrow$$ $$imT=V$$ and $$kerT=\{0\}$$

$$\therefore \forall v \in V$$, $$v \in imV$$,

and therefore the sum $$V=imT + kerT$$ can be expressed as the sum of each element of $$V$$ and the zero vector, $$v+0$$, which for all $$v \in V$$ is unique as $$v+0=v$$.

Is this proof correct? I feel like I can't move on to statement (b) until I've proved statement (a), but any tips on (b) would also be appreciated.

Thanks

• Do you know the rank-nullity theorem? Your proof of (a) is not correct, consider, for example $\begin{bmatrix}1&0\\0&0\end{bmatrix}$. – Michael Burr Mar 4 '19 at 0:04

This questions is most easily solved with the rank-nullity theorem. Even if you don't know this theorem, observe that if $$A$$ is the standard matrix for $$T$$, then $$\dim\operatorname{im}(T)$$ is equal to the number of pivots in $$A$$ and $$\dim\operatorname{ker}(T)$$ is equal to the number of free variables in $$A$$. Therefore, $$\dim\operatorname{im}(T)+\dim\operatorname{ker}(T)$$ is equal to the number of columns of $$A$$, i.e., $$m$$. At this point, the two statements become dimension arguments (using a bit of inclusion/exclusion).