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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

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

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