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.


  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.