Guys I'm stuck with a problem of linear algebra. I have a linear transformation $L:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ and I need to prove that $\mathbb{R}^3 = \text{ker}(L) \oplus \text{im}(L) $. I don't really seem able of tackling this problem by using the definitions... Any hint or explanation would be greatly appreciated!

Edit: I have forget to write that $\mathbb{R}^3 = \text{ker}(L) \oplus \text{im}(L) $ means that for all $\text{v} \in \mathbb{R}^3$ there exists $\text{x}\in\text{ker}(L)$ and $\text{y}\in\text{im}(L)$ such that $\text{v}=\text{x}+\text{y}$ and $\text{ker}(T) \cap \text{im}(T) = \{0\}$

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    $\begingroup$ This could only be true relative to a particular $L$. So, define $L$ and we've got a decent chance of helping you. $\endgroup$ – user24142 Sep 7 '15 at 22:03
  • $\begingroup$ I have edited the post. $L$ is not defined, only the explanation above is given $\endgroup$ – james42 Sep 7 '15 at 22:18
  • $\begingroup$ Well, the assertion is not true in general, so you must have some more information about $L$. $\endgroup$ – egreg Sep 7 '15 at 22:28
  • $\begingroup$ As the current answer shows, the statement is not true in general. You need to add some more information other than the definition of the direct sum (which is well known). What sort of linear transformation is $L$? There must be some information. $\endgroup$ – M. Vinay Sep 11 '15 at 10:28
  • $\begingroup$ I have no information about $L$, this is a question I've posed to myself while self studying linear algebra... Anyway, the answer given is satisfactory and helped me understand a lot of things! $\endgroup$ – james42 Sep 11 '15 at 10:32

Your equality isn't always true without further assumption: take for example the matrix

$$A=\begin{pmatrix}0&0&1\\ 0&0&0\\ 0&0&0\end{pmatrix}$$ We have $A^2=0$ so $\operatorname{Im}A\subset \ker A$ and $\ker A\ne \Bbb R^3$.

  • $\begingroup$ I don't really get your point. By the computations that I've made, $\text{ker}A=(1,1,0)$ and $\text{im}A=(0,0,z)$ why the above relationship is true? $\endgroup$ – james42 Sep 8 '15 at 0:32
  • $\begingroup$ No the image of $A$ is the span of $(1,0,0)$. $\endgroup$ – user260717 Sep 8 '15 at 4:01

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