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I was wondering if $V=\ker(T) \oplus \mathrm{im}(T)$ if $T:V \to V$. I know the theorem that if $T:V\to W$ is linear then $\dim(V) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$. This should imply $V=\ker(T) \oplus \mathrm{im}(T)$ because if $\dim(V) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$ it mean that a basis of $\ker(T)$ plus a basis of $\mathrm{im}(T)$ forms a basis of $V$ which is the same like $V=\ker(T) \oplus \mathrm{im}(T)$. Is it correct?

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    $\begingroup$ No, what if $im(T)\cap ker(T)\neq \{0\}?$ E.g. $T:\Bbb R^2\to \Bbb R^2: \begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}y\\0\end{pmatrix}$ $\endgroup$ – walcher Sep 2 '13 at 16:08
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    $\begingroup$ Related (arguably a duplicate of): When does $V = \ker(f) \oplus \operatorname{im}(f)$? $\endgroup$ – Zev Chonoles Sep 2 '13 at 17:28
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Sometimes, it is true that $V=\ker(T)\oplus\mathrm{im}(T)$. But consider the vector space $V=\mathbb{R}^2$ and the linear map $T:V\to V$ defined by $$T(a,b)=(b,0).$$ Both the kernel and the image of $T$ are $\{(t,0):t\in\mathbb{R}\}$. Thus the dimensions still add up: $$\dim(\ker(T))+\dim(\mathrm{im}(T))=1+1=2=\dim(V)$$ but it is impossible for $\ker(T)$ and $\mathrm{im}(T)$ to span $V$, much less for $V$ to be the direct sum of them.

It seems to me that you're (perhaps unconsciously) committing the fallacy of considering all vector spaces that have the same dimension to be the same. In my example above, it is very important to distinguish these two subspaces of $\mathbb{R}^2$: $$\{(t,0):t\in\mathbb{R}\}\quad\text{and}\quad \{(0,t):t\in\mathbb{R}\}$$ even though they are both one-dimensional (and hence isomorphic to each other).

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If $V$ is a finite dimensional vector space it is true up to isomoprhism. Let $T:V\rightarrow V$ be a linear map. Then there exists the short exact sequence

$$0\longrightarrow \ker(T)\xrightarrow{\;\mathrm{id}\;} V\xrightarrow{\;\pi\;} V/\ker(T)\longrightarrow 0$$

The sequence splits because $V/\ker(T)$ is a free module. So $V\cong \ker(T)\oplus V/\ker(T)$. Finally by first isomorphism theorem $V/\ker(T)\cong\mathrm{im}(T)$. So the statement is true if you replace the $=$ with $\cong$.

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Your mistake is in saying that the basis of kernel and the basis of image will combine to give a basis of the vector space. This is not true as when you combine the two basis, they may not be linearly independent and as the dimension is the same, they cannot span the entire space.

I think this question comes after you have studied Rank-Nullity Theorem. What it says that $\dim(V) = \dim(\ker(T)) + \dim(\mathrm{im}(T))$ but that does not mean that these spaces are independent.

At this point, I strongly urge you to review the proof of Rank-Nullity Theorem. It may be found for example, in Linear Algebra by Hoffman Kunze.

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