Is it true that if $T$ is a linear operator on a finite-dimensional vector space $V$ then $V=\ker T\oplus \operatorname{im}T$? $\newcommand{\im}{\operatorname{im}}$I am trying to prove or disprove the following statement:
Let $V$ and $W$ be finite-dimensional vector spaces. If $T:V\rightarrow W$ is a linear transformation then $V=\ker T\oplus \im T$. (By the symbol $\oplus$ I mean the direct sum of two vector spaces.)
This statement cannot be true if $V\neq W$ because a vector space can only be a direct sum of its subspaces. However, I am not sure about the case when $V=W$, i.e., when $T$ is a linear operator.
I want to use the following proposition:
$\textbf{Proposition.}$ Let $V$ be a finite-dimensional vector space and let $U$ and $W$ be subspaces of $V$. Then $V=U\oplus W$ if and only if $V=U+W$ and $U\cap W=\left\{ 0 \right\}$.
First I want to show that $V=\ker T + \im T$. I just don't have a clue how to possibly do this, which leads me to believe there must be a counterexample. I believe that $\ker T\cap \im T=\left\{0\right\}$ since $T(0)=0$ for any linear transformation and it is not possible for $Tv\neq 0$ if $v\in \ker T$. Some help?
Thank you in advance.
 A: You have the right ideas.  Indeed, your claim is not true. Consider, for example, the transformation 
$$
T = \pmatrix{0&1\\0&0}
$$
Verify that im$(T) = \ker(T)$, and that both of these are one-dimensional subspaces of $\Bbb R^2$.
Notably, however, the statement will hold for any self-adjoint (symmetric) operator $T:V \to V$.
A: This is not true in general. I will restrict the question to the case where $V = W$, i.e. $T$ is a linear operator on $V$.
Consider the linear operator over $V = \mathbb{R}^3$ where the matrix is written with respect to the standard basis: $(e_1, e_2, e_3)$
$\begin{bmatrix}
0 & 1& 1 \\
0 & 0& 0\\
0 & 0& 0 \\
\end{bmatrix}$
It is easy to see that $\mathrm{Null}(T) = span(e_1, e_2 - e_3)$, and $\mathrm{Im}(T) = span(e_1)$. Clearly $V \neq \mathrm{Null}(T) \oplus \mathrm{Im}(T)$.
However, this is an interesting question, because by adding a few additional hypothesis, we can get a few true statements that similar to this, and these become very important in linear algebra.
$1.$ For a finite dimensional complex vector space (or over an algebraically closed field), $V$, where the dimension of $V$ is $n$, for any linear operator $T$, $V = \mathrm{Null}(T^n) \oplus \mathrm{Im}(T^n)$.
$2.$ If we assume $V$ is complex (or over an algebraically closed field), finite dimensional, and we have a linear transformation such that $T^2 = T$, we do indeed have: $V = \mathrm{Null}(T) \oplus \mathrm{Im}(T)$. 
A: It is true if $T$ is a projection i.e. $T^2=T$.But in general you may take the counterexample $T(\epsilon_1)=\epsilon_2$ and $T(\epsilon_2)=0$,on $\mathbb R^2$ over the field $\mathbb R$.Here $\operatorname {Range}T=\operatorname {Ker}T$.Although for any finite dimensional vector space $V$ there is a $k$ such that $V=R(T^k)\oplus N(T^k)$.
