$V = \operatorname{Im} T + \ker T $ then $ \operatorname{Im} T \cap \ker T = \{0\}$ Let $F$ be a field, let $V$ be a vector space with finite dimension over $F$ and let $T$ be a linear operator on $V$. Prove that:
a) If $V = \operatorname{Im} T + \ker T $ then $\operatorname{Im} T \cap \ker T = \{0\}$;
b) If $\operatorname{Im} T \cap \ker T = \{0\}$ then $V = \operatorname{Im} T \bigoplus \ker T $
I'm really stuck with this problem, some help to solve this please.
 A: Hint: $\dim V = \dim(Im T) + \dim(Ker T)$
Second hint: $\dim(U+W)+\dim(U\cap W)=\dim(U)+\dim(W)$
A: I will just provide the details to (i) using the hints suggested by vadim123.  You can work on the other part yourself.
i) Suppose that $\def\Im{\operatorname{Im}}V = \Im(T) + \ker(T)$, then by the dimension theorem, $\dim(V) = \dim(\Im(T)) + \dim(\ker(T))$ and so $\dim(V) = \dim(\Im(T) + \ker(T))$.  Hence, by the second hint we have that
$\dim(\Im(T) \cap \ker(T)) = 0 \implies \Im(T) \cap \ker(T) = \{0\}$, as needed.
A: These results hold only in finite dimensional vector spaces, so you must use a dimension argument somewhere. The rank-nullity theorem tells you that $\dim(V)=\operatorname{rk}(T)+\dim(\ker T)$, where $\operatorname{rk}(T)$ is by definition the dimension of the image of $T$. This reduces the question to the following (after which you should forget $T$)

If $U,W$ are subspaces of a space $V$ of finite dimension$~n$, and if $\dim U+\dim W=n$ then show that

*

*if $U+W=V$ then $U\cap W=\{0\}$, and


*if $U\cap W=\{0\}$ then $U\oplus W=V$.

If you know that in finite dimension one has $\dim(U+W)=\dim U+\dim W-\dim(U\cap W)$ for any subspaces $U,W$ (a formula notable for not extending in the obvious way to more than $2$ subspaces) then things become easy. For point 1.  the formula gives you $\dim(U\cap W)=n-(\dim U+\dim W)=0$ which allows you to conclude, and for point 2. it similarly gives you that $\dim(U+W)=\dim U+\dim W=n$.
In case you need a proof for the formula, the basic method is to choose a basis for $U\cap W$, extend this basis separately to bases of $U$ and of $W$, and then prove that the $\dim U+\dim W-\dim(U\cap W)$ vectors in the union of those bases together form a basis of $V$; this proof is easy though not trivial. A more slick proof consists of forming the vector space $X=U\times W=\{\,(u,w)\mid u\in U,w\in W\,\}$, which has $\dim X=\dim U+\dim V$, and then apply the rank-nullity theorem to $f: X\to V$ defined by $f(u,w)=u-w$, using that $\ker(f)=\{\, (v,v)\mid v\in U\cap W\,\}$ whence $\dim(\ker f)=\dim U\cap W$.
