$\ker(T) + \ker(S) = V \implies {\rm Im}(T + S) ={\rm Im}(T) +{\rm Im}(S)$ Let $V$ be a vector space. Let $T, S$ be two linear operators $T:V \rightarrow V$, $S: V \rightarrow V$, such that $$\ker(T) + \ker(S) = V$$, then we must have $${\rm Im}(T+S) = {\rm Im}(T) + {\rm Im}(S)$$.
If the statement is true, then give the proof. If the statement is false, give an counterexample.
Attempt: So far, I am trying to use the $rank-nullity$ theorem and the theorem $\dim(U + W) + \dim(U \cap W) = \dim(U) + \dim(W)$, but didn't get any luck and I am getting stuck for this question for hours. Could someone tell me how to solve this question?
Thanks!
 A: Any vector $v\in V$ may be written $v=a+b$ with $a\in$ker$(S)$, $b\in$ker$(T)$.  Now write out $(S+T)v=(S+T)(a+b)$. Simplify.  Is the result the sum of something in im$(S)$ and something in im$(T)$?
Then write out $Sv=S(a+b)$ and $Tv=T(a+b)$. Simplify. Can you write the results in the form $(S+T)x$ for some suitable $x$'s?
A: True.
Proof: Clearly $\text{Im}(T+S) \subseteq \text{Im}(T)+\text{Im}(S)$, so we just need to show $\text{Im}(T)+\text{Im}(S) \subseteq \text{Im}(T+S)$. As another answer suggested, it suffices to show $\text{Im}(T) \subseteq \text{Im}(T+S)$ and $\text{Im}(S) \subseteq \text{Im}(T+S)$, because in general if $U, W$ are contained in a subspace $X$, then $U+W$ is also contained in $X$.
Thus, let $v \in V$ be any vector, hence $v = k_t + k_s$ for some $k_t \in \ker(T), k_s \in \ker(S)$. Then $T(v) = T(k_t) + T(k_s) = T(k_s) = T(k_s) + S(k_s) = (T+S)(k_s)$, 
showing that $\text{Im}(T) \subseteq \text{Im}(T+S)$. The case $\text{Im}(S) \subseteq \text{Im}(T+S)$ is similar. $\square$
