$dim(T+S)\; \; $, $T,S:V \mapsto V$ Are linear transformations So I had the following question:
$T,S:V \mapsto V$ Are linear transformations
$dim(V) = n\; $, $rank(T) = n\; $, $rank(S) = k$.
Prove that $rank(T+S) \geq n-k$
Now, I know how to put many sub-conclusions to point such as:
$T$ is one to one and onto, which means $dimKer(T) = 0, \; dimIm(T) = n\; $.
Also, $dimIm(S) = k\; $, $Im(S)\subseteq \;  Im(T)$
But I'm having a difficult time putting this proof together.
Also, could you please give me some tips and point of views on how to look at such questions regarding the dimension of summation \ scalar multiplication \ composition of linear transformations?
 A: Hint: We have $\mathrm{rank}(A)+\mathrm{rank}(B)\ge\mathrm{rank}(A+B)$ for any linear transformations $A,B:V\to V$. 
Prove it then apply it with $A=T+S,\ B=-S$. 
A: If $y\in \operatorname{im}(\textsf{T})$ then $y=\textsf{T}(x)$ for some $x \in \textsf{V}$. Notice that $y$ can be written as
$$y=-\textsf{S}(x)+\big(\textsf{T}(x)+\textsf{S}(x)\big) =\textsf{S}(-x)+(\textsf{T}+\textsf{S})(x)$$
then $y\in  \operatorname{im}(\textsf{S})+\operatorname{im}(\textsf{T}+\textsf{S})$. Therefore
$$\operatorname{im}(\textsf{T}) \subseteq \operatorname{im}(\textsf{S})+\operatorname{im}(\textsf{T}+\textsf{S})$$
From this we can conclude that
$$\begin{align}
\operatorname{rank}(\textsf{T}) &= \dim\big(\operatorname{im}(\textsf{T})\big) \\
& \leq \dim\big (\operatorname{im} (\textsf{S})+\operatorname{im}(\textsf{T}+\textsf{S})\big) \\
&=\dim\big(\operatorname{im} (\textsf{S})\big) + \dim\big(\operatorname{im}(\textsf{T}+\textsf{S}) \big) - \dim\big( \operatorname{im}(\textsf{S})\cap \operatorname{im}(\textsf{T}+\textsf{S})\big) \\
& \leq \dim\big(\operatorname{im} (\textsf{S})\big) + \dim\big(\operatorname{im}(\textsf{T}+\textsf{S}) \big) \\
& = \operatorname{rank}(\textsf S) + \operatorname{rank}(\textsf{T}+\textsf{S}) \\
\end{align}$$
So,
$$n\leq k+ \operatorname{rank}(\textsf{T}+\textsf{S})$$
or as you wanted to show : $\operatorname{rank}(\textsf{T}+\textsf{S}) \geq n-k$. 
