# $dim(T+S)\; \;$, $T,S:V \mapsto V$ 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?

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 $$A=T+S,\ B=-S$$.
• The proof is basically the same as the beginning of the other answer. Note that $(A+B)x\in \mathrm{im}(A) +\mathrm{im}(B)$, so $\mathrm{im}(A+B)\ \subseteq\ \mathrm{im}(A)+\mathrm{im}(B)$, and calculating dimensions we arrive to the statement. Jul 16, 2019 at 18:46
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 : $$\operatorname{rank}(\textsf{T}+\textsf{S}) \geq n-k$$.