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?


2 Answers 2


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 $A=T+S,\ B=-S$.

  • $\begingroup$ Could you please provide a proof? $\endgroup$
    – GoodWilly
    Jul 16, 2019 at 17:35
  • $\begingroup$ 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. $\endgroup$
    – Berci
    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 : $\operatorname{rank}(\textsf{T}+\textsf{S}) \geq n-k$.


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