$V$ is a vector space. $S,T: V \to V$, $\dim(V) = rk(T) = n$, and $rk(S)=k$.
To prove: $$ rk(T+S) \ge n-k. $$ I tried to use rank nullity theorem and that $\dim Im(T) = rk(T)$, couldn't find anything.
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4$\newcommand{\rank}{\mathrm{rk}}$Are you able to show that $\color{blue}{\rank(A+B)\le \rank(A) + \rank(B)}$ for any linear maps $A,B:V\to V$? (For example see here for a hint.) If you can show this, then you can use the fact that $\rank(-U) = \rank(U)$ for all linear maps $U:V\to V$, and set $A = T+S, B = -S$ (so $A+B = T$). – Minus One-Twelfth Feb 26 '19 at 14:18
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@MinusOne-Twelfth I do, but I have no idea how -U is available. – vvildchild Feb 26 '19 at 14:26
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$\newcommand{\rk}{\mathrm{\rank}}$What do you mean by available? You can let $B=-S$, so $\rk(B) =\rk(S) = k$. Also let $A=T+S$. Then use $\color{blue}{\rk(A+B)\le\rk(A) + \rk(B)}$. (Remember $A+B = T$.) – Minus One-Twelfth Feb 26 '19 at 14:31