# $k > \operatorname{rank}(A)+\operatorname{rank}(B) \Rightarrow \dim\{Au_1-Bv_1,Au_2-Bv_2,\dots,Au_k-Bv_k\} < k$

Let $$A, B \in M_n, \operatorname{Im}(A)\cap\operatorname{Im}(B)=\{0\}$$ and 2 subsets of $$\mathbb{R}^n$$ are $$\{u_1,u_2,\dots,u_k\}, \{v_1,v_2,\dots,v_k\}$$.
Prove that: If $$k>\operatorname{rank}(A)+\operatorname{rank}(B)$$, there are always $$\lambda_1,\lambda_2,\dots,\lambda_k\in\mathbb{R}, \lambda_1^2+\lambda_2^2+\dots+\lambda_k^2>0$$ such that
$$\lambda_1Au_1+\lambda_2Au_2+\dots+\lambda_kAu_k=\lambda_1Bv_1+\lambda_2Bv_2+\dots+\lambda_kBv_k$$

My attemp: rewrite equality $$\lambda_1(Au_1-Bv_1)+\lambda_2(Au_2-Bv_2)+\dots+\lambda_k(Au_k-Bv_k)=0$$ My idea is proving that $$\dim\{Au_1-Bv_1,Au_2-Bv_2,\dots,Au_k-Bv_k\} < k$$ from the inequality $$k>\operatorname{rank}(A)+\operatorname{rank}(B)$$
I think that we may use Sylvester's inequality $$\operatorname{rank}(A)+\operatorname{rank}(B)=\operatorname{rank}(A)+\operatorname{rank}(-B) \ge \operatorname{rank}(A-B)$$ but nothing works for me now.
Hope to see your suggestions. Thank you.

Hint: Note that $$\{Au_1-Bv_1,Au_2-Bv_2,\dots,Au_k-Bv_k\} \subset \operatorname{Im}(A) + \operatorname{Im}(B)$$, and $$\dim(\operatorname{Im}(A) + \operatorname{Im}(B)) = \operatorname{rank}(A) + \operatorname{rank}(B)$$.
• ohhhh, I forgot $\operatorname{Im}(A)\cap\operatorname{Im}(B) = \{0\} \Rightarrow k> \dim\left(\operatorname{Im}(A) \oplus \operatorname{Im}(B)\right)$. Thank you. Feb 2, 2020 at 12:02