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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.

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1 Answer 1

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Hint: Note that $\{Au_1-Bv_1,Au_2-Bv_2,\dots,Au_k-Bv_k\} \subset \operatorname{Im}(A) + \operatorname{Im}(B)$, and $\dim(\operatorname{Im}(A) + \operatorname{Im}(B)) = \operatorname{rank}(A) + \operatorname{rank}(B)$.

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  • $\begingroup$ 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. $\endgroup$
    – trung
    Feb 2, 2020 at 12:02

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