Answer by user1551 for A more general question regarding $\operatorname{rank}$
Yes. Pick a basis of the ($k$-dimensional) column space of $A$ and put those basis vectors together to form a matrix $X$. Since the $j$-th column $\mathbf a_j$ of $A$ is a linear combination of these...
View ArticleA more general question regarding $\operatorname{rank}$
I have recently posted this problem and I've been thinking about a generalization: Let $A$ be a $n \times n$ matrix with complex entries and $\operatorname{rank}A=k.$ Prove that there are two $n \times...
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