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Let $A$ be a complex $n\times n$ matrix. Can we always find two vectors $a,b\in \mathbb C^n$ such that $A=a\otimes b^T$?

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    What's the rank of $ab^T$?2012-11-06

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In general, no. For example, you will never be able to write the identity matrix as such a product. The rank of the resulting matrix is $1$ (or $0$), this is a consequence of the inequality $\mathrm{rank}(AB) \le \min\left(\rm{rank}(A),\ \rm{rank}(B)\right)$ So necessarily all matrices with rank $2$ or more will not be expressible. However, all rank one matrices can be written as such a product. This is a special case of the more general rank factorization.