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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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Have you tried it for, say, $n=2$? – Gerry Myerson Nov 6 '12 at 11:40
It seems like you should write either $ab^T$ or $a \otimes b$ instead of $a \otimes b^T$. At least it looks that way here: – Dan Shved Nov 6 '12 at 11:41
On second thought, if you mean the Kronecker product, then it's OK, but it does look strange to find this $\otimes$ sign in such a simple vector-and-vector situation. – Dan Shved Nov 6 '12 at 11:46
What's the rank of $ab^T$? – wj32 Nov 6 '12 at 12:19

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.

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