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This question already has an answer here:

Let $x,y$ be arbitrary non-zero column vectors in $\mathbb R^n$ , then how do we prove that $\det (xy^t)=0$ ?

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marked as duplicate by leo, Daniel Fischer, user147263, RE60K, Najib Idrissi Oct 1 '14 at 18:04

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    $\begingroup$ All the columns of $xy^{t}$ are proportional, this matrix has rank $1$, so it is non-invertible and its determinant is $0$. $\endgroup$ – Ewan Delanoy Oct 1 '14 at 15:08
  • $\begingroup$ This is only true if $n>1$. $\endgroup$ – user147263 Oct 1 '14 at 16:48
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The statement is true only when $n\ge2$. There may be several approaches

  • $n\ge2$ so $\exists z\ne 0:y^tz=0$ so $(xy^t)z=0$ so $\det (xy^t)=0$

  • $\operatorname{rank}(xy^t)=1$ and $n>1$ so $\det(xy^t)=0$

  • If all $x_i=0$ then $xy^t=0$ then $\det(xy^t)=0$. Otherwise if $x_r\ne 0$ since $n\ge 2$ we can take the $j$th row minus $x_j/x_r$ the $r$th row to get a row with all $0's$ without changing the determinant. So $\det(xy^t)=0$

  • $\det(xy^t)=\sum_\sigma \operatorname{sign}(\sigma)\prod_i(xy^t)_{i\sigma(i)}=\sum_\sigma\operatorname{sign}(\sigma) \prod_ix_iy_{\sigma(i)}=(\prod_ix_iy_i)\sum_\sigma \operatorname{sign}(\sigma)=0$

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$xy' = (x_{i} \cdot y_{j})_{i,j}$ which yields row 1 and row 2 are proportional.

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