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I represented $x$ as $[x_1\ x_2\ ... x_n]^T$, and $y^T$ as $[y_1\ y_2\ ... y_n]$.

Multiplying them produces a matrix $n$x$n$: $$ \begin{pmatrix}x_1y_1&x_1y_2&\dots& x_1y_n\\ x_2y_1&x_2y_2&\dots& x_2y_n\\ \\ \\ \\ \\ x_ny_1& x_ny_2&\dots& x_ny_n \end{pmatrix}$$

(I apologize, I don't know how to format a matrix using LaTeX). :)

Obviously, for $n=2$, $n=3$ the determinant is $0$. I just don't know how to prove it for $n$.


marked as duplicate by user1551, egreg, Algebraic Pavel, Grigory M, Davide Giraudo May 6 '14 at 11:27

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  • 2
    $\begingroup$ To create a matrix enclosed by parentheses, you can use the \begin{pmatrix}...\end{pmatrix} environment. $\endgroup$ – Olivier Bégassat May 6 '14 at 9:55
  • $\begingroup$ Can you see that all rows/columns are scalar multiples of the others? What does it say about its rank? $\endgroup$ – Algebraic Pavel May 6 '14 at 11:16

Hint. when both $x,y$ are nonzero, this matrix has rank 1, so cannot be invertible when $n>1$.

  • $\begingroup$ How do you know that rank is 1? $\endgroup$ – Eutherpy May 6 '14 at 9:59
  • 1
    $\begingroup$ By the linear dependence of the columns. $\endgroup$ – Martín-Blas Pérez Pinilla May 6 '14 at 10:04

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