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Suppose $\forall x_1,\ldots,x_n,\in R^n$, denote $x_1=(x_1^1,x_1^2,\ldots,x_1^n),x_2=(x_2^1,\ldots,x^n_2)$. Define $\psi:V\times V\times \dots\times V\to R$ as follows: $$\psi (x_1,\ldots,x_n)=\det \begin {pmatrix} x^1_1 & \dots &x^1_n \\ \vdots & \ddots &\vdots \\ x^n_1 &\dots &x^n_n\end{pmatrix}.$$ Then is $\psi$ multilinear?

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    $\begingroup$ What have you tried? What do you think? Have you explored the simpler case of $n = 2$? $\endgroup$ – Tom Oct 27 '14 at 1:25
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    $\begingroup$ how do you define the determinant? $\endgroup$ – James S. Cook Oct 27 '14 at 1:29
  • $\begingroup$ Sum over the permutation for entries along the diagonal multiplied by signature, $\endgroup$ – pxc3110 Oct 27 '14 at 2:11
  • $\begingroup$ I think Tom had good advice, write out what you know for $n=2$. Or, think about $\psi(x_1+cy, \dots , x_n)$ and see if you can simplify it. Then, exploit the symmetry of the determinant in term of column exchange... $\endgroup$ – James S. Cook Oct 27 '14 at 2:42
  • $\begingroup$ Well, I think exchanging the column yields a minus sign isn't it? $\endgroup$ – pxc3110 Oct 27 '14 at 22:05
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You probably mean $\psi \colon V^n \to R$.

Yes, $\psi$ is multilinear. This fact is one of the fundamental properties of determinants.

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  • $\begingroup$ sure, but, if he has the misfortune of thinking the determinant is defined by the cofactor expansion then there is something for him to prove... $\endgroup$ – James S. Cook Oct 27 '14 at 1:29
  • $\begingroup$ Yes, the approach taken to prove it will depend on the definition of determinants used. $\endgroup$ – user187373 Oct 27 '14 at 1:32
  • $\begingroup$ exactly, if we used my preference for determinant the multilinearity is manifest in the definition. $\endgroup$ – James S. Cook Oct 27 '14 at 2:43

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