# Is $\psi (x_1,…,x_n)=\det \begin {pmatrix} x^1_1 & \dots &x^1_n \\ \vdots & &\vdots \\ x^n_1 & \dots & x^n_n\end{pmatrix}$ multilinear?

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?

• What have you tried? What do you think? Have you explored the simpler case of $n = 2$? – Tom Oct 27 '14 at 1:25
• how do you define the determinant? – James S. Cook Oct 27 '14 at 1:29
• Sum over the permutation for entries along the diagonal multiplied by signature, – pxc3110 Oct 27 '14 at 2:11
• 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... – James S. Cook Oct 27 '14 at 2:42
• Well, I think exchanging the column yields a minus sign isn't it? – pxc3110 Oct 27 '14 at 22:05

You probably mean $\psi \colon V^n \to R$.
Yes, $\psi$ is multilinear. This fact is one of the fundamental properties of determinants.