Postby **Cauchy** » Sun Jun 04, 2017 12:02 am UTC

Don't you do that determinant thing that gets you the cross product?

For n-1 vectors in R^n, look at the matrix

[v_11 v_12 ... v_1n]

[v_21 v_22 ... v_2n]

.

.

.

[v_{n-1}1 v_{n-1}2 ... v_{n-1}n]

[x_1 x_2 ... x_n]

where the x_i is 1 in the ith position and 0 everywhere else, and take its determinant. It's an abuse of notation, I know, but it works.

More formally, let v_i, 1 <= i <= n-1 be the differences, so the hyperplane is spanned by the v_i. For 1 <= k <= n, let A_k be the determinant of the n-1 by n-1 matrix

[v_11 v_12 ... v_1{k-1} v_1{k+1} ... v_1n]

.

.

.

[v_{n-1}1 v_{n-1}2 ... v_{n-1}{k-1} v_{n-1}{k+1} ... v_{n-1}n]

where the kth component of each vector has been stripped out. These are the minors of the first matrix I wrote when you expand along the bottom row. Let A = {A_1, A_2, ..., A_k}. Then, for arbitrary w, the determinant of

[v_1]

[v_2]

.

.

.

[v_{n-1}]

[w]

is A dot w, by expanding the determinant along the bottom row. It follows that A dot w = 0 when w is any of the v_i, because a row in the determinant is repeated, so A is orthogonal to each of the v_i. A is also not the 0 vector because... handwave involving the v_i being linearly independent.

(∫|p|^{2})(∫|q|^{2}) ≥ (∫|pq|)^{2}

Thanks, skeptical scientist, for knowing symbols and giving them to me.