# An interesting equality from linear algebra

It's a bit mind-blowing when you want to understand this equality geometrically:

we have vectors $x_1, ..., x_n$ in n-dimensional space

Let's take an orthonormal basis in this space $e_1, ..., e_n$ and compute such vectors obtained with via scalar product:

$$ z_i = (< e_i, x_1 >, < e_i, x_2 >, ..., < e_i, x_n >) $$

So the theorem is: $$ \det_{i,j} < z_i, z_j > = \det_{i,j} < x_i, x_j > $$

And the proof is very simple, let's introduce the matrix $A$: $ A_{ij} = <x_i, e_j> $ $$ \det_{i,j} < z_i, z_j > = \det A^T A = \det A A^T = \det_{i,j} < x_i, x_j >$$

Voila! Beautiful, but completely unclear.

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