Proof that orthogonalization preserves determinant

Question

Prove that the Gram-Schmidt orthogonalization (without normalization) preserves the determinant of the $n\times n$ matrix formed by the $n$ linear independent vectors in $\mathbb{R}^n$.

Answer 1

Looking closer at the orthogonalization process proves that the determinant is not changed. Let $$[v_1,v_2,\dots,v_n]$$ be the $n\times n$ column-matrix of the $n$ linear independent vectors. The orthogonalized matrix has columns $$[u_1,u_2,\dots,u_n]$$ where the individual $u_i$ are inductively calculated starting from calculating $u_1$ and ending at calculating $u_n$. The calculation of $u_i$ using the Gram-Schmidt process (definition) requires $v_i$ and the already calculated $u_k$ with $k=1,\dots, i-1$

$$u_i=v_i-\sum_{k=1}^{i-1}\frac{\langle v_i,u_k \rangle}{\langle u_k,u_k \rangle}u_k$$

The inner product $\langle \cdot,\cdot \rangle$ is a scalar. The orthogonalization process consists therefore of repeated adding of multiples of already calculated columns (therefore called process). Adding a multiple of another column does not change the determinant (proof).

If the terms are explicitely written one sees that the orthogonal matrix results from the original matrix where multiples of the original columns are added to another column and one even better sees why the determinant is preserved. The terms quickly grow and are nested so only $u_1,u_2,u_3$ are explicitely shown

$$[u_1,u_2,u_3,\dots] =\left[v_1,v_2-\frac{\langle v_2,u_1 \rangle}{\langle u_1,u_1 \rangle}u_1,v_3-\frac{\langle v_3,u_1 \rangle}{\langle u_1,u_1 \rangle}u_1-\frac{\langle v_3,u_2 \rangle}{\langle u_2,u_2 \rangle}u_2,\dots\right] =\left[v_1,v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1,v_3-\frac{\langle v_3,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1-\frac{\langle v_3,u_2 \rangle}{\langle u_2,u_2 \rangle}u_2,\dots\right] =\left[v_1,v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1,v_3-\frac{\langle v_3,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1-\frac{\left\langle v_3,v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1 \right\rangle}{\left\langle v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1,v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1 \right\rangle}\left(v_2-\frac{\langle v_2,v_1 \rangle}{\langle v_1,v_1 \rangle}v_1\right),\dots\right] =[v_1,v_2+a\cdot v_1,v_3+b\cdot v_1+c \cdot v_2,\dots]$$ with $a,b,c$ scalars.

Answer 2

It is well-known that adding a row to another, with an arbitrary factor, does not change the value of the determinant (a determinant "cancels" the linear dependencies). And this is precisely what Gram-Schmidt does.

Note that if you performed the normalizations, that would result in the determinant being divided by the product of the norms of the successive vectors. As the determinant of an orthogonal matrix is plus or minus unity, the determinant can come as a byproduct.