Proof that the absolute determinant equals the product of norms of orthogonal vectors

Question

$A$ is a $n\times n$ matrix of $n$ orthogonal vectors $x_1, \dots, x_n$ in $\mathbb{R}^n$ that are not necessarily normalized. Prove that the absolute determinant equals the product of vector norms

$$|\det(A)| = \prod_{k=1}^n |x_k|.$$

Answer 1

The matrix $$\begin{bmatrix} \frac{x_1}{\|x_1\|} & \frac{x_2}{\|x_2\|} & \cdots &\frac{x_n}{\|x_n\|}\end{bmatrix}$$ has orthonormal columns so it is orthogonal. Therefore its determinant is $\pm 1$ so $$\pm 1 = \det \begin{bmatrix} \frac{x_1}{\|x_1\|}& \cdots& \frac{x_n}{\|x_n\|}\end{bmatrix} = \frac1{\|x_1\|\cdots \|x_n\|} \det \begin{bmatrix} x_1 & \cdots & x_n\end{bmatrix} = \frac{\det A}{\|x_1\|\cdots \|x_n\|}.$$

We conclude $$\left|\det A\right| = \|x_1\|\cdots \|x_n\|.$$

Answer 2

If you multiply $A^TA$ you get a diagonal matrix, which has the norms squared of the vectors in its diagonal. This is because to compute the entry $(i,j)$ you will multiply $x_i^Tx_j$, which is $0$ if $i\neq j$ and equal to $x_i^Tx_i=|x_i|^2$ if $i=j$. In the complex case we can use $A^*=\overline{A^T}$ instead.

Then $$|\det(A)|^2=\det(A^T)\det(A)=\det(A^TA)=\prod_{i=1}^{n}|x_i|^2$$

Taking the positive square root, you get your result.