Let $V$ be a vector space and $\lbrace v_1,\ldots , v_n\rbrace$ be a basis of $V$. Via the Gram-Schmidt process we get a matrix $G$ which takes any $v_i$ to a vector $e_i$ in such a way that $\lbrace e_1,\ldots , e_n\rbrace$ is an orthonormal basis of $V$. Is it true that $\det G>0$?
[Math] Determinant in Gram-Schmidt process
linear algebralinear-transformations
Best Answer
The determinant of a matrix is invariant under conjugation, so w.l.o.g. let’s examine what $G$ looks like relative to the original basis. In the $k$th iteration of the Gram-Schmidt process, the algorithm cranks out the vector $v_k$ plus a linear combination of the orthonormal basis vectors generated up to that point, all divided by a positive number (the normalization step). However, each of these already-generated basis vectors is itself of this form, so the result of step $k$ is a linear combination of $v_1$ through $v_k$, with $v_k$’s coefficient positive.
Recalling that the columns of a transformation matrix are the images of the basis vectors, this means that relative to the original basis, $G$ is upper-triangular with positive entries along its main diagonal, therefore $\det G\gt0$.
One way of describing this result is that the Gram-Schmidt process preserves orientation.