# Diagonalize a quadratic form using Gram-Schmidt

I'd like to know how to use the Gram-Schmidt method to diagonalize a quadratic form.

I tried to look on the internet for this method but I got no results. My teacher told me I could use Gram-Schmidt method to diagonalize a quadratic form, instead of using the eigenvalues method which is much slower. Is there anybody here who knows this method? Please, can you explain me how does it work? I really need it. • Either your teacher is confused or you've misunderstood what your teacher told you. – Brian Borchers Aug 26 '16 at 22:04
• There are no the Gram-Schmidt method of diagonalization of a quadratic form. The QR method of finding eigenvalues is closely related to the Gram-Schmidt orthogonalization. – Pawel Kowal Aug 26 '16 at 22:08
• I've added a photo of what he means with "Gram Schmidt diagonalization of a quadratic form". – PeppeDAlterio Aug 27 '16 at 3:46

Your teacher is clearly wrong. If it would be possible to diagonalize a positive definite quadratic form over $\mathbb{Q}$ with the Gram-Schmidt process only, all the eigenvalues of the associated matrix would be algebraic numbers over $\mathbb{Q}$ with degree $2^m$, given the operations involved in the GS process. If, for instance, the characteristic polynomial of our matrix with coefficients in $\mathbb{Q}$ is an irreducible polynomial with degree $3$, we are doomed.
What actually holds is that we may use the Gram-Schmidt process to find a QR decomposition, but that's a different story, a QR decomposition is not a diagonalization ($Q^{-1} D Q$).
However, we may use the $QR$ algorithm to find an approximate diagonalization. The idea is the following one: we may use GS to write $M$ as $QR$. Then we may consider $M_1=RQ$ and apply GS to $M_1$ to get $M_1=Q_1 R_1$. If we set $M_2=R_1 Q_1$ and continue this way, $R_n$ converges to $D$ and $Q_n$ converges to the matrix $Q$ giving the diagonalization $M=Q^{-1}D Q$.
• Q is not orthogonal. Given an inner product $<x,y> = x^t A y$, Gram-Schmidt finds a basis in which the inner product looks like the standard inner product, i.e. $<Qx,Qy> = x^t y$. From this we see $Q^t A Q = I$. For indefinite inner products it's not the identity but a matrix with only $1,0,-1$ on the diagonal. – Jules Feb 6 '18 at 11:50
• @Jules: a change of basis is something of the form $M\to J^{-1} N J$. Gram-Schmidt cannot be used for diagonalization, I have already explained why. Consider a real, symmetric, positive definite $3\times 3$ random matrix and apply GS to its column space (or anything else): you do not get the eigenvectors of such a matrix, simply because you cannot through $+,-,\cdot, /,\sqrt{}$ only. – Jack D'Aurizio Feb 6 '18 at 12:00