My professor states the following lemma: In a suitable basis, the symmetric matrix of a bilinear form is diagonal.
In other words, "$\mathcal{V}$ has a $Q$-orthogonal basis", where $\mathcal{V}$ is the vector space for the symmetric bilinear form, $Q$.
Is a "$Q$-orthogonal basis" different from an orthogonal basis? Can you give me an example of where this is the case?
Thanks.
The main distinction is this: given a symmetric real matrix $A=A^T,$ we can find eigenvalues and so on, leading to the (possibly very difficult) task of finding an orthogonal matrix $R$ that solves $R^T A R = D_e,$ diagonal with the eigenvalues as diagonal entries. This is always possible.
However, the analogous task $P^T A P = D,$ where we make no restriction on $P,$ allow $P P^T \neq I,$ is typically quite easy; there are algorithms, if $A$ has all integer or rational entries, it turns out that $P$ also has rational entries. In the most favorable case, $P$ is also upper triangular, but that cannot be guaranteed ahead of time. However, the columns of $P$ are "$A$-orthogonal," as the off-diagonal entries of $P^T A P$ are zero. In symbols, where a $1$ by $1$ matrix is just identified with its (single) real entry, and is its own transpose as well, when $p$ and $q$ are distinct columns of $P,$ then $$ p^T A q = q^T A p = 0. $$
Let's see, in quadratic forms books this whole topic is called Hermite Reduction. The thing that surprised me was that people worked out an algorithm, the whole purpose of which is to do Hermite reduction without having to invert any matrices. Go Figure.
I collected some related questions by link,
reference for linear algebra books that teach reverse Hermite method for symmetric matrices
Find the transitional matrix that would transform this form to a diagonal form.
