Product of matrices associated to bilinear forms

Question

Let $V$ be a real vector space of finite dimension where $f,g: V \times V \to \mathbb{R}$ are two symmetric positive semidefinite bilinear forms. For a given basis $B$ of $V$, there are two symmetric positive semidefinite matrices $F$ and $G$ that represent $f$ and $g$, respectively. The matrix product $H=FG$ represents a positive semidefinite bilinear form $h$ on $V$.

I think the definition of $h$ does not depend on the basis $B$ and therefore neither on the associated matrices. How do I define the operation $(f,g)\mapsto h$ without making reference to associated matrices?

Answer 1

If you apply a basis transformation to both factors, so that $x=Su$ and $y=Tv$, the matrix of the bilinear form $x^\top Fy=u^\top S^\top FTv=u^\top(S^\top FT)v$ becomes $S^\top FT$, and the matrix of the bilinear form $x^\top Gy=u^\top S^\top GTv=u^\top(S^\top GT)v$ becomes $S^\top GT$. Multiplying these two new matrices yields $S^\top FTS^\top GT$. Now $h$ would be defined as $u^\top(S^\top FTS^\top GT)v=(u^\top S^\top)(FTS^\top G)(Tv)$ $=x^\top(FTS^\top G)y$, which will generally not be equal to $x^\top FGy$, the value defined for the original bases.

The reason this doesn’t work is that multiplying matrices of bilinear forms isn’t a natural thing to do – you typically multiply matrices when they represent linear operations that you can compose, but the matrix that represents a bilinear form doesn’t represent a linear operation.

Answer 2

Suppose $g$ is definite and $B$ is a $g$-orthonormal basis. Then $G$ is the identity and $H = F$ implying that $h = f$. But clearly there are $f, g, B$ such that $h \ne f$.

Consider the simple case of one real dimension with $f(x,y) = 2xy$ and $g(x,y) = 5xy$. Then $g$ has orthonormal basis $1/\sqrt 5$ and $h(x,y) = f(x,y) = 2xy$ in this basis. But now choose $3$ as a basis; then $$ h(3,3) = f(3,3)g(3,3) = 9\cdot 90 \implies h(1,1) = 90 \implies h(x,y) = 90xy. $$ Thus $h$ is highly basis dependent.

Answer 3

I think that if $F,G$ are real symmetric positive semidefinite matrices, then $F^{\frac{1}{2}}GF^{\frac{1}{2}}$ is also symmetric positive semidefinite.

Consider the example by @user8675309, in which

$$ F = \begin{pmatrix} 13 & 43 \\ 43 & 145 \end{pmatrix} \qquad G = \begin{pmatrix} 117 & 87 \\ 87 & 145 \end{pmatrix} $$ Then $$ F^{\frac{1}{2}} = \frac{1}{\sqrt{170}} \begin{pmatrix} 19 & 43 \\ 43 & 151 \end{pmatrix} \qquad G^{\frac{1}{2}} = \frac{1}{\sqrt{257}} \begin{pmatrix} 150 & 87 \\ 87 & 74 \end{pmatrix} $$ and $$ F^{\frac{1}{2}}GF^{\frac{1}{2}} = \frac{1}{17} \begin{pmatrix} 45250 & 144754 \\ 144754 & 465226 \end{pmatrix} $$ The eigenvalues of $F^{\frac{1}{2}}GF^{\frac{1}{2}}$ are $\lambda_1 = 2(7507 - \sqrt{56270485})$ and $\lambda_2 = 2(7507 + \sqrt{56270485})$, which are both positive.

I found stated in this thread that $FG$ and $F^{\frac{1}{2}}GF^{\frac{1}{2}}$ have the same nonzero eigenvalues, and this example confirms it.

Let $P$ be an orthogonal matrix (a change from one orthogonal basis to another). Then $(P^{\top}F^{\frac{1}{2}}P)(P^{\top}GP)(P^{\top}F^{\frac{1}{2}}P) = P^{\top}(F^{\frac{1}{2}}GF^{\frac{1}{2}})P$. In other words the operation $(f, g) \mapsto h$ given by the following steps:

1. pick an orthogonal basis $B$ of $V$;
2. represent $f$ as a matrix $F$ w.r.t. $B$;
3. represent $g$ as a matrix $G$ w.r.t. $B$;
4. compute $H = F^{\frac{1}{2}}GF^{\frac{1}{2}}$;
5. lef $h$ be the bilinear form associated to $H$ w.r.t. $B$.

is well-defined because it does not depend on the basis $B$. This is really the operation I cared about. Is it possible to define it without using the isomorphism between forms and matrices?