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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?

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    $\begingroup$ "The matrix product 𝐻=𝐹𝐺 represents a positive semidefinite bilinear form ℎ on 𝑉"... consider multiplying the following two symmetric PD matrices $F=\left[\begin{matrix}13 & 43\\43 & 145\end{matrix}\right],G=\left[\begin{matrix}117 & 87\\87 & 74\end{matrix}\right]$ -- the product is an indefinite matrix, contradicting your statement $\endgroup$ Dec 29, 2023 at 6:22
  • $\begingroup$ Perhaps you would be interested in the matrix $H=F M^{-1} G$, where $M$ is the mass matrix. That is, the matrix with entries $M_{ij} = (b_i, b_j)$ where $b_i$ are the basis vectors in $B$. $\endgroup$
    – Nick Alger
    Dec 29, 2023 at 9:54

3 Answers 3

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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.

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    $\begingroup$ A small note; The matrices $F$ and $G$ do represent the linear operations of currying (partial application of the bilinear forms to only one vector). But that maps $V$ to it’s dual, $V’$, so you can’t compose them. For the operation to make sense you would need to put an inverse mass matrix between $F$ and $G$ to convert the dual vector output of $F$ into its Riesz representation in $V$, which can then be input into $G$. $\endgroup$
    – Nick Alger
    Dec 29, 2023 at 3:37
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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.

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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?

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  • $\begingroup$ You should probably open a new question with this information and follow up question. $\endgroup$ Dec 29, 2023 at 17:41
  • $\begingroup$ I'm also not really convinced, either that it's basis independent or that it's what you want. Apply the same argument as I did in my answer: assume $g$ is positive definite and choose $B$ to be $g$-orthonormal. Then again $G = 1$ and $H = F$. So either (1) this construction is really basis independent and we just have $h = f$ so it's uninteresting; or (2) it is not basis independent. $\endgroup$ Dec 29, 2023 at 17:45

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