# Product of matrices associated to bilinear forms

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?

• "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 Dec 29, 2023 at 6:22
• 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 $B$. Dec 29, 2023 at 9:54

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.

• 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 $G$. Dec 29, 2023 at 3:37

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.

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?

• You should probably open a new question with this information and follow up question. Dec 29, 2023 at 17:41
• 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. Dec 29, 2023 at 17:45