35 - Gram Matrix

Gram Matrix

Let $V$ be a vector space over $\mathbf{F}$
Let $B$ be a bilinear form om $V$

For $v_1,\cdots ,v_n\in V$
The Gram matrix of $B$ with respect to $v_1,\cdots ,v_n$ is the $n\times n$ matrix

$$B_{GM}=(B(v_i,v_j))\in {\mathcal{M}}_{n\times n}(\mathbf{F})$$

That is the $(i,j)th\text{ coordinate of }{B}_{GM}\text{ is }B(v_i,v_j)$

Bilinear Form of a Gram Matrix

Let $V$ be a finite-dimensional vector space over $\mathbf{F}$
Let $v_1,\cdots ,v_n$ be a basis for $V$
Let $B$ be a bilinear form on $V$
Let $A\in {\mathcal{M}}_{n\times n}(\mathbf{F})$ be the associated Gram matrix

For $X,Y\in V$
Let $\mathbf{x}=(x_1,\cdots ,x_n)\in {\mathbb{F}}^n$
Let $\mathbf{y}=(y_1,\cdots ,y_n)\in {\mathbf{F}}^n$
where $\mathbf{x},\mathbf{y}$ are the unique coordinate vectors such that

$$X=x_1{v}_1+\cdots +x_n{v}_n\text{ and }Y=y_1{v}_1+\cdots +y_n{v}_n$$

Then

$$B(X,Y)=\mathbf{x}A{\mathbf{y}}^T$$

Proof

$$\begin{array}{rl}B(X,Y)& =B\left(\sum _{i=1}^n{x}_i{v}_i,\sum _{j=1}^n{y}_j{v}_j\right)\\ & =\sum _{i=1}^n{x}_{i}B\left(v_i,\sum _{j=1}^n{y}_j{v}_j\right)\\ & =\sum _{i=1}^n{x}_i\sum _{j=1}^n{y}_{j}B(v_i,v_j)\\ & =\sum _{i=1}^n\sum _{j=1}^n{x}_i{a}_{ij}{y}_j\\ & =xA{y}^T\end{array}$$