Bilinear form matrix

Question

So I read on one of the answers here that β(x,y)=x^TAy where A is a matrix.

So if I'm reading this right then the matrix of the bilinear form corresponding to the dot product would be the identity Matrix?

Answer 1

You're correct!

To see this, let $\beta$ be the bilinear form on a vector space $V$ with basis $\{\vec{e}_i\}$. Let $\vec{v},\vec{w}\in V$, where $\vec{v}=\sum_{j=1}^n a_j\vec{e}_j$ and $\vec{w}=\sum_{k=1}^m b_k\vec{e}_k$. Then by bilinearity of $\beta$, \begin{align*} \beta(\vec{v},\vec{w})&=\beta\left(\sum_{j=1}^n a_j\vec{e}_j,\sum_{k=1}^m b_k\vec{e}_k\right)\\ &=\sum_{j=1}^n\beta\left(a_j\vec{e}_j,\sum_{k=1}^m b_k\vec{e}_k\right)\\ &=\sum_{j=1}^n\sum_{k=1}^m a_jb_k\beta(\vec{e}_i,\vec{e}_j)\\ &=\begin{bmatrix}a_1 & \ldots & a_n\end{bmatrix}A\begin{bmatrix}b_1 \\ \vdots \\ b_m\end{bmatrix}\\ &=\vec{v}^\top A\vec{w} \end{align*}

where $A=[a_{i,j}]$ for $a_{i,j}=B(\vec{e}_i,\vec{e}_j)$.

Let $D(\vec{v},\vec{w})$ denote the bilinear form that is the dot product. Then since $D(\vec{e}_i,\vec{e}_j)=\vec{e}_i\cdot\vec{e}_j=\begin{cases} 1 & i = j\\ 0 & i\neq j\end{cases}$, the matrix $A$ associated to $D$ as defined above is just the identity matrix!