# About change of basis

This is proof Lemma 1.2.1 from Scharlau's book Quadratic and Hermitian Forms.

Lemma 2.1. $$B_{b,\mathfrak E'}=T^TB_{b,\mathfrak E}T'.$$

Proof. We compute $$b(e_i',e_j')=b(\sum e_kt_{ki}, \sum e_lt_{lj})=\sum_{k,l} t_{ki} b(e_k,e_l)t_{lj}$$ yielding the $(i,j)$ entry of $T^TB_{b,\mathfrak E}T'$. I don't know why bilinear form splits like this. In first step they just take new basis as linear combination of standard basis. Then next step is unclear to me. Can someone please tell in explicit way assuming n basis

you simply use the definition of bilinear functions, i.e. linearity in every component: $b(\sum_{i=1}^{n} t_ie_i,y)=\sum_{i=1}^{n}t_ib(e_i,y)$ for a fixed $y$ in the second component and similarly $b(x,\sum_{j=1}^{n} t_je_j)=\sum_{j=1}^{n}t_jb(x,e_j)$. Then just use them both together to get above result.

Given a bilinear form $B$ in the standard basis

$$x^tBy$$

by a change of basis with matrix $T$ we have

• $x=Tu$
• $y=Tv$

and therefore

$$x^tBy=(Tu)^tBTv=u^tT^tBTv$$