- #1
nightingale123
- 25
- 2
I'm having trouble understanding a step in a proof about bilinear forms
Let ## \mathbb{F}:\,\mathbb{R}^{n}\times\mathbb{R}^{n}\to \mathbb{R}## be a bilinear functional.
##x,y\in\mathbb{R}^{n}##
##x=\sum\limits^{n}_{i=0}\,x_{i}e_{i}##
##y=\sum\limits^{n}_{j=0}\;y_{j}e_{j}##
##F(x,y)=F(\sum\limits^{n}_{i=1}\,x_{i}e_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j})=\sum\limits^{n}_{1=i,j}F(x_{i}e_{i},y_{j}e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}F(e_{i},e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}a_{ji}##
##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##
Could someone explain to me how we got that
##F(e_{i},e_{j})=a_{ji}##
I couldn't find the explanation anywhere in my notebook
thank you
Edit: ( added some stuff I missed )
##A=[a_{ij}]_{i,j=1,...,n}##
##<Ax,y>=<A(\sum\limits^{n}_{i=1}x_{i}e_{i}),\sum\limits^{n}_{j=1}y_{j}e_{j}>=\sum\limits^{n}_{i=1}x_{i}Ae_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j}>=\sum\limits^{n}_{i,j=1}x_{i}y_{j}<Ae_{i},e_{j}>##
##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##
I understand why ##\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}=<Ax,y>##,
however I don't understand why ##<Ax,y>=F(x,y)##
Edit(Edit).
In other words I don't understand why there exists a Matrix A so that ##F(x,y)=<Ax,y>##
Let ## \mathbb{F}:\,\mathbb{R}^{n}\times\mathbb{R}^{n}\to \mathbb{R}## be a bilinear functional.
##x,y\in\mathbb{R}^{n}##
##x=\sum\limits^{n}_{i=0}\,x_{i}e_{i}##
##y=\sum\limits^{n}_{j=0}\;y_{j}e_{j}##
##F(x,y)=F(\sum\limits^{n}_{i=1}\,x_{i}e_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j})=\sum\limits^{n}_{1=i,j}F(x_{i}e_{i},y_{j}e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}F(e_{i},e_{j})=\sum\limits^{n}_{1=i,j}x_{i}y_{j}a_{ji}##
##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##
Could someone explain to me how we got that
##F(e_{i},e_{j})=a_{ji}##
I couldn't find the explanation anywhere in my notebook
thank you
Edit: ( added some stuff I missed )
##A=[a_{ij}]_{i,j=1,...,n}##
##<Ax,y>=<A(\sum\limits^{n}_{i=1}x_{i}e_{i}),\sum\limits^{n}_{j=1}y_{j}e_{j}>=\sum\limits^{n}_{i=1}x_{i}Ae_{i},\sum\limits^{n}_{j=1}\;y_{j}e_{j}>=\sum\limits^{n}_{i,j=1}x_{i}y_{j}<Ae_{i},e_{j}>##
##=\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}##
I understand why ##\sum\limits^{n}_{ji}x_{i}y_{j}a_{ji}=<Ax,y>##,
however I don't understand why ##<Ax,y>=F(x,y)##
Edit(Edit).
In other words I don't understand why there exists a Matrix A so that ##F(x,y)=<Ax,y>##
Last edited: