# I Bilinear forms

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1. Jul 3, 2017

### nightingale123

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 dont 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: Jul 3, 2017
2. Jul 3, 2017

### Orodruin

Staff Emeritus
$F(e_i, e_j)$ is a number. Define that number to be $a_{ij}$.

3. Jul 3, 2017

### StoneTemplePython

I assume that $\mathbf e_k$ refers to standard basis vectors, i.e.

$\mathbf I = \bigg[\begin{array}{c|c|c|c} \mathbf e_1 & \mathbf e_2 &\cdots & \mathbf e_{n} \end{array}\bigg]$

I would start real simple. Suppose we had the constraint that $\mathbf x := \mathbf e_k$ and $\mathbf y := \mathbf e_r$.

What happens when you do $\mathbf e_r^T \mathbf A \mathbf e_k = \mathbf y^T \mathbf A \mathbf x$? This equation "grabs" one cell from $\mathbf A$, which you should verify by inspection.

From here recall that the standard basis vectors in fact form a basis, so let's loosen up our restriction and let $\mathbf x$ be anything (while keeping its dimension and field the same, of course.)

$\mathbf x = \gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n$

$\mathbf y^T \mathbf A \mathbf x = \mathbf y^T \mathbf A\big(\gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n\big)$

now loosen up the constraint on $\mathbf y$ so it too can be anything

$\mathbf y = \eta_1 \mathbf e_1 + \eta_2 \mathbf e_2 + ... + \eta_n \mathbf e_n$

$\mathbf y^T \mathbf A \mathbf x = \big(\eta_1 \mathbf e_1 + \eta_2 \mathbf e_2 + ... + \eta_n \mathbf e_n \big)^T \mathbf A\big(\gamma_1 \mathbf e_1 + \gamma_2 \mathbf e_2 + ... + \gamma_n \mathbf e_n\big) = \sum \sum \eta_j \gamma_i a_{j,i} = \sum \sum \eta_j \gamma_i F(e_{i},e_{j}) = \sum \sum y_j x_i F(e_{i},e_{j})$

If you work through that, any bi-linear form your want (over reals with finite dimension) is just "grabbing" and scaling stuff from $\mathbf A$ and then summing them up so you get a scalar result.

4. Jul 3, 2017

### WWGD

If A is linear, then <Ax,y> is bilinear and so it has a representation as a matrix, as a quadratic form. Is that your question?

5. Jul 3, 2017

### StoneTemplePython

follow-up thought:

my posting had $\mathbf y^T \mathbf {Ax} = \mathbf y^T\big( \mathbf {Ax}\big) = <\mathbf y, \mathbf{Ax}>$

$< \mathbf{Ax}, \mathbf y>$

inner products over Reals are symmetric, so

$< \mathbf{Ax}, \mathbf y> = <\mathbf y, \mathbf{Ax}>$

(over complex numbers inner products are symmetric, plus complex conjugation at the end)

note that Chapter 7 of Linear Algebra Done Wrong has a very good discussion of bilinear and quadratic forms. Freely available by the author here: