0
$\begingroup$

First I'll define what I talk about:

A bilinear form on a vector space V is a mapping:

$F: V \times V \rightarrow \mathbb{R}, (a,b) \mapsto F(a,b)$

which is linear in every argument:

$a, b, c \in V$ and $\lambda, \mu \in \mathbb{R}$:

  • $F(\lambda a + \mu b, c) = \lambda F(a, c) + \mu F(b, c)$
  • $F(a, \lambda b + \mu c) = \lambda F(a, b) + \mu F(a, c)$

If I get an expression which could be a bilinear form, I check those two. This can be quite long.

A bilinear form F is symmetric, if:

$\forall a, b \in V: F(a, b) = F(b, a)$

Now my question: If I know that a mapping is symmetric, can I make the checks for bilinearity shorter? Something like that:

$F(\lambda a + b, c) = \lambda F(a, c) + F(b, c)$?

If it is not possible, do you have counterexamples where it doesn't work?

  • 0
    Thanks, I thought I missed something.2012-05-04

1 Answers 1

0

I think this is a proof that it works:

Pre.:

$f:V \times V \to \mathbb{R}, (a,b) \mapsto f(a,b)$

(I) Symmetry: $f(a,b) = f(b,a)$

(II) Short biliniearity: $f(\lambda a + b,c) = f( \lambda a, c) + f(b,c) = \lambda f(a, c) + f(b,c)$

statement: f is a bilinear form

proof:

part 1: $f(\lambda a + \mu b, c) = \lambda f(a, c) + \mu f(b, c)$

$f(\lambda a + \mu b, c) = (II) f(\lambda a, c) + f(\mu b, c) = \lambda f(a,c) + \mu f(b,c)$

part 2: $f(a, \lambda b + \mu c) = \lambda f(a, b) + \mu f(a, c)$

$f(a, \lambda b + \mu c) = (I) f(\lambda b + \mu c, a) = (II) f(\lambda b, a) + f(\mu c, a) = \lambda f(b, a) + \mu f(c, a) = (I) $ $= \lambda f(a,b) + \mu f(a, c) \blacksquare$