# Bilinear Forms

• Jan 11th 2010, 08:47 AM
Bilinear Forms
f:VxV -> F
we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V.

Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V?

(BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)
• Jan 11th 2010, 10:54 AM
tonio
Quote:

f:VxV -> F
we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V.

Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V?

(BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)

Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!!

Tonio
• Jan 11th 2010, 11:28 AM
Quote:

Originally Posted by tonio
Perhaps I'm missing something, but I'm afraid your question is nonsensical. You're given a definition (If A then B) and then you ask: could it be that not B but still A? Of course not! The definition tells you that if A (=exists a non zero vector s.t. f(u,v) = 0 ...etc.) then B(=the bil. form is called withered)...!!

Tonio

Please notice the difference, it's f(v,u) once, then f(u,v).
• Jan 11th 2010, 11:15 PM
NonCommAlg
Quote:

we call f a withered form if there's a vector $v\ne 0$ for which f(v,u)=0 for all u in V.
Could f be not-withered, while there still is a $v\ne 0$ for which f(u,v)=0 for all u in V?
(BTW : $f(u,v)=[u]^t_{B}[f]_B[v]_B$ for any base B of V)
the answer is no. to see this, suppose $\dim_F V = n$ and identify $V$ with $F^n.$ then $f(X,Y)=X^TAY,$ for some $A \in \mathbb{M}_n(F)$ and all $X,Y \in F^n.$ let $E_i$ be the $n \times 1$ vector with $1$ in the $i$th row
and $0$ everywhere else. see that $E_i^T A$ is the $i$th row of $A$ and, for any $Y \in F^n,$ the $i$th row of $AY$ is $E_i^TAY.$ now suppose there exists $0 \neq Y_0 \in F^n$ such that $X^TAY_0 = 0$ for all $X \in F^n.$ so
$E_i^T A Y_0=0,$ for all $i.$ thus $AY_0 = 0$ and so $\det A = 0.$ but then $\det A^T =0$ and hence $A^T X_0 = 0,$ for some $0 \neq X_0 \in F.$ thus $X_0^T A=0$ and so $X_0^TAY=0,$ for all $Y \in F^n,$ i.e. $f$ is withered.