A Particular Symmetric Bilinear Map

Question

Does there exist symmetric bilinear map on $\mathbb{C}^n$ such that

$$<v, v> \neq 0 \quad \forall v \in \mathbb{C}^n-\{0\}?$$

Answer 1

Nope. E.g. bilinearity means linearity in the first argument and thus: $$ \langle 0, 0 \rangle = \langle 0\cdot 0, 0 \rangle = 0 \langle 0, 0 \rangle = 0 $$ and $0 \in \mathbb{C^n}$.

Answer 2

Nop, because if $f$ is a bilinear function, then $f(0, 0) = 0$

Answer 3

No, because $\langle 0 , 0 \rangle \not=0$ would prevent the map from being both bilinear and well-defined. In other words, such a map would be multi-valued.

$$\langle v , v \rangle = \langle v + 0 , v +0 \rangle = \langle v , v+0 \rangle + \langle 0, v+ 0 \rangle = \langle v , v \rangle + \langle v, 0 \rangle + \langle 0, v \rangle + \langle 0, 0 \rangle \\ = \langle v, v \rangle + \langle0, 0\rangle + 2\langle v, 0 \rangle \not= \langle v , v \rangle$$

as long as $2 \langle v, 0 \rangle \not= \langle 0, 0 \rangle$ for some $v$ (which must be the case, take for example $v$=0; if $\langle 0,0 \rangle \not=0$, then 2$\langle 0,0\rangle \not= \langle 0, 0 \rangle$).

A similar argument also applies in the case of conjugate symmetry:

$$\langle v, v \rangle + \langle 0,0\rangle + 2Re[\langle v, 0 \rangle]$$ with the term $2Re[\langle v, 0 \rangle]$ not possibly equalling $\langle 0, 0 \rangle$ for all $v \in \mathbb{C}^n$.

The basic idea of the proof is that a linear map needs to "inherit" some of the properties of a field, namely needs inherit the additive identity (i.e. such that $f(a+b) = f(a)$ always for the unique $b$ such that $a+b=a$ always -- particularly we must have $f(0)=0$ -- relaxing this restriction gives us in general an affine function but not a linear one), and the condition imposed prevents this from being possible.

Answer 4

Others have pointed out the trivial problem that the zero vector is a problem, and we should also point out that if $n=1$ and the zero vector is exempted, the naive dot product is such a bilinear form. But furthermore the answer is negative if the zero vector is exempted and $n>1$.

If $\langle v,v \rangle\neq 0$ for all nonzero $v$, then clearly the bilinear form must be nondegenerate. Then in some basis the symmetric bilinear form diagonalizes to the naive dot product. In case $n=2$ you can see that $(1,i)$ is a null vector, and by padding with zeros you can get a null vector in any $n> 2$.

Answer 5

There is no polynomial function of the complex coordinates of $v \in C^n$ ($n>1$) for which $f(v)=0$ implies $v=0$.

One polynomial equation cannot cut out a locus of codimension higher than than $1$.