Does an Inner Product Always Induce a Metric?

I am working with linear algebra over finite fields, specifically $F_2$. In class my professor has explained that every inner product induces a norm, $\sqrt{\left < v,v \right>}$ which in turn induces a metric. All of this has seemed pretty obvious to me.

Considering the vector space $V={F_2}^2$, with the standard sum of the product of each coordinate inner product, it seems to me that this is not true (probably because I am not understanding something right). The inner product is a function from $V \times V \rightarrow V$ which means that the under the induced metric, the magnitude of $\left(1,1 \right)^T$ is zero, though it is not the zero vector, contradicting the definition of a metric.

Can somebody please clear up my misunderstanding?

• Did your professor intend for her remarks to be applied to finite fields? Mar 17, 2013 at 23:03
• On $\Bbb F_2^2$ the product is not an inner product, as it is not positive-definite. (Positive definite implies for instance that $\langle x, x\rangle = 0 \Leftrightarrow x = 0$) Mar 17, 2013 at 23:05
• Inner products are not really defined for finite fields. You need some form of order for positivity to work and that immediately requires characteristic $0$.
– EuYu
Mar 17, 2013 at 23:05
• You might find this useful: math.stackexchange.com/questions/185403/… also this: math.stackexchange.com/questions/49348/… Mar 17, 2013 at 23:07
• Inner products need a real or complex base field. Complex conjugacy is right in the definition of an inner product. Mar 17, 2013 at 23:08

An inner product on a real or complex vector space $V$ is a bilinear map $V\times V\to K$ (where $K$ is the ground field, either $\bf R$ or $\bf C$), that satisfies conjugate symmetry and positive-definiteness; for the second property to make sense we have to realize that $\langle x,x\rangle$ is real for all $x\in V$ (this follows from the first property actually), so it makes sense to say it is nonnegative.
The map $\|x\|=\langle x,x\rangle^{1/2}$ will in fact be a vector space norm, and this norm induces a metric via the formula $d(u,v)=\|u-v\|$. The nontrivial part of checking these facts is using Cauchy-Schwarz for establishing the triangle inequality (fix an orthogonal basis to do it in).
It is not possible to define an inner product on a vector space over a field of positive characteristic, by definition. It is, however, possible to define bilinear forms $(\cdot,\cdot):{\bf F}_q^n\times{\bf F}_q^n\to{\bf F}_q$, and two vectors are orthogonal with respect to it if $(a,b)=0$. In most cases of positive characteristic and dimension greater than one it is possible to find an $x$ which is orthogonal to itself under the usual coordinate-determined dot product (this is actually an interesting number-theoretic question: over which finite fields and numbers $n$ do there exist $n$ scalars not all zero whose squares sum to zero?)
It is also not possible to define a metric $X\times X\to {\bf F}$ where $\bf F$ is a field of positive characteristic: by definition in the first place it would have to take real values, but furthermore it could not satisfy the triangle inequality since there can be no ordering in positive characteristic.