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?

  • 3
    $\begingroup$ Did your professor intend for her remarks to be applied to finite fields? $\endgroup$ Mar 17, 2013 at 23:03
  • 1
    $\begingroup$ On $\Bbb F_2^2$ the product is not an inner product, as it is not positive-definite. (Positive definite implies for instance that $\langle x, x\rangle = 0 \Leftrightarrow x = 0$) $\endgroup$
    – Arthur
    Mar 17, 2013 at 23:05
  • 7
    $\begingroup$ Inner products are not really defined for finite fields. You need some form of order for positivity to work and that immediately requires characteristic $0$. $\endgroup$
    – EuYu
    Mar 17, 2013 at 23:05
  • $\begingroup$ You might find this useful: math.stackexchange.com/questions/185403/… also this: math.stackexchange.com/questions/49348/… $\endgroup$ Mar 17, 2013 at 23:07
  • $\begingroup$ Inner products need a real or complex base field. Complex conjugacy is right in the definition of an inner product. $\endgroup$ Mar 17, 2013 at 23:08

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.