By Gram-Schmidt process, we can get an orthogonal basis for such a $\mathbb{Q}-$vector space $V$. However, if we normalize each vector, the corresponding scalar may not be a rational number but a root of a rational problem. Is this a problem? Is there a way to get an orthonormal basis for $V$?
Can we get an orthonormal basis for a $\mathbb{Q}$-vector space
linear algebra
Best Answer
Consider $V:=\Bbb Q[i\pi]\subset \Bbb C$ with scalar product $\langle z,w\rangle :=\overline zw$. If $a+ib\pi$ (with $a,b\in\Bbb Q$) has length $1$, then $a^2+b^2\pi^2=1$. As $\pi $ is trancendental, we must have $b^2=0$, and then $a=\pm1$. It follows that $V$ does not admit even a normal basis (i.e., consisting of unit vectors that need not be orthogonal).
(You can replace $\pi$ with$\sqrt[3]2$.)