Problem 11.3.2

This will be very short. We are asked for an orthogonal basis for

$$ \mathrm{span} \left\{ \left[ \begin{array}{r} 0\\ 1\\ 1\\ 0 \end{array} \right],\ \left[ \begin{array}{r} 3\\ 0\\ 2\\ -1 \end{array} \right] \right\} $$ This, you can simply ask SymPy for: search for GramSchmidt on that page.

from sympy import *
init_printing(use_latex='mathjax')

GramSchmidt([Matrix([0,1,1,0]), Matrix([3,0,2,-1])])

$\displaystyle \left[ \left[\begin{matrix}0\\1\\1\\0\end{matrix}\right], \ \left[\begin{matrix}3\\-1\\1\\-1\end{matrix}\right]\right]$

Those two vectors are supposed to be orthogonal. Why don't we check, just to make sure:

Matrix([0,1,1,0]).dot(Matrix([3,-1,1,-1]))

$\displaystyle 0$

By the way, you'll notice the two orthogonal vectors by applying GramSchmidt are not orthonormal: their lengths will not be 1, generally. You can fix that by also passing an extra second parameter True to the GramSchmidt function:

GramSchmidt([Matrix([0,1,1,0]), Matrix([3,0,2,-1])], True)

$\displaystyle \left[ \left[\begin{matrix}0\\\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\\0\end{matrix}\right], \ \left[\begin{matrix}\frac{\sqrt{3}}{2}\\- \frac{\sqrt{3}}{6}\\\frac{\sqrt{3}}{6}\\- \frac{\sqrt{3}}{6}\end{matrix}\right]\right]$

This gives us the same vectors as before, but also scaled to length 1.