Find an orthonormal basis for the subspace of $\mathbb R^4$

Question

Find an orthonormal basis for the subspace of $\mathbb{R}^4$ that consists of vectors perpendicular to $u = (1, -1, -1, 1)$.

I know the components of the vector $u$ is $u_1 = 1, u_2 = -1, u_3 = -1, u_4 = 1$.

I managed to find a vector $v$ that is perpendicular to the $u$.

This is done by $u_1v_1+ u_2v_2+ u_3v_3 + u_4v_4 = 0$.

This means $v_1 = -1, v_2 = 1, v_3= 1, v_4 = -1$ because $(1)(-1) + (-1)(1) + (-1)(1) + (1)(-1) = 0$.

So I found a vector $v$ perpendicular to the given vector $u$ which is $(-1, 1, 1, -1)$.

So my question would I be able to call $(-1, 1, 1, -1)$ my basis and perform Gram Schmidt process on it?

Answer 1

A vector $v=(v_1,v_2,v_3,v_4)$ is orthogonal to $u$ iff $v_1-v_2+v_3-v_4=0$, or in other words $v_4=v_1-v_2+v_3$.

Then

$$ v=\left(\begin{matrix} v_1 \\ v_2 \\ v_3 \\ v_1-v_2+v_3 \\ \end{matrix}\right)= v_1\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 1 \\ \end{matrix}\right)+ v_2\left(\begin{matrix} 0 \\ 1 \\ 0 \\ -1 \\ \end{matrix}\right)+ v_3\left(\begin{matrix} 0 \\ 0 \\ 1 \\ 1 \\ \end{matrix}\right) $$

So you can start Gram-Schmidt on those three vectors.

Answer 2

Here's a way without Gram-Schmidt.

Form the matrix $A$ with row $u$. Find one vector $v$ in the nullspace of $A$.

Form the matrix $B$ with rows $u,v$. Find one vector $w$ in the nullspace of $B$.

Form the matrix $C$ with rows $u,v,w$. Find a vector $x$ in the nullspace of $C$.

Now $v,w,x$ form an orthogonal basis of the space of vectors orthogonal to $u$. To make it orthonormal, just divide each by its length.