Question is out of Strang's Intro Linear Algebra:
(a) Find orthonormal vectors $ q_1, q_2, q_3 $ such that $ q_1, q_2 $ span the column space of:
$$
A=\left[\begin{array}{rr}1 & 1 \\ 2 & -1 \\ -2 & 4\end{array}\right]
$$
(b) Which of the four fundamental subspaces contains $ q_3$?
In finding $ q_1, q_2 $ we solve the problem through Gram-Schmidt. I set $q_1$ as is, except for normalizing and dividing by length, and set that first column of A to $q_1$. To find $q_2$, I subtract from the column 2, its own projection onto column 1. After normalizing I get:
$$
\boldsymbol{q}_1 = \frac{1}{3}(1,2,-2) \text { and } \boldsymbol{q}_{2}=\frac{1}{3}(2,1,2)
$$
I'm unclear how to generally find $ q_3$. Is there a formula or method for this? The answer is:
$$
\boldsymbol{q}_{3}=\frac{1}{3}(2,-2,-1)
$$
Beyond that, the answer to (b) is that $q_3$ lies in the nullspace of A transpose. I do not understand this. Isn't $q_3$ supposed to be in the column space of A?
Best Answer
Good, you have already found $q_1, q_2$ by Gram-Schmidt.
Now, we want $q_3$ to be a unit vector orthogonal to $q_1$ and $q_2$. For some vector $q_3 = (x,y,z)$ to be orthogonal to $q_1$ and $q_2$ means $(x,y,z) \cdot q_1 = 0$ and $(x,y,z) \cdot q_2 = 0$. This determines two equations in the three variables $x, y, z$. In fact, it is a homogeneous system, and hence always has a solution (in fact, infinitely many), which can be found by row-reduction. Then, divide by the length to get a unit vector. I leave all this to you to do.
$q_3$ is not in the column space of $A$: The column space of $A$ equals the span of $q_1$ and $q_2$ (recall that Gram-Schmidt does not change the span of the original vectors), and $q_3$ is linearly independent from them (recall that nonzero orthogonal vectors are linearly independent). As you may know, the nullspace of $A^T$ is the orthogonal complement of the column space of $A$. Thus $q_3$ is in the nullspace of $A^T$.