How would I be able to find an orthogonal basis for the null space of A.
Where A is a 3×3 matrix with rank 1 and $(2,1,2)^{T}$ belongs to the Row space of A.
According to my teacher a possible answer is $Nul(A) = \{ (-2,2,1)^{T}, (1, 2, -2)^{T} \} $ . I do not know how one could reach that answer.
My intuition was thinking of $(2,1,2)^{T}$ as the orthogonal vector of a surface, such that
$2x + y + 2z = 0$, which (after solving for y) you get that it is generated by the vectors: $\{ (1,-2,0)^{T}, (0,-2,1)^{T} \}$
However those aren't orthogonal with each other (but they are orthogonal with $(2,1,2)^{T}$. But you could do the dot product of one of them against $(2,1,2)^{T}$ in order to find another orthogonal vector (which also belongs to the nullspace).
So if you compute: $(2,1,2)^{T} \times (1,-2,0)^{T}$ you get $(4,2,-5)^{T}$
Therefore an orthogonal basis for the Nullspace of A would be: $\{ (1,-2,0)^{T}, (4,2,-5)^{T} \}$
Is my reasoning correct? Are both my teacher's and my answer valid? If my answer is valid, how could I find the teacher's answer?
Best Answer
Your reasoning is correct. Both answers are valid. As you may know, every real vector space (of nonzero dimension) has infinitely many bases, even infinitely many orthogonal bases. How your teacher came up with that particular basis is anybody's guess. Only your teacher knows for sure.