Science:Math Exam Resources/Courses/MATH307/December 2012/Question 03 (e)/Solution 1

First, we show that Q is a projection, that is Q² = Q:

$\displaystyle Q^{2}=(I-P)(I-P)=I-P-P+P^{2}=I-P-P+P=I-P=Q$

since P² = P because P is a projection.

Next we show that N(Q) = R(P). So let x be in the nullspace of Q. Then 0 = Qx = (I-P)x = x-Px and hence Px = x which implies in particular that x is in the range of P. Next, let x be in the range of P, and choose y such that x = Py. Multiplying both sides with P yields Px = P²y = Py = x, which we can rewrite as (I-P)x = 0 and thus x is in the nullspace of Q. Therefore N(Q) = R(P). But P was chosen such that R(P) = S and hence

$\displaystyle N(Q)=R(P)=S$

Interchanging the roles of P and Q, after all Q = I-P implies that P = I-Q, we obtain that R(Q) = N(P). We claim that N(P) = S^T. This is quick since Px = 0 is equivalent to saying that x has no component in S which is equivalent to saying that x is orthogonal to S. From part (a) we remember that the vector [1, 1, 1]^T spans the orthogonal complement of S and hence

$\displaystyle R(Q)=S^{T}=\mathrm {span} \left\{{\begin{bmatrix}1\\1\\1\end{bmatrix}}\right\}$