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

MATH307 December 2012

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Question 03 (e)
Let $S=\left\{[x_{1},x_{2},x_{3}]^{T}:\,x_{1}+x_{2}+x_{3}=0\right\}$ be the subspace of vectors in $\mathbb {R} ^{3}$ whose components sum to zero. (e) Let Q = I - P. What kind of matrix is Q? What are N(Q) and R(Q)?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Use that P2 = P to show that Q2 = Q, that is, Q is also a projection.

Hint 2
In fact, Q is called the complementary projection of P. As such Q swaps the range and nullspace of P (why?).

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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) = ST. 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 $\displaystyle R(Q)=S^{T}=\mathrm {span} \left\{{\begin{bmatrix}1\\1\\1\end{bmatrix}}\right\}$

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