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

MATH307 December 2012

Work in progress: this question page is incomplete, there might be mistakes in the material you are seeing here.

• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q3 (d) • Q3 (e) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q4 (e) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q6 (a) • Q6 (b) • Q6 (c) • Q6 (d) • Q6 (e) • Q6 (f) •

Other MATH307 Exams

• December 2012 • April 2013 • April 2012 • April 2006 • December 2008 • December 2010 •

[hide] 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.

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

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

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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