Science:Math Exam Resources/Courses/MATH307/April 2005/Question 03 (b)

MATH307 April 2005

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Question 03 (b)
Let P be a real symmetric n x n matrix such that P2 = P. Also let R = In - 2P be the matrix that reflects a vector across a plane. (b) If P is the matrix of the projection of $\mathbb {R} ^{n}$ onto a subspace V and if Q is the matrix of the projection of $\mathbb {R} ^{n}$ onto the orthogonal complement $V^{\perp }$, explain (either algebraically or geometrically) why PQ = 0 and RQ = 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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

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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. Why is $PQ=0$ $Px$ is in the subspace$V$ and $Qx$ will be in the subspace orthogonal to $V$. So it is easy to imagine that if $y$ is in $V$ then $Qy$ will be the zero vector meaning $Qy=0$. To get a vector in $V$ from any vector in$R^{n}$ we simply use $y=Px$. Substituting this into the first equation we get $Q(Px)=QPx=0$ and because x does not have to be a zero vector $QP=0$. The same logic is true for the opposite case so $PQ=0$. To see RQ = Q we can solve algebraically $RQ=(I-2P)Q=Q-2PQ$ but as seen above $PQ=0$ so $Q-2PQ=Q$