Question

linear Algebra:
Recall that P is a projection matrix if and only if P^T=P and P^2=P. Also, recall that R is a reflection matrix if and only if R^T=R and R^2=I. Finally, recall that A is an orthogonal matrix if and only if AA^T=I=A^TA.
Let P be a projection matrix. Determine whether each of the following statements is true or false.
1. If is an eigenvalue for P, then =0 or =1.

2. −1 is an eigenvalue for P.

3. det(P)=0 or det(P)=1.

4. −P is a projection matrix.

5. P^2=I

6. P is a reflection matrix.

Answer 1

@joemath314159

Answer 2

any ideas?

Answer 3

I see this was posted a while ago. My bad, im in the middle of packing >.<

Answer 4

I am sorry about that I think 1 and 3

Answer 5

@joemath314159

Answer 6

You are correct. those are the only ones that are true.

Answer 7

thanks a lot

Answer 8

what about -P isnt it a projection?

Answer 9

@eliassaab do u have anything to add?

Answer 10

You can check if -P satisfies the condition:\[P^2 = P\]or not.

Answer 11

well it does -P=(-P)^2

Answer 12

\[(-P)^2 = (-P)(-P) = (-1)(-1)(P^2) = P^2 =P\ne -P\]

Answer 13

oh I see so the right answers are 1 and 3

Answer 14

yep

Answer 15

I agree 1 and 3

Answer 16

Let x be an eigenvector of a a projection matrix, and c its eigenvalue. Then:\[Px = cx\]Applying P one more time gives:\[P(Px)=P(cx)\Longrightarrow P^2x=c(Px)\]but P^2=P, and Px = cx, so we obtain:

Answer 17

\[P^2x=c(Px)\Longrightarrow Px=c(cx)\Longrightarrow cx=c^2x\Longrightarrow c=c^2\]which only has the solutions 0 or 1.

Answer 18

For 3, note that the det(P) is the product of its eigenvalues, So it will be a product of only 1's, only 0s, or 1's and 0s. So the det will either be 1 or 0.

Answer 19

ok thanks a lot for the explanation, would mind if u take a look at my new post
http://openstudy.com/study#/updates/4fdbf607e4b0f2662fd226b2

Answer 20

You can also say if \( \lambda\) is an eigenvalue and X is its eigenvector, then
\[
PX= \lambda X\\
P^2 X = \lambda^2 X=PX=\lambda X\\
\lambda = \lambda^2 \\
\lambda = 0 \text { or } \lambda =1
\]