Question

Recall that P is a projection matrix if and only if PT=P and P2=P. Also, recall that R is a reflection matrix if and only if RT=R and R2=I. Finally, recall that A is an orthogonal matrix if and only if AAT=I=ATA.
Let R be a reflection matrix. Determine whether each of the following statements is true or false.
1. If R is invertible, then R=I.

2. I−R is a projection matrix.

3. R−I is a projection matrix.

4. det(R)=1 or det(R)=−1.

5. If S is also a projection matrix, then so is R+S.

6. 1 is an eigenvalue for R.

Answer 1

any ideas?

Answer 2

Can you clarify what P2=P means? Also, what R2=I means? By these, do you mean:
\(P^2=P\) and \(R^2=I\)?

Answer 3

yup

Answer 4

Well, look at \(1\). Can you derive that from your given statements? If so, it's true. If not, it's false.

Answer 5

it is false

Answer 6

Correct. Do you know why?

Answer 7

cus we can find many reflection matrices and they are not the identity

Answer 8

That's a great way of looking at it. Mine was that if we have \(R^2=I\), then multiplying by the inverse (we can, since \(R\) is invertible), we have \(R=R^{-1}\).

Question: Does \(P^{T}\) represent the transpose of \(P\)?

Answer 9

Rephrase that: Are you talking about \(P^{T}\) when you say PT?

Answer 10

yes

Answer 11

second one is also false cus (R-I)^2 is not equal to (R-I)

Answer 12

I'm not so certain. I'm still thinking on that one and the other two.

Answer 13

Oops, the others, I should say. :p

Answer 14

\[(I-R) \text{ is a projection matrix} \Rightarrow (I-R)^{T}=I-R\]
Is the second part true? You can try to figure it out for yourself by working out general matrices.

Answer 15

2 is false because the send condition of projection matrices says P^2=P

Answer 16

\[
\begin{bmatrix}
1 & 0 & \dots & 0\\
0 & 1 & \dots &0\\
\vdots & \vdots & \ddots &\vdots\\
0 & 0 &\dots & 1
\end{bmatrix}
-
\begin{bmatrix}
x_{11} & x_{21} & \dots & x_{n1}\\
x_{12} & x_{22} & \dots & x_{n2}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1n} & x_{2n} & \dots & x_{nn}
\end{bmatrix}
=
\begin{bmatrix}
1-x_{11} & x_{21} & \dots & x_{n1}\\
x_{12} & 1-x_{22} & \dots & x_{n2}\\
\vdots & \vdots & \ddots & \vdots\\
x_{1n} & x_{2n} & \dots & 1-x_{nn}
\end{bmatrix}
\]

I'm not seeing how this could be equal to the transpose, so I see 2 as false, too.

Answer 17

Hmm.. Wait...

Answer 18

Okay. #2 is true. Do you know why?

Answer 19

nope

Answer 20

Start working out what \(I-R\) is. Then create the transpose. Employing the properties of \(R\), you can thus show that \(I-R=(I-R)^{T}\). Give it a shot.

Answer 21

however to complete that u should prove that (I-R)^2=(I-R)

Answer 22

Hmm, good point.

Answer 23

Yeah, I had forgot that condition. So I suppose it is false. I cannot see any way in which \((I-R)^2=(I-R)\).

Answer 24

I think 3 is also false for nearly the same reasons. What do you think?

Answer 25

yes it is false

Answer 26

@eliassaab