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

i think the right answer is 4 and 6

Answer 2

yes i believe that is correct.

Answer 3

ok thanks a million

Answer 4

thats not true there is something wrong

Answer 5

hmmm... let me think about it.

Answer 6

1 is false for sure. There are plenty of reflections such that R is not I and R is invertible.

Answer 7

2 is false because:\[(I-R)^2=I-2R+R^2=2(I-R)\ne I-R\]Same goes for 3.

Answer 8

4 is true since:\[R^2=I\Longrightarrow \det(R^2)=\det(I)\Longrightarrow \det(R)^2=1\Longrightarrow \det(R)=\pm 1\]

Answer 9

5 is false since:\[(R+S)^2=R^2+RS+SR+S^2=I+RS+SR+S\ne R+S\]

Answer 10

6 is the only one im iffy about. 1 and -1 can both be eigenvalues.

Answer 11

it says *an* so it is also true

Answer 12

yeah thats what i thought too.

Answer 13

well I have no idea what is wrong with the system

Answer 14

how many times can you try the problem?

Answer 15

2 and i did one

Answer 16

ugh

Answer 17

well do u have any idea about this question
Consider the sixth roots of unity and choose ALL correct responses

A. The argument of any sixth root of unity is a multiple of 3
B. If z is a sixth root of unity, then z2 is also a sixth root of unity
C. If z is a sixth root of unity, then z is also a 12th root of unity
D. If z is a sixth root of unity, then z is a cube root of unity
E. If z is a sixth root of unity, then −z is also a sixth root of unity

Answer 18

Not sure about a
b is true since if z^6=1, then (z^2)^6=(z^6)^2=1^2=1.
c is true for the same reason as b-ish. if z^6=1 then z^12=(z^6)^2=1^2=1
d is false. \[\frac{1}{2}+i\frac{\sqrt{3}}{2}\]is a sixth root of unity, but is not a cube root of unity.
e is true. since if z^6=1, then (-z)^6=(-1)^6z^6=1.

Answer 19

1 is pi/3

Answer 20

if we are talking in degrees (not radians), then the argument of a the sixth roots of unity will be:\[n\cdot \frac{360}{6}=60n,n=1,2,3,4,5,6\]i guess that makes a true? not sure, i dont really understand what they mean by arguement being a multiple of 3, since normally i think of the argument in radians.

Answer 21

it is not 3 it is pi/3

Answer 22

oh oh oh oh oh. then yes its true, since in radians, the arguments of the sixth roots of unity are:\[n\cdot \frac{2\pi}{6}=n\cdot \frac{\pi}{3}\]

Answer 23

I really appreciate ur help

Answer 24

no prob :)

Answer 25

u deserve the green color

Answer 26

i got demoted a while ago! i just got it back lol. apparently if you do nothing for a couple of days you level down >.> guess i'll be lvl 80 when i come back from my trip lol >.<

Answer 27

well have a wonderful trip