Question

Linear Algebra:
Let R be the matrix that reflects a vector in R^3, in the plane x+y+z=0\[\huge R= \left[\begin{matrix}a & b &c \\ d & e& f \\ g & h& i\end{matrix}\right]\]

Answer 1

\[\huge R= \left[\begin{matrix}a & b &c \\ d & e& f \\ g & h& i\end{matrix}\right]\]

**Side note How do you do latex in the question part?? What would i needed to do in order for R^3 to be\[\huge R^3\]

Answer 2

Was looking at this: http://tutorial.math.lamar.edu/Classes/LinAlg/LinearTransExamples.aspx because i'm quite lost. Usually we would reflect something about an axis or two, IE: the xy plane or the xz plane but that is not specified here in this question.

So i'm struggling to picture what i'm even supposed to do.

Answer 3

answering the side note (not sure I can do the other :P) to retrieve the tex input do
right click_>show math as->tex commands
(you may need to click at the bottom of the screen "show as tab")
\huge R= \left[\begin{matrix}a & b &c \\ d & e& f \\ g & h& i\end{matrix}\right]
then put it in the Q part and enclose with brackets like normal

Answer 4

I messed up in copying the tex commands, but refresh and it should be right

Answer 5

now as far as your Q that is a tricky one, not sure I can answer that...
gotta think about it for a while

Answer 6

the plane is going to have a normal vector of\[\langle1,1,1\rangle\] and so be slicing the xy, xz, and yz planes at an angle of pi/4 I think... (trying to picture the situation)

Answer 7

hard to draw, easier to picture in my mind
http://www.wolframalpha.com/input/?i=x%2By%2Bz%3D0&t=crmtb01

Answer 8

Was thinking of it as we have a plane with a vector pointing like in the first picture but we need to rotate it so the vector is now 'reflected' like in the second picture. but that's all i have.

Answer 9

oh *in* the plane
I thought it said *across* the plane
ok let me think again how I might approach this...

Answer 10

ok, now I am realizing that I do not understand what it means to "reflect a vector *in* a plane"
reflect it about or across what?

Answer 11

Exactly, thats what confuses me.

Answer 12

actually there is no question in your question now that I reread it, so I'm pretty lost...

Answer 13

I might just need to consult my professor for a clarification on what he means.

Answer 14

@Spacelimbus here for a vector problem?

Answer 15

(calling on people who might know what this means, but I think we may have a typo somewhere)

Answer 16

I know vector reflection, but I have only seen such things when they give you an axis, do they want to reflect a vector on an entire plane? Let me see real quick.

Answer 17

that is what we are both confused about, read the wording carefully and see if you can tell us what it means please

Answer 18

I believe we are concerned about reflecting the Vector here.

Answer 19

but as spacelimbus and I are saying, reflecting about what axis/plane ?

Answer 20

it says reflect *in* the plane
I don't know what they mean by that...

Answer 21

I guess it is as you drew it more or less...

Answer 22

Only thing i could think of was a reflection about the origin. :/

Answer 23

I think they may want a linear transformation that will reverse any vectors direction in the x+y+z=0 plane

Answer 24

What I do know about the problem is that none of the answers are in terms of x, y or z. none of the values in the matrix are 0 or 1. So it's all purely numerical.

Answer 25

well it can't be like a rotation all in terms of trig functions I don't think

Answer 26

Forgot that the following was shown at the bottom of the problem.
-------
Hint: There are at least three ways of doing this problem:

1. Remember our discussion of reflection matrices

2. Ask what the reflection does to the unit vectors, and then use the reflection of those vectors as columns of R.

3. Ask what the eigenvectors and eigenvalues are, and then construct R.

Answer 27

hm... I didn't expect eigenvalues to come into it

Answer 28

http://www.geom.uiuc.edu/docs/reference/CRC-formulas/node45.html Interesting, do a search on the page for "reflection in a plane" 3x3 with the situation i'm in.

Answer 29

I just read that @agentc0re, what seems logical to me is that the diagonal in our case would be unitary, so it makes some sense, but I don't know about the remaining coefficients.

Answer 30

a=b=c=1

Answer 31

Thats how i interpreted it. I attempted to enter that in as an answer and it was wrong. My college uses a program called 'webwork' if you are familiar with it for, for our homework. It's nice since i'll know when i have the right answer. :D

Answer 32

*grins* too bad, I am pretty sure about the diagonal, but the remaining entries are a myth to me, also the site says reflection "in" the plane, which I would call a projection, so I am a bit confused there with the terminology.

Answer 33

\[ R =\left[\begin{matrix}-1 & -2 & -2 \\ -2 & -1 & -2 \\-2 & -2 & -1\end{matrix}\right] \]

Answer 34

have you tried this one already?

Answer 35

I had tried the first row before but i just tried entering the whole matrix in, none of the entries are a correct answer.

Answer 36

Out of my depth here. But ( http://en.wikipedia.org/wiki/Reflection_(mathematics)). I would do this by dotting the vector with the normal to the plane, then subtracting that from the vector to find the projection in the plane, then adding back together -1 times the dot product + the projection. But I don't know how to turn that into a matrix..

Answer 37

Also, it looks like @spacelimbus R above has this property:
R N = -5 N
So you're in the right direction for this special case but stretching it.

Answer 38

Can you try \[(\sqrt{3} - 2) / \sqrt{3}\] for the diagonal entries and \[-2\div \sqrt{3}\] for the off diagonal entries. I think this does what you want to the unit vectors..

Answer 39

Sorry, none of those worked for answers. :(

Answer 40

Probably I'm doing something stupid. Still method #2 makes a lot of sense to me. Find the part of the unit vector i that's in the same direction as N and reverse it, and add that back to the other part of i that's in the plane x + y + z = 0. Use that as the first column of R.

Answer 41

And I think you want unit N.

Answer 42

Just got a reply via email from my professor about it and this is what he had to say:

The matrix reflects in the plane x+y+z=0. If a vector is perpendicular to the plane it gets reflected onto its negative
(like in a mirror) if it's in the plane it gets reflected onto itself. The normal to that plane is (1,1,1).

Answer 43

Yes. Exactly what we thought. I'm just having a bit of trouble with the actual calculation. Someone else?

Answer 44

So lets try and break this down again for me. I can give it a try as well, i do understand how to do this it's just that this question has put me in lala land if you know what i mean. :D

1) so we have a perpendicular vector, (1,1,1) to our plane x+y+z=0, correct?
2) We use the Identity matrix in some fashion for the problem?

How did you go about the calculations you did?

Answer 45

Sorry I was away.
Start with i, unit vector in x-direction. We need to find the components of i that are parallel to and N, then subtract to find the part that is perpendicular to N.
Then, the reflection part means to reverse the sign of the part that is parallel to N, pushing it through the plane an equal distance behind it.
Then add them back together for the result.
If you do this for each unit vector in turn and put the results in a matrix, I think it should work. I was just having trouble with magnitude of the projection, I'm a little rusty on linear algebra.

Answer 46

parallel to N, then subtract

Answer 47

The Matix
\[
P=\left(
\begin{array}{ccc}
\frac{2}{3} & -\frac{1}{3} & -\frac{1}{3}
\\
-\frac{1}{3} & \frac{2}{3} & -\frac{1}{3}
\\
-\frac{1}{3} & -\frac{1}{3} & \frac{2}{3}
\\
\end{array}
\right)
\]
is the projection matrix on the given plane.
The reflector matrix is 2P-I
\[
2P -I=\left(
\begin{array}{ccc}
\frac{1}{3} & -\frac{2}{3} & -\frac{2}{3}
\\
-\frac{2}{3} & \frac{1}{3} & -\frac{2}{3}
\\
-\frac{2}{3} & -\frac{2}{3} & \frac{1}{3}
\\
\end{array}
\right)
\]
See http://www.stanford.edu/class/cs205a/homework/hw2_solutions.pdf

Answer 48

thank you so much. I'm going to rework this out myself. Thank you for the reference to the pdf and conformation that it does have something to do with the householder transformation

Answer 49

yw