Proving Reflections in R^n are Symmetric Matrices

Click For Summary
The discussion revolves around proving that the matrix of any reflection in R^n is symmetric. It is established that reflections can be represented as F(x) = Ax + b, where A is an orthogonal matrix and b is a vector in R^n. The key argument is that since f(f(x)) = x, it leads to the conclusion that A² = I, implying A = A^t, thus proving symmetry. The conversation clarifies that every reflection in an n-1 dimensional hyperplane through the origin results in a symmetric matrix. The proof is confirmed through a series of logical deductions about the properties of orthogonal matrices and reflections.
Pearce_09
Messages
71
Reaction score
0
done

Hi there,
I am have trouble with a proof. I have some steps done, but I am not sure if I am aproaching this correctly.
the question is:

Show that the matrix of any reflection in R^n is a symmetric matrix.

I know that F(x) = Ax + b is an isometry
where A is an Orthoganol matrix , and vector b is in R^n
this implies that A^t A = I (identity matrix) iff A^-1 = A^t

and if f is a reflection then f(f(x)) = x
then f(f(x)) = A(Ax + b) + b = x
_________
i need to somehow prove that A = A^t which then means the matrix is symmetric

thank you for you time and help
regards,
adam
 
Last edited:
Physics news on Phys.org
Not every reflection in Rn can be represented by a matrix. Now, do you mean that (1) every reflection in an n-1 dimensional hyperplane passing through the origin in Rn is a symmetric matrix, or do you mean (2) the matrix part of every reflection in Rn is symmetric? If f is a reflection in an n-1 dimensional hyperplane L, then f(x) = Ax + b, where A is the matrix corresponding to a reflection in the n-1 dimensional hyperplane passing through the origin, parallel to the hyperplane L, and if d is the shortest vector from 0 to L, then b = 2d. You know that A is orthogonal, so you know that A-1 = AT. It remains to show that A-1 = A. But this is true since:

A(Ax + b) + b = x
A²x + Ab + b = x
(A² - I)x = -(Ab + b)

Now you could use the fact that Ab = -b, so the right side is:

-(Ab + b) = -(-b + b) = -0 = 0

for all x, hence A² - I = 0, hence A² = I, and thus A = A-1. But suppose you didn't know what Ab was. Regardless, you still know that the right side is:

-(Ab + b) for all x

but this is a constant, so you have (A² - I)x = y, where y is a constant vector, and this is true for all x. If you plug in x = 0, then you get 0 = y. In other words, the only linear function that is constant is the 0 function. So without knowing explicitly what Ab is, you can deduce still that the right side must be 0 (and so, if you wanted, could then deduce that Ab = -b), and since the right side is 0 for all x, A² - I = 0, and the desired result follows.
 
Oh wow, that really clears things up. thank you very much akg..
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
12
Views
2K
  • · Replies 0 ·
Replies
0
Views
411
  • · Replies 9 ·
Replies
9
Views
3K