Proving Reflections in R^n are Symmetric Matrices

In summary, the conversation discusses the proof that the matrix of any reflection in R^n is a symmetric matrix. The conversation goes into detail about the steps and equations involved in proving this, including the use of orthogonal matrices and the fact that a reflection is represented by the equation f(x) = Ax + b, where A is the matrix corresponding to a reflection in a hyperplane passing through the origin. The conversation concludes by explaining how the desired result can be deduced even without knowing the explicit value of Ab.
  • #1
Pearce_09
74
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
 
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  • #2
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.
 
  • #3
Oh wow, that really clears things up. thank you very much akg..
 

1. What are reflections in R^n?

Reflections in R^n are transformations that flip an object across a line, plane, or hyperplane in n-dimensional space. These transformations preserve the distance between points and are often used in geometry and linear algebra.

2. How can reflections be represented as matrices?

In R^n, reflections can be represented as n x n matrices known as reflection matrices. These matrices have a special property where they are equal to their own transpose, making them symmetric.

3. What is a symmetric matrix?

A symmetric matrix is a square matrix where the elements above and below the diagonal are reflections of each other. This means that the matrix is equal to its own transpose, and the order of the elements does not matter.

4. How can we prove that reflections in R^n are symmetric matrices?

We can prove this by using the definition of a reflection matrix and the properties of symmetric matrices. Since a reflection matrix is equal to its transpose, it satisfies the definition of a symmetric matrix.

5. Why is proving that reflections in R^n are symmetric matrices important?

Proving this enables us to understand the properties of reflections and their relationship to symmetric matrices. This knowledge is essential in various fields, including geometry, linear algebra, and computer graphics.

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