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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.
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