Proving A Must Be of Rank 2: The 2x2 Matrix Dilemma

LukasMont
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Homework Statement
4.1 Show that one may express any second rank matrix as the sum of a symmetric
and an antisymmetric matrix.
Relevant Equations
I was able to proof that any matrix could be constructed by adding a symmetric and antisymmetric matrix:

A= A/2 + A/2 + A'/2 - A'/2,
A= (A/2 + A'/2) + (A/2 - A'/2), where A' is the transposed matrix. Now,

A/2 + A'/2 is symmetric, since (A/2 +A'/2)' = A'/2 + A/2 (equal) and
A/2 - A'/2 is antisymmetric, since (A/2 - A'/2)' = - A'/2 + A/2= -(A/2 - A'/2).
My trouble is being to show A must be of rank 2. Any ideas?
 
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You don't have to show that A has rank 2. You are given that as a premise.
I don't know why they give you that as a premise though, because the theorem is true for any matrix, not just any rank-2 matrix, as your proof shows.
Anyway, you have proven what they asked you to.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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