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Prove the theorem for the matrix

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that every square real matrix X can be written in a unique way as the sum of a symmetric matrix A and a skew-symmetric matrix B.

    2. Relevant equations

    X = A + B
    A = [tex]\frac{X+X^{T}}{2}[/tex]
    B = [tex]\frac{X-X^{T}}{2}[/tex]
    X = [tex]\frac{X+X^{T}}{2}[/tex] + [tex]\frac{X-X^{T}}{2}[/tex]


    3. The attempt at a solution

    So I tried to solve [tex]\frac{X+X^{T}}{2}[/tex] + [tex]\frac{X-X^{T}}{2}[/tex] and it gives out X as a solution. However, how can I know that A is a symmetric and B is a skew-symmetric? Any idea?
     
  2. jcsd
  3. Sep 10, 2008 #2
    Take the transpose of A and B. You also need to prove uniqueness which I would do by contradiction.
     
  4. Sep 10, 2008 #3
    Can this be done symbolically? Also, what do you mean by contradiction? could you give some examples?

    Thank you for your time.
     
  5. Sep 10, 2008 #4
    Yes why not, if you take transpose of A, you will get A again. And B is skew because of the negative sign.

    Example of uniqueness. Let e be a number (in reals) such that[tex]a \cdot a^{-1}=e[/tex] and [tex]a\cdot e=a \quad \forall a \in \mathbb{R}[/tex]. e is unique.

    Proof:
    Fix a in reals and assume e is not unique. You have [tex] a\cdot e=a[/tex] and [tex] a\cdot e'=a[/tex] for [tex]e\neq e'[/tex] (same for inverses). Now you have
    [tex] a\cdot e \cdot e'=a \cdot e'=a[/tex]
    taking inverses gives the result that [tex] e \cdot e'=e[/tex] and [tex] e \cdot e'=e'[/tex]
    thus [tex]e=e'[/tex] which contradicts the assumption, thus e must be unique.

    Hope that helps
     
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