1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook