# Prove the theorem for the matrix

1. Sep 10, 2008

### frostshoxx

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 = $$\frac{X+X^{T}}{2}$$
B = $$\frac{X-X^{T}}{2}$$
X = $$\frac{X+X^{T}}{2}$$ + $$\frac{X-X^{T}}{2}$$

3. The attempt at a solution

So I tried to solve $$\frac{X+X^{T}}{2}$$ + $$\frac{X-X^{T}}{2}$$ 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. Sep 10, 2008

### Focus

Take the transpose of A and B. You also need to prove uniqueness which I would do by contradiction.

3. Sep 10, 2008

### frostshoxx

Can this be done symbolically? Also, what do you mean by contradiction? could you give some examples?

4. Sep 10, 2008

### Focus

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$$a \cdot a^{-1}=e$$ and $$a\cdot e=a \quad \forall a \in \mathbb{R}$$. e is unique.

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

Hope that helps