Question: Let $A$ be any square matrix
(a) Show that $A+A^T$ is symmetric and $A-A^T$ is skew-symmetric.
(b)Prove that there is one and only one way to write A as the sum of a symmetric matrix and a skew-symmetric matrix
For part(a), my ans:
since $(A+A^T)^T=A^T+(A^T)^T=(A^T+A)$ , thus $A+A^T$ is symmetric
since $(A-A^T)^T=A^T-(A^T)^T=A^T-A=-(A-A^T)$, thus $(A-A^T)$ is skew symmetric
For part (b),
Assume there is more than one way to write A as the sum of symmetric and skew_symmetric matrix
Let $B_1$ and $B_2$ be symmetric matrix , $C_1$ and $C_2$ be skew-symmetric matrix, such that
$$A=B_1+C_1$$
$$A=B_2+C_2$$
the difference of two-equation is $$C_2-C_1=B_1-B_2$$
Note that $C_2-C_1$ is a skew-symmetric matrix and $B_1-B_2$ is a symmetric.
The contradiction occurs since there does not exist matrix that are both skew-symmetric and symmetric [zero matrix is an exceptional case since it implies A=B+C is the one and only one solution. ]
Therefore, there is one and only one way to write A as sum of symmetric and skew-symmetric.
Is it a correct proof?
To satisfy the sticklers, you'll need to use or prove the fact that a square matrix is symmetric and skew-symmetric iff it's a zero matrix, and to show there's exactly one suitable decomposition you must provide it as $A=\frac{A+A^T}{2}+\frac{A-A^T}{2}$.
Another strategy that doesn't require a proof by contradiction is to note symmetric $B$ and skew-symmetric $C$ satisfy $A=B+C$ iff $A=B+C$ and $A^T=B-C$, i.e. iff $B=\frac{A+A^T}{2}$ and $C=\frac{A-A^T}{2}$. This has the advantage that you prove both parts (at most one, at least one) in a single step.