Unique decomposition of square matrices theorem My professor mentioned a theorem for decomposing square matrices into a symmetric and anti-symmetric matrix. Paraphrasing,

If $A$ is square, then $A$ has the unique decomposition $A=U+V$, for $U$ symmetric and $V$ anti-symmetric.

But I'm not quite sure how to prove this, because it seems like there should be a formula for $U$ and $V$ respectively if they are unique, but I haven't been able to find one in terms of $A$. Where would I begin to find this?
 A: $A+A^T$ is symmetric and $A-A^T$ is anti-symmetric, which is easily proven by the definitions of those terms. You would get $2A$ if you added these directly, so take $U=\frac{1}{2}(A+A^T)$ and $V=\frac{1}{2}(A-A^T).$
However, after this you still need to prove that they are unique. Suppose $U_1+V_1=U_2+V_2$ with $U_1,U_2$ symmetric and $V_1,V_2$ anti-symmetric. Then, $U_1-U_2=V_2-V_1,$ but the LHS is symmetric whereas the RHS is anti-symmetric, a contradiction unless $U_1=U_2$ and $V_1=V_2,$ proving the solution is unique.
A: $A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$ where $\frac{A + A^T}{2}$ is symmetric and $\frac{A - A^T}{2}$ is antisymmetric. If $A = U + V$ where $U$ is symmetric and $V$ is antisymmetric, then $A^T = U^T + V^T = U - V$. Then $$A + A^T = (U+V) + (U-V) = 2U \Rightarrow U = \frac{A + A^T}{2}$$ and $$A - A^T = (U+V) - (U-V) = 2V \Rightarrow V = \frac{A-A^T}{2},$$
so the decomposition is unique.
A: If you don't know the “formula”, you can deduce it. You trust your teacher, don't you? So, suppose $A=U+V$ with $U$ symmetric and $V$ antisymmetric.
Then you want to use their properties, which have to do with the transpose, so let's do the transpose:
$$
A^T=U^T+V^T
$$
But now we can apply the properties, namely, $U^T=U$ and $V^T=-V$, so we get
$$
A^T=U-V
$$
OK, now we can sum:
$$
A+A^T=U+V+U-V=2U
$$
and we found that $U$ must be $\frac{1}{2}(A+A^T)$. Is this matrix actually symmetric? If it is, you have also proved uniqueness.
How can you get $V$?
