Any square matrix can be expressed as the sum of anti/symmetric matrices

[Eclair_de_XII]

TL;DR Summary

es. Show that any square matrix can be expressed as the sum of a symmetric matrix and an anti-symmetric matrix of the same size.

Important: I would like critique on the style of my proof more than its content. That being said, I am more than open to critique on the latter, as well. In my academic career, I've often been told that my proofs are very long and needlessly excessive. This is the very first proof-based exercise in my linear algebra book, and it took about a page for me to write.

Let $A$ be a matrix of size $(n,n)$. Denote the entry in the i-th row and the j-th column of $A$ by $a_{ij}$, for some $i,j\in\mathbb{N}$. For brevity, we call $a_{ij}$ entry $(i,j)$ of $A$.

Define the matrix $X$ to be of size $(n,n)$, and denote entry $(i,j)$ of $X$ as $x_{ij}$, where $x_{ij}=\frac{1}{2}\left(a_{ij}+a_{ji}\right)$. We note that $x_{ji}=\frac{1}{2}\left(a_{ji}+a_{ij}\right)=\frac{1}{2}\left(a_{ij}+a_{ji}\right)=x_{ij}$. Since matrices are uniquely identified by their entries, we conclude from this string of equalities that $X^T=X$.

Define the matrix $Y$ to be of size $(n,n)$. Denote entry $(i,j)$ of $Y$ as $y_{ij}$, where $y_{ij}=\frac{1}{2}\left(a_{ij}-a_{ji}\right)$. We note that $y_{ji}=\frac{1}{2}\left(a_{ji}-a_{ij}\right)=-\frac{1}{2}\left(a_{ij}-a_{ji}\right)=-y_{ij}$. Since matrices are uniquely identified by their entries, we conclude from this string of equalities that $Y^T=Y$.

\begin{eqnarray}
x_{ij}+y_{ij}&=&\frac{1}{2}\left(a_{ij}+a_{ji}\right)+\frac{1}{2}\left(a_{ij}-a_{ji}\right)\\
&=&\frac{1}{2}a_{ij}+\frac{1}{2}a_{ji}+\frac{1}{2}a_{ij}-\frac{1}{2}a_{ji}\\
&=&\left(\frac{1}{2}a_{ij}+\frac{1}{2}a_{ij}\right)+\left(\frac{1}{2}a_{ji}-\frac{1}{2}a_{ji}\right)\\
&=&a_{ij}+0\\
&=&a_{ij}
\end{eqnarray}

Since matrices are uniquely identified by their entries, we conclude that $A=X+Y$.

 

[hutchphd]

I guess the absolute minimalist version of this is
Note that for a square matrix A $$2\mathbf A= [\mathbf A+{\mathbf A}^t]+ [ \mathbf A-{\mathbf A}^t] $$

 

[Eclair_de_XII]

Wouldn't you need to prove that the sum of any matrix and its transpose is symmetric, and the difference of any matrix and its transpose is anti-symmetric? Or is that just a proposition that is too obvious to prove --- which is to say, it could be proven in just a handful of lines of algebra?

 

[hutchphd]

I would not feel the need but I am not a mathematician. Do you need to define what "square" means?
I understand your angst and it used to drive me crazy..."where do I start??""what do I assume?"
I suggest that you figure out the really hard part first and write it down. Then anything else is is just the decoration.. maybe you need it
I would appreciate other opinions.

 

[fresh_42]

Wouldn't you need to prove that the sum of any matrix and its transpose is symmetric, and the difference of any matrix and its transpose is anti-symmetric? Or is that just a proposition that is too obvious to prove --- which is to say, it could be proven in just a handful of lines of algebra?

It is too obvious to prove. What are $\left(A^\tau\right)^\tau$ and $(A+B)^\tau ?$

 

[Eclair_de_XII]

Oh. So set $B=\pm A^T$.

 

[fresh_42]

Oh. So set $B=A^T$.

Yes. You use linearity of transposition and that it is inverse to itself. If you transpose a product, then you must change the order of the factors, but this isn't used here.

 

[StoneTemplePython]

I guess the absolute minimalist version of this is
Note that for a square matrix A $$2\mathbf A= [\mathbf A+{\mathbf A}^t]+ [ \mathbf A-{\mathbf A}^t] $$

This is how I'd write it, with a footnote saying that in the case of characteristic 2, this isn't meaningful and OP's claim isn't true. (Skew symmetric matrices are a subset of symmetric matrices in characteristic 2, so by a dimension argument it can't be true or just exhibiting a single non-symmetric matrix gives a counterexample.)