- #1

Eclair_de_XII

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- 91

- 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##.

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##.