Question

By explicit construction of the matrices in question, show that any matrix \(\M T\), can be written as:

a) the sum of a symmetric matrix \(\M S\) and antisymmetric matrix \(\M A\);
\[\boxed{
\newcommand \M [1] {\mathbf{#1}}
\newcommand\m[1]{\begin{bmatrix}#1\end{bmatrix}}}\]

Answer 1

\[\M S = \m{s_{11}&s_{12}&s_{13}\\
s_{12}&s_{22}&s_{23}\\
s_{13}&s_{23}&s_{33}}\]
\[\M A = \m{0& a_{12}&a_{13}\\
-a_{12}& 0&a_{23}\\
-a_{13}&-a_{23}&0}\]

Answer 2

\[\M T = \M S + \M A
= \m{s_{11}&s_{12}+a_{12}&s_{13}+a_{13}\\
s_{12}-a_{12}&s_{22}&s_{23}+a_{23}\\
s_{13}-a_{13}&s_{23}-a_{23}&s_{33}}
= \m{t_{11}&t_{12}&t_{13}\\
t_{21}&t_{22}&t_{23}\\
t_{31}&t_{32}&t_{33}}\]

Answer 3

\[
t_{ij} = \begin{cases}
s_{ij} & i=j\\
s_{ij}+a_{ij}&i< j\\
s_{ij}-a_{ij}&i> j\\
\end{cases}\]

Answer 4

\[s_{ij}=\frac{t_{ij}+t_{ji}}2\]
\[a_{ij}=\frac{t_{ij}-t_{ji}}2\]

Answer 5

is this right? does it make sense?

Answer 6

@hartnn

Answer 7

yes, thats correct!
Took me a while since we used different terminology when we did it.

Answer 8

\[s_{ij}=\frac{t_{ij}+t_{ji}}2=s_{ji}\]
\[a_{ij}=\frac{t_{ij}-t_{ji}}2=-a_{ji}\]

Answer 9

\(t_{ij} = s_{ij} +a_{ij} \)

fir every i,j

since \(a_{ij} = - a_{ij}\)

Answer 10

Maybe this is clearer:

\begin{align*}
\M T
&= \M S + \M A\\
&= \m{s_{11}&s_{12}&s_{13}\\
s_{12}&s_{22}&s_{23}\\
s_{13}&s_{23}&s_{33}}
+ \m{0& a_{12}&a_{13}\\
-a_{12}& 0&a_{23}\\
-a_{13}&-a_{23}&0}\\
&= \m{s_{11}&s_{12}+a_{12}&s_{13}+a_{13}\\
s_{12}-a_{12}&s_{22}&s_{23}+a_{23}\\
s_{13}-a_{13}&s_{23}-a_{23}&s_{33}}\\[2ex]
\tilde{\M T}
&= \m{s_{11}&s_{12}-a_{12}&s_{13}-a_{13}\\
s_{12}+a_{12}&s_{22}&s_{23}-a_{23}\\
s_{13}+a_{13}&s_{23}+a_{23}&s_{33}}\\[2ex]
\M S
&= \frac{\M T + \tilde{\M T}}2\\[2ex]
\M A
&= \frac{\M T - \tilde{\M T}}2\\
\end{align*}

Answer 11

yes thats correct,
and T = S + A

the point was that you can generalize that for all i,j
we need not have 3 different break-ups for i =j, i<j, i >j

Answer 12

and t = s + a