UnkleRhaukus
  • UnkleRhaukus
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 \(\M A\); \[\boxed{ \newcommand \M [1] {\mathbf{#1}} \newcommand\m[1]{\begin{bmatrix}#1\end{bmatrix}}}\]
Linear Algebra
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SOLVED
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chestercat
  • chestercat
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UnkleRhaukus
  • UnkleRhaukus
\[\M S = \m{s_{11}&s_{12}&s_{13}\\ s_{12}&s_{22}&s_{23}\\ s_{13}&s_{23}&s_{33}}\] \[\M A = \m{0& a_{12}&a_{13}\\ -a_{12}& 0&a_{23}\\ -a_{13}&-a_{23}&0}\]
UnkleRhaukus
  • UnkleRhaukus
\[\M T = \M S + \M A = \m{s_{11}&s_{12}+a_{12}&s_{13}+a_{13}\\ s_{12}-a_{12}&s_{22}&s_{23}+a_{23}\\ s_{13}-a_{13}&s_{23}-a_{23}&s_{33}} = \m{t_{11}&t_{12}&t_{13}\\ t_{21}&t_{22}&t_{23}\\ t_{31}&t_{32}&t_{33}}\]
UnkleRhaukus
  • UnkleRhaukus
\[ t_{ij} = \begin{cases} s_{ij} & i=j\\ s_{ij}+a_{ij}&i< j\\ s_{ij}-a_{ij}&i> j\\ \end{cases}\]

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More answers

UnkleRhaukus
  • UnkleRhaukus
\[s_{ij}=\frac{t_{ij}+t_{ji}}2\] \[a_{ij}=\frac{t_{ij}-t_{ji}}2\]
UnkleRhaukus
  • UnkleRhaukus
is this right? does it make sense?
UnkleRhaukus
  • UnkleRhaukus
@hartnn
hartnn
  • hartnn
yes, thats correct! Took me a while since we used different terminology when we did it.
ganeshie8
  • ganeshie8
\[s_{ij}=\frac{t_{ij}+t_{ji}}2=s_{ji}\] \[a_{ij}=\frac{t_{ij}-t_{ji}}2=-a_{ji}\]
hartnn
  • hartnn
\(t_{ij} = s_{ij} +a_{ij} \) fir every i,j since \(a_{ij} = - a_{ij}\)
UnkleRhaukus
  • UnkleRhaukus
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*}
hartnn
  • hartnn
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, ij
anonymous
  • anonymous
and t = s + a

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