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 Exercise [13.23] 
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Post Exercise [13.23]
The following solution only takes (3\times 3) matrices into consideration. I intend to generalize this, in time, to (m\times n) matrices, if necessary.

T^a{}_b=\left(
\begin{array}{ccc}
 T^1{}_1 & T^1{}_2 & T^1{}_3 \\
 T^2{}_1 & T^2{}_2 & T^2{}_3 \\
 T^3{}_1 & T^3{}_2 & T^3{}_3
\end{array}
\right)

S^a{}_b=\left(
\begin{array}{ccc}
 S^1{}_1 & S^1{}_2 & S^1{}_3 \\
 S^2{}_1 & S^2{}_2 & S^2{}_3 \\
 S^3{}_1 & S^3{}_2 & S^3{}_3
\end{array}
\right)

Addition of (3\times 3) matrices is defined by

R^a{}_b=T^a{}_b+S^a{}_b=\left(
\begin{array}{ccc}
 \left[T^1{}_1+S^1{}_1\right] & \left[T^1{}_2+S^1{}_2\right] & \left[T^1{}_3+S^1{}_3\right] \\
 \left[T^2{}_1+S^2{}_1\right] & \left[T^2{}_2+S^2{}_2\right] & \left[T^2{}_3+S^2{}_3\right] \\
 \left[T^3{}_1+S^3{}_1\right] & \left[T^3{}_2+S^3{}_2\right] & \left[T^3{}_3+S^3{}_3\right]
\end{array}
\right)

The trace is defined as \text{trace}\,\, R^a{}_b=R^a{}_a=T^1{}_1+T^2{}_2+\text{...}+T^n{}_n . The trace can be thought of as the sum of the diagonal elements.

\text{trace}\,\, R^a{}_b=\left(T^1{}_1+S^1{}_1\right)+\left(T^2{}_2+S^2{}_2\right)+\left(T^3{}_3+S^3{}_3\right)
\text{trace}\,\, R^a{}_b=\left(T^1{}_1+T^2{}_2+T^3{}_3\right)+\left(S^1{}_1+S^2{}_2+S^3{}_3\right)
\text{trace}\,\, T^a{}_b=\left(T^1{}_1+T^2{}_2+T^3{}_3\right)
\text{trace}\,\, S^a{}_b=\left(S^1{}_1+S^2{}_2+S^3{}_3\right)

Therefore

\text{trace}\,\, R^a{}_b=\text{trace}\,\, T^a{}_b+\text{trace}\,\, S^a{}_b

_________________
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29 May 2008, 15:04
   [ 1 post ]