Exercise [13.23]

Shaun Culver · 29 May 2008, 15:04

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

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

Exercise [13.23]

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