9.10.1 Transposition Matrix as Outer Product

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Transposition Matrix as Outer Product

Rewrite the 9.10 Transposition Matrix as an 9.1 Outer Product of Vectors.

P_{35}, n = 5

Solution. Because this is a transposition matrix, we are swapping the $i$ th and $k t h$ rows. In this case, $i = 3$ and $k = 5$ .

\begin{aligned} P_{35} & = [\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}] - \underset{swap row 3 with 5}{\underset{⏟}{[\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]}} - \underset{swap row 5 with 3}{\underset{⏟}{[\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}] + [\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array}]}} \\ = I_{5} - [\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}] [\begin{array}{c} 0 & 0 & 1 & 0 & 0 \end{array}] + [\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}] [\begin{array}{c} 0 & 0 & 0 & 0 & 1 \end{array}] - [\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}] [\begin{array}{c} 0 & 0 & 0 & 0 & 1 \end{array}] + [\begin{array}{c} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{array}] [\begin{array}{c} 0 & 0 & 1 & 0 & 0 \end{array}] \\ ⟹ P_{35} & = I_{5} - {\vec{e}}_{3} \cdot {\vec{e}}_{3}^{T} + {\vec{e}}_{3} \cdot {\vec{e}}_{5}^{T} - {\vec{e}}_{5} \cdot {\vec{e}}_{5}^{T} + {\vec{e}}_{5} \cdot {\vec{e}}_{3}^{T} \\ ⟹ P_{i k} & = I_{n} - {\vec{e}}_{i} \cdot {\vec{e}}_{i}^{T} - {\vec{e}}_{k} \cdot {\vec{e}}_{k}^{T} + {\vec{e}}_{i} \cdot {\vec{e}}_{k}^{T} + {\vec{e}}_{k} \cdot {\vec{e}}_{i}^{T} \end{aligned}