9.8 Shear Matrix

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Shear Matrix

Let natural number $n \in N$ . For indices $i \neq k$ , we define an $n \times n$ shear matrix as

S_{i k} (c) = I_{n} + c \cdot {\vec{e}}_{i} \cdot {\vec{e}}_{k}^{T}

Remark. Notice how shear matrices are rank one updates to the 8.6 Identity Matrix. Shear matrices are also the identity matrix with entry in row $i$ and column $k$ equal to $c$ .
$⧫$

Shear Matrix

Let's consider $S_{31} (- 5)$ in the $n = 3$ case.

We can interpret this by using our definition from above.

\begin{aligned} S_{31} (- 5) & = I_{3} + (- 5) \cdot {\vec{e}}_{3} \cdot {\vec{e}}_{1}^{T} & Recall shear matrix \\ = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] + (- 5) \cdot {\vec{e}}_{3} \cdot {\vec{e}}_{1}^{T} & Recall identity matrix \\ = {[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]}_{3 \times 3} + (- 5) \cdot {[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}]}_{3 \times 1} {[\begin{array}{c} 1 & 0 & 0 \end{array}]}_{1 \times 3} & Recall rank one update \\ = {[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]}_{3 \times 3} + - 5 \underset{rank one}{{[\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}]}_{3 \times 3}} \\ = {[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]}_{3 \times 3} + {[\begin{array}{c} 0 & 0 & 0 \\ 0 & 0 & 0 \\ - 5 & 0 & 0 \end{array}]}_{3 \times 3} & Scalar-matrix multiplication \\ = {[\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 5 & 0 & 1 \end{array}]}_{3 \times 3} \end{aligned}

& Notice how $- 5$ appears in row $i = 3$ , column $k = 1$ .

We can determine the shear matrix without computation

Consider $S_{34} (7)$ in the $n = 5$ case. Without computing, we will see that

S_{34} (7) = [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix}]

Let's formally compute this to verify our conjecture.

\begin{aligned} S_{34} (7) & = S_{i k} (c) \\ = I_{n} + c \cdot {\vec{e}}_{i} \cdot {\vec{e}}_{k}^{T} \\ = I_{5} + 7 \cdot {\vec{e}}_{3} \cdot {\vec{e}}_{4}^{T} \\ = {[\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}]}_{5 \times 5} + 7 {[\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}]}_{5 \times 1} {[\begin{array}{c} 0 & 0 & 0 & 1 & 0 \end{array}]}_{1 \times 5} \\ = {[\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}]}_{5 \times 5} + 7 {[\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]}_{5 \times 5} \\ = {[\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}]}_{5 \times 5} + {[\begin{array}{c} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 7 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}]}_{5 \times 5} \\ ⟹ S_{34} (7) & = {[\begin{array}{c} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 7 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}]}_{5 \times 5} \end{aligned}

Our conjecture seems to be correct. However, it is important to understand WHY it is correct.