9.9 Dilation Matrix

Properties: N/A
Sufficiencies: N/A
Questions: N/A

Dilation Matrix

Let natural number $n \in N$ and scalar $c \in R$ . For $j \in {1, 2, \dots, n}$ , we define an $n \times n$ dilation matrix

D_{j} (c) = I_{n} + (c - 1) \cdot {\vec{e}}_{j} \cdot {\vec{e}}_{j}^{T}

Remark. Similar to the 9.8 Shear Matrix, the dilation matrix also uses the 9.7 Rank-one Update.
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Dilation Matrix

Let's consider $D_{2} (- 3)$ in the $n = 4$ case.

We can interpret this by using our definition from above.

\begin{aligned} D_{2} (- 3) & = I_{4} + (- 3 - 1) \cdot {\vec{e}}_{2} \cdot {\vec{e}}_{2}^{T} \\ = {[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]}_{4 \times 4} + (- 3 - 1) {[\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array}]}_{4 \times 1} {[\begin{array}{c} 0 & 1 & 0 & 0 \end{array}]}_{1 \times 4} \\ = {[\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}]}_{4 \times 4} + (- 3 - 1) [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \end{aligned}

Notice how we kept $- 3$ and $- 1$ separate. Using distributivity of scalar addition, we see

\begin{aligned} ⟹ D_{2} (- 3) & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] + - 3 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] + - 1 [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \\ = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] + \underset{Rearranged ussing commutativity}{\underset{⏟}{[\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] + [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}]}} \end{aligned}

Let's add the first two matrices.

\begin{aligned} D_{2} (- 3) & = [\begin{array}{c} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}] + [\begin{array}{c} 0 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}] \end{aligned}

& Look at the boxed entry on the first matrix. Notice how we are "replacing" this spot with the number $c$ . Let's finish our computation.

D_{2} (- 3) = [\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & - 3 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]

We can determine the dilation matrix without computation

With our newfound intuition, the dilation matrix $D_{1} (5)$ at $n = 6$ will be

D_{1} (5) = [\begin{matrix} 5 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix}]

Let's formally compute this.p