13.2 Inverses of Elementary Matrices

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: 13.1 Inverse of a Square Matrix

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

Inverses of Elementary Matrices

Let $i$ , $k$ , $n \in N$ such that $1 \leq i, k \leq n$ with $i \neq k$ and let $c \in R$ with $c \neq 0$ . Then, the inverses of each of the elementary matrices are given below:

Shear matrices: ${(S_{i k} (c))}^{- 1} = S_{i k} (- c)$
Transposition matrices: ${(P_{i k})}^{- 1} = P_{i k}^{T}$
Dilation matrices: ${(D_{i} (c))}^{- 1} = D_{i} (\frac{1}{c})$

\begin{proof}[Proof 1.]
!1000
\end{proof}

Dilation Matrix

$[D_{2} (400)]_{3 \times 3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 400 & 0 \\ 0 & 0 & 1 \end{matrix}]$

Can we guess what the inverse of $D_{2}$ looks like?

[D_{2} (400)]^{- 1} = [\begin{matrix} 1 & 0 & 0 \\ 0 & \frac{1}{400} & 0 \\ 0 & 0 & 1 \end{matrix}] = D_{2} (\frac{1}{400})

Let's make a conjecture.

Conjecture

Let $n \in N$ and $j \in {n}$ with nonzero $c \in R$ . Then

[D_{j} (c)]^{- 1} = D_{j} (c^{- 1}) = D_{j} (\frac{1}{c})

\begin{proof}[Proof 2.]
Let $n \in N$ , $c \in R$ with $c \neq 0$ and $j \in {n}$ . We want to show that

[D_{j} (c)]^{- 1} = D_{j} (c^{- 1}) = D_{j} (\frac{1}{c})

Consider

\begin{aligned} D_{j} (c) \cdot D_{j} (\frac{1}{c}) & = (I_{n} + (c - 1) \cdot {\vec{e}}_{j} \cdot {\vec{e}}_{j}^{T}) \cdot (I_{n} + (\frac{1}{c} - 1) {\vec{e}}_{j} \cdot {\vec{e}}_{j}^{T}) \\ = (I_{n} + (c - 1) {\vec{e}}_{j} {\vec{e}}_{j}^{T}) \cdot I_{n} + (I_{n} + (c - 1) {\vec{e}}_{j} {\vec{e}}_{j}^{T}) (\frac{1}{c} - 1) {\vec{e}}_{j} {\vec{e}}_{j}^{T} \\ = I_{n} + (c - 1) \cdot {\vec{e}}_{j} {\vec{e}}^{T} + (\frac{1}{c} - 1) {\vec{e}}_{j} {\vec{e}}_{j}^{T} + (c - 1) \cdot (\frac{1}{c} - 1) \underset{1}{\underset{⏟}{{\vec{e}}_{j} {\vec{e}}_{j}^{T} {\vec{e}}_{j} {\vec{e}}_{j}^{T}}} \\ = I_{n} + (c - 1) \cdot {\vec{e}}_{j} {\vec{e}}^{T} + (\frac{1}{c} - 1) {\vec{e}}_{j} {\vec{e}}_{j}^{T} + (2 - c - \frac{1}{c}) {\vec{e}}_{j} {\vec{e}}_{j}^{T} \\ = I_{n} + (c - 1 + \frac{1}{c} - 1 + 2 - c - \frac{1}{2}) {\vec{e}}_{j} \cdot {\vec{e}}_{j}^{T} \\ = I_{n} + 0_{n \times n} \\ = I_{n} \end{aligned}

\end{proof}

Transposition Matrix

Let $P_{24} \in R^{5 \times 5}$ with

P_{24} = [\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix}]

Then,

P_{24}^{- 1} \cdot P_{24} = I_{5} ⟺ P_{24}^{- 1} = P_{24} = P_{24}^{T}

& Note: this property is famous and called an orthogonal matrix.

Let's make a conjecture from our example.

Conjecture

$P_{i k} \in R^{n \times n}, then P_{i k}^{- 1} = P_{i k}^{T}$