13.1 Inverse of a Square Matrix

Types: N/A
Examples: N/A
Constructions: N/A
Generalizations: N/A

Properties: 13.4 Properties of Matrix Inverses
Sufficiencies: N/A
Questions: N/A

Inverse of a Square Matrix

Let $A \in R^{n \times n}$ be a square matrix. We say that $A$ is invertible iff there exists a matrix $C \in R^{n \times n}$ such that

A \cdot C = C \cdot A = I_{n}

where $I_{n}$ is the $n \times n$ 8.6 Identity Matrix.

$C = A^{- 1}$ is the inverse of $A$ .
An invertible matrix is called nonsingular while a matrix that is not invertible is called singular
! Not all square matrices are invertible!

Remark. Matrix inverses are the inverse of 11 Matrix-Matrix Multiplication. Only square matrices can be two-side inverses.
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Inverse of a Square Matrix

Consider matrices

A = [\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}], C = [\begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{matrix}]

Notice that $C$ is the inverse of $A$ , denoted as $C = A^{- 1}$ , because

A \cdot C = [\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix}] [\begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]

Inverse of Shear Matrix

Consider the matrices

S_{31} (- 3) = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 3 & 0 & 1 \end{matrix}], C = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{matrix}]

We see that $C = {(S_{31} (- 3))}^{- 1}$ because

\begin{aligned} S_{31} (- 3) \cdot C & = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - 3 & 0 & 1 \end{array}] [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{array}] \\ = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \end{aligned}

Conjecture

Note if $D \in R^{n \times n}$ is diagonal with $d_{i i} \neq 0$ for all $i \in {n}$ , then

D = [\begin{matrix} d_{11} & 0 & \dots & 0 \\ 0 & d_{21} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & d_{n n} \end{matrix}]

is invertible with

D^{- 1} = [\begin{matrix} d_{11}^{- 1} & 0 & \dots & 0 \\ 0 & d_{21}^{- 1} & ⋱ & ⋮ \\ ⋮ & ⋱ & ⋱ & 0 \\ 0 & \dots & 0 & d_{n n} \end{matrix}]

Therefore,

\begin{aligned} nonsingular & ⟺ linearly independent columns and square \\ ⟺ linearly independent rows and square \\ ⟺ there is a C \in R^{n \times n} with A \cdot C = C A = I_{n} and G = A^{- 1} \end{aligned}