\(\DeclarePairedDelimiterX{\Set}[2]{\{}{\}}{#1 \nonscript\;\delimsize\vert\nonscript\; #2}\) \( \DeclarePairedDelimiter{\set}{\{}{\}}\) \( \DeclarePairedDelimiter{\parens}{\left(}{\right)}\) \(\DeclarePairedDelimiterX{\innerproduct}[1]{\langle}{\rangle}{#1}\) \(\newcommand{\ip}[1]{\innerproduct{#1}}\) \(\newcommand{\bmat}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\barray}[1]{\left[\hspace{2.0pt}\begin{matrix}#1\end{matrix}\hspace{2.0pt}\right]}\) \(\newcommand{\mat}[1]{\begin{matrix}#1\end{matrix}}\) \(\newcommand{\pmat}[1]{\begin{pmatrix}#1\end{pmatrix}}\) \(\newcommand{\mathword}[1]{\mathop{\textup{#1}}}\)
Needs:
Linear Transformations
Function Inverses
Linear Transformation Products
Needed by:
Linear Isomorphisms
Links:
Sheet PDF
Graph PDF

Invertible Linear Transformations

Motivating result

Suppose $T: V \to W$ is linear and $T^{-1}$ exists. Then $T^{-1}$ is linear.
We show that $T^{-1}$ is additive and homogenous. Let $w_1, w_2 \in W$ and define $v_1$ and $v_2$ so that

\[ v_1 = T^{-1}(w_1) \quad \text{ and } \quad v_2 = T^{-1}(w_2) \]

In other words,

\[ Tv_1 = w_1 \quad \text{ and } \quad Tv_2 = w_2 \]

and so by the linearity of $T$,
Recall that we can use the terminology the inverse because inverses are unique (if they exist; see Function Inverses).

Definition

A linear map $T \in \mathcal{L} (V, W)$ is invertible if there is a linear map $S \in \mathcal{L} (W, v)$ so that $ST$

Copyright © 2023 The Bourbaki Authors — All rights reserved — Version 13a6779cc About Show the old page view