I-AB is invertible if and only if I-BA is invertible IABIsInvertibleIfAndOnlyIfIBAIsInvertible 2004-10-17 07:40:26 2007-12-07 20:03:09 Theorem linear isomorphism % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{mathrsfs} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions % % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\sR}[0]{\mathbb{R}} \newcommand{\sC}[0]{\mathbb{C}} \newcommand{\sN}[0]{\mathbb{N}} \newcommand{\sZ}[0]{\mathbb{Z}} \usepackage{bbm} \newcommand{\Z}{\mathbbmss{Z}} \newcommand{\C}{\mathbbmss{C}} \newcommand{\R}{\mathbbmss{R}} \newcommand{\Q}{\mathbbmss{Q}} \newcommand*{\norm}[1]{\lVert #1 \rVert} \newcommand*{\abs}[1]{| #1 |} \newtheorem{thm}{Theorem} \newtheorem{defn}{Definition} \newtheorem{prop}{Proposition} \newtheorem{lemma}{Lemma} \newtheorem{cor}{Corollary} \PMlinkescapeword{invertible} \PMlinkescapeword{inverse} In this entry $A$ and $B$ are endomorphisms of a vector space $V$. If $V$ is finite dimensional, we may choose a basis and regard $A$ and $B$ as square matrices of equal dimension. {\bf Theorem -} Let $A$ and $B$ be endomorphisms of a vector space $V$. We have that \begin{enumerate} \item $I-AB$ is \PMlinkname{invertible}{LinearIsomorphism} if and only if $I-BA$ is invertible, and moreover \item $I-AB$ is injective if and only if $I-BA$ is injective. \end{enumerate} {\bf Proof :} \begin{enumerate} \item Suppose that $I-AB$ is invertible. We shall prove that $B(I-AB)^{-1}A +I$ is the inverse of $I-BA$. In fact \begin{eqnarray*} \Big(I-BA\Big)\Big(B(I-AB)^{-1}A +I\Big) &=& B(I-AB)^{-1}A + I - BAB(I-AB)^{-1}A - BA\\ &=& B \Big((I-AB)^{-1} - AB(I-AB)^{-1}\Big)A + I - BA\\ &=& B \Big((I-AB)(I-AB)^{-1}\Big)A + I -BA\\ &=& BA + I - BA\\ &=& I \end{eqnarray*} A similar computation shows that $\Big(B(I-AB)^{-1}A +I\Big)\Big(I-BA\Big) = I$, i.e. $I - BA$ is invertible. Exchanging the roles of $A$ and $B$ we can prove the "if" part. So $I-AB$ is invertible if and only if $I - BA$ is invertible. \item Let us first recall that a linear map between vector spaces is invertible if and only if its kernel $\operatorname{ker}$ is the zero vector (see \PMlinkname{this page}{KernelOfALinearTransformation}). Suppose $I - AB$ is not injective, i.e. there exists $u \neq 0$ such that $(I-AB)u=0$. Then \begin{displaymath} (I-BA)Bu = B(I-AB)u = 0 \end{displaymath} i.e. $Bu \in \operatorname{ker}(I-BA)$. Notice that $Bu \neq 0$ because $u = ABu$ (by definition of $u$), so $I-BA$ is also not injective. Similarly, if $I-BA$ is not injective then $I - AB$ is not injective. $\square$ \end{enumerate} {\bf Remark -} It is known that for finite dimensional vector spaces a linear endomorphism is invertible if and only if it is injective. This does not remain true for infinite dimensional spaces, hence 1 and 2 are two different statements. \subsection{Comments} The result stated in 1 can be proven in a more general context --- If $A$ and $B$ are elements of a ring with unity, then $I-AB$ is invertible if and only if $I-BA$ is invertible. See the entry on techniques in mathematical proofs, in which this result is proven using several different techniques. This entry is based on \PMlinkexternal{this discussion on PM}{http://planetmath.org/?op=getmsg&id=5088}.