ABAndBAAreAlmostIsospectral 2004-10-17 14:13:09 2007-10-24 13:11:28 Corollary I-AB is invertible if and only if I-BA is invertible % 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} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions \usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \PMlinkescapeword{parent} \PMlinkescapeword{similar} \PMlinkescapeword{argument} \subsection{General case} Let $A$ and $B$ be endomorphisms of a vector space $V$. Let $\sigma(AB)$ and $\sigma(BA)$ denote, respectively, the \PMlinkname{spectra}{spectrum} of $AB$ and $BA$. The next result shows that $AB$ and $BA$ are ``almost'' isospectral, in the sense that their spectra is the same except possibly the value $0$. {\bf Theorem -} Let $A$ and $B$ be as above. We have \begin{enumerate} \item $\sigma(AB) \cup \{0\} = \sigma(BA) \cup \{0\}$, and moreover \item $AB$ and $BA$ have the same eigenvalues, except possibly the zero eigenvalue. \end{enumerate} {\bf Proof :} Let $\lambda \ne 0$. \begin{enumerate} \item If $\lambda \in \sigma(AB)$ then $\lambda^{-1} AB - I$ is not invertible. By the result in the parent entry, this implies that $\lambda^{-1} BA - I$ is not invertible either, hence $\lambda \in \sigma(BA)$. A similar argument proves that every non-zero element of $\sigma(BA)$ also belongs to $\sigma(AB)$. Hence $\sigma(AB) \cup \{0\} = \sigma(BA) \cup \{0\}$. \item If $\lambda$ is an eigenvalue of $AB$, then $I-\lambda^{-1}AB$ is not injective. By the result in the parent entry, this implies that $I-\lambda^{-1}BA$ is also not injective, hence $\lambda$ is an eigenvalue of $BA$. A similar argument proves that non-zero eigenvalues of $BA$ are also eigenvalues of $AB$. $\square$ \end{enumerate} {\bf Remark :} Note that for infinite dimensional vector spaces the spectrum of a linear mapping does not consist solely of its eigenvalues. Hence, 1 and 2 above are two different statements. \subsection{Finite dimensional case} When the vector space $V$ is finite dimensional we can strengthen the above result. {\bf Theroem -} $AB$ and $BA$ are isospectral, i.e. they have the same spectrum. Since $V$ is finite dimensional, this means that $AB$ and $BA$ have the same eigenvalues. {\bf Proof :} By the above result we only need to prove that: $AB$ is invertible if and only if $BA$ is invertible. Suppose $AB$ is not invertible. Hence, $A$ is not invertible or $B$ is not invertible. For finite dimensional vector spaces invertibility, injectivity and surjectivity are the same thing. Thus, the above statement can be rewritten as: $A$ is not injective or $B$ is not surjective. Either way $BA$ is not invertible. A similar argument shows that if $BA$ is not invertible, then $AB$ is also not invertible, which concludes the proof. $\square$ \subsection{Comments} The first theorem can be proven in a more general context : If $A$ and $B$ are elements of an arbitrary unital algebra, then \begin{center} $\sigma(AB) \cup \{0\} = \sigma(BA) \cup \{0\}$. \end{center} This humble result plays an important role in the spectral theory of operator algebras.