Eigenspace 2007-07-08 12:30:19 2007-07-10 22:00:35 Definition eigenvalue \usepackage{amssymb,amscd} \usepackage{amsmath} \usepackage{amsfonts} \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 \usepackage{amsthm} % making logically defined graphics \usepackage{xypic} \usepackage{pst-plot} \usepackage{psfrag} % define commands here \newtheorem{prop}{Proposition} \newtheorem{thm}{Theorem} \newtheorem{ex}{Example} \newcommand{\real}{\mathbb{R}} \newcommand{\pdiff}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\mpdiff}[3]{\frac{\partial^#1 #2}{\partial #3^#1}} Let $V$ be a vector space over a field $k$. Fix a linear transformation $T$ on $V$. Suppose $\lambda$ is an eigenvalue of $T$. The set $\lbrace v\in V\mid Tv=\lambda v\rbrace$ is called the \emph{eigenspace} (of $T$) corresponding to $\lambda$. Let us write this set $W_{\lambda}$. Below are some basic properties of eigenspaces. \begin{enumerate} \item $W_{\lambda}$ can be viewed as the kernel of the linear transformation $T-\lambda I$. As a result, $W_{\lambda}$ is a subspace of $V$. \item The dimension of $W_{\lambda}$ is called the geometric multiplicity of $\lambda$. Let us denote this by $g_{\lambda}$. It is easy to see that $1\le g_{\lambda}$, since the existence of an eigenvalue means the existence of a non-zero eigenvector corresponding to the eigenvalue. \item $W_{\lambda}$ is an invariant subspace under $T$ ($T$-invariant). \item $W_{\lambda_1}\cap W_{\lambda_2}=0$ iff $\lambda_1\ne \lambda_2$. \item In fact, if $W_{\lambda}'$ is the sum of eigenspaces corresponding to eigenvalues of $T$ other than $\lambda$, then $W_{\lambda}\cap W_{\lambda}'=0$. \end{enumerate} From now on, we assume $V$ finite-dimensional. Let $S_T$ be the set of all eigenvalues of $T$ and let $W=\oplus_{\lambda \in S} W_{\lambda}$. We have the following properties: \begin{enumerate} \item If $m_{\lambda}$ is the algebraic multiplicity of $\lambda$, then $g_{\lambda}\le m_{\lambda}$. \item Suppose the characteristic polynomial $p_T(x)$ of $T$ can be factored into linear terms, then $T$ is diagonalizable iff $m_{\lambda}=g_{\lambda}$ for every $\lambda\in S_T$. \item In other words, if $p_T(x)$ splits over $k$, then $T$ is diagonalizable iff $V=W$. \end{enumerate} For example, let $T:\mathbb{R}^2\to \mathbb{R}^2$ be given by $T(x,y)=(x,x+y)$. Using the standard basis, $T$ is represented by the matrix \begin{center}$M_T= \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}.$ \end{center} From this matrix, it is easy to see that $p_T(x)=(x-1)^2$ is the characteristic polynomial of $T$ and $1$ is the only eigenvalue of $T$ with $m_1=2$. Also, it is not hard to see that $T(x,y)=(x,y)$ only when $y=0$. So $W_1$ is a one-dimensional subspace of $\mathbb{R}^2$ generated by $(1,0)$. As a result, $T$ is not diagonalizable.