\documentclass{article} %\pagestyle{empty} \setlength{\topmargin}{0in} \setlength{\oddsidemargin}{0in} \setlength{\textheight}{8in} \setlength{\textwidth}{6in} \usepackage{amssymb} \usepackage{amsthm} \usepackage{enumerate} \usepackage{amscd} \newcommand{\stbar}{\;|\;} \newcommand{\R}{\mathbb{R}} \newcommand{\ra}{\rightarrow} \newcommand{\dotprod}{\bullet} \newcommand{\proj}{\mathrm{Proj}} \renewcommand{\span}{\mathrm{Span}} \theoremstyle{definition} \newtheorem{theorem}{Theorem} \newtheorem{example}[theorem]{Example} \newtheorem{claim}[theorem]{Claim} \newtheorem{fact}[theorem]{Fact} \newtheorem{defn}[theorem]{Definition} \newtheorem{exercise}{Exercise} \newtheorem{maple}{Maple Exercise} \begin{document} \title{Diagonalization} \author{Blake Thornton} \date{November 20, 2002} \maketitle \section{Definitions and Theorem} \begin{defn} An $n\times n$ matrix $A$ is \emph{diagonalizable} if there is some matrix $Q$ such that $Q^{-1}AQ$ is diagonal. \end{defn} \begin{theorem} \label{n evecs} An $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has a set of $n$ linearly independent eigenvectors. \end{theorem} What this theorem means is that if you can find a basis for $\R^n$, consisting of eigenvectors, then you can diagonalize the matrix. \begin{defn} Let $A$ be a $n \times n$ matrix and let $\lambda_1, \ldots, \lambda_k$ be all the different eigenvalues of $A$. Let $n_1, \ldots, n_k$ be the multiplicities of the eigenvalues (so $n_i$ is the multiplicity of the eigenvalue $\lambda_i$). Then $A$ is \emph{deficient} if for any of the eigenvalues, the eigenspace associated to the eigenvector $\lambda_i$ has dimension less then $n_i$. \end{defn} Basically what this definition is saying is that you should be able to ``see'' the dimension of the eigenspace by looking at the characteristic polynomial and looking at the multiplicity of the roots. If the eigenspace turns out to have not big enough dimension, then the matrix is deficient. \begin{theorem} \label{not deficient} A matrix is diagonalizable if and only if it is not deficient. \end{theorem} \begin{theorem} \label{n different} Let $A$ be a $n \times n$ matrix with eigenvalues $\lambda_1, \ldots, \lambda_k$. Let $Q_1, \ldots, Q_k$ be eigenvectors associated to the eigenvalues. Then $\{ Q_1, \ldots, Q_k \}$ are independent. \end{theorem} What this is saying is that different eigenvalues give different eigenvectors. Or, more precisely, different eigenvectors give linearly independent eigenvectors. \section{Procedure} Here I will outline how to go about diagonalizing a matrix. There are a couple of points that need addressed. Here are the main issues that you want to address: \begin{enumerate}[I.] \item Can you diagonalize the matrix? \item What will the diagonalized matrix look like? (What is $D$?) \item What is the diagonalizing matrix? (What is the matrix $Q$?) \end{enumerate} Here is the procedure. You should notice that these are the steps that I followed in class and also the steps that the book follows (although probably not as clearly laid out for you). We will assume that $A$ is a $n\times n$ matrix. \begin{enumerate} \item Find all the eigenvalues of $A$. \\ When you do this, you should notice: \begin{enumerate} \item What are the multiplicities of each eigenvalues. (You should be able to see this if you factored the characteristic polynomial.) If you can't see this, it is not the end of the world, but this will tell you want dimension you are expecting each eigenspace to be. \end{enumerate} \item Find a basis for each eigenspace. \\ When doing this notice: \begin{enumerate} \item Is the matrix deficient? You should be able to see this by either comparing the dimension of each eigenspace to the multiplicity of eigenvector found above. Otherwise, you can just see if you found $n$ independent eigenvectors. \end{enumerate} \item Is the matrix diagonalizable? \\ Ways to answer this question: \begin{enumerate} \item Are there $n$ different eigenvalues? If so, then by Theorem~\ref{n different} $A$ is diagonalizable. \item Is $A$ deficient? If so, then by Theorem~\ref{not deficient} $A$ is not diagonalizable. If not, then $A$ is diagonalizable. \item Is there a basis for $\R^n$ consisting of eigenvectors? If so, then by Theorem~\ref{n evecs} $A$ is diagonalizable. If not, then $A$ is not diagonalizable. \end{enumerate} \item Find the diagonalizing matrix $Q$: \begin{enumerate} \item Let $Q$ be the $n \times n$ matrix with the $n$ linearly independent eigenvectors as columns. \end{enumerate} \item Find the diagonal matrix $D$. \begin{enumerate} \item Then $D=Q^{-1}AQ$ is the diagonal matrix. $D$ should be the matrix with eigenvalues on the diagonal (in the same order that you put the eigenvectors into the matrix $Q$). \item I find it confusing to remember if $D=Q^{-1}AQ$ or if $D=QAQ^{-1}$. I think the easiest way to remember this is the remember the chart I drew on the board on November 18 and to remember what the matrix $Q$ does. \\ Let $S$ be the standard basis in $\R^n$ and let $E$ be the basis of eigenvectors in $\R^n$. Let $(\R^n,S)$ be $\R^n$ written in $S-coordinates$ and $(R^n,E)$ be $\R^n$ in $E$-coordinates: \[ \begin{CD} (\R^n, S) @> {A} >> (\R^n, S) \\ {Q}@AAA Q@AAA \\ (\R^n,E) @> {D} >> (\R^n, E) \\ \end{CD} \] \end{enumerate} \end{enumerate} \section{Exercises} Diagonalize the following matrices. This means that you must do the following: \begin{enumerate}[(a)] \item Find all eigenvectors and eigenvalues. \item Determine if the matrix is diagonalizable. \item Find the diagonalizing matrix $Q$. \item Find the diagonal matrix $D$. \item (Using the results from matrix exponentials, find the exponential of each matrix too) \end{enumerate} \begin{enumerate} \item \[ \left[ \begin{array}{rr} 1 & 1 \\ 1 & 1 \end{array} \right] \] \item \[ \left[ \begin{array}{rrr} 1 & 0 & 0 \\ -1 & 3 & 0 \\ 3 & 2 & -2 \end{array} \right] \] \item \[ \left[ \begin{array}{rrr} 3 & -11 & 16 \\ 2 & -8 & 8 \\ 1 & -3 & 2 \end{array} \right] \] \item \[ \left[ \begin{array}{rrr} 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \end{array} \right] \] \item \[ \left[ \begin{array}{rrrr} 3 & -1 & 0 & 2\\ 0 & 3 & 1 & 0 \\ 0 & 0 & -1 & 2\\ 0 & 0 & 1 & 0 \end{array} \right] \] \item \[ \left[ \begin{array}{rrr} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1 \end{array} \right] \] \end{enumerate} \section{Application - Matrix exponentials} I am not going to motivate this too deeply by giving lots of reasons to care about matrix exponentials. You are welcome to come talk to me about it, but matrix exponentials are often used in: \begin{itemize} \item Differential equations. \item Geometry (this is where I have used them) \end{itemize} The basis point is that given a matrix $A$, compute $\exp(A)$ where \begin{eqnarray*} \exp(A) & = & I + A + \frac{1}{2!}A^2 + \frac{1}{3!}A^3 + \cdots \\ & = & \sum_{n=0}^\infty \frac{1}{n!}A^n \end{eqnarray*} Here is how to compute $\exp(A)$ (there are some gaps in this process that you need to fill in): \begin{enumerate} \item Diagonalize $A$ (if possible). \item Write $A=QDQ^{-1}$ and plug into the formula for $\exp(A)$: \begin{eqnarray*} \exp(A) & = & \sum_{n=0}^\infty \frac{1}{n!}A^n \\ & = & \sum_{n=0}^\infty \frac{1}{n!}(QDQ^{-1})^n \\ & = & Q \left(\sum_{n=0}^\infty \frac{1}{n!}D^n \right)Q^{-1} \\ & = & Q(\exp(D))Q^{-1} \end{eqnarray*} \item But, $\exp(D)$ should be easy since $D$ is diagonal -- the exponential of a diagonal is the diagonal matrix made up of the exponential of the diagonal elements of $D$. \end{enumerate} \subsection{Example} \[ A = \left[ \begin{array}{rr} 3 & 1 \\ 1 & 3 \end{array} \right] \quad D = \left[ \begin{array}{rr} 2 & 0 \\ 0 & 4 \end{array} \right] \quad Q = \left[ \begin{array}{rr} -1 & 1 \\ 1 & 1 \end{array} \right] \] \[ \exp(D) = \left[ \begin{array}{rr} e^2 & 0 \\ 0 & e^4 \end{array} \right] \] \begin{eqnarray*} \exp(A) & = & Q\exp(D) Q^{-1} \\ & = & \left[ \begin{array}{rr} \frac{1}{2}(e^4+e^2) & \frac{1}{2}(e^4-e^2) \\ \frac{1}{2}(e^4-e^2) & \frac{1}{2}(e^4+e^2) \end{array} \right] \end{eqnarray*} \end{document}