FindingEigenvalues 2006-04-27 09:23:48 2008-12-05 19:43:02 Example eigenvalue eigenvalue eigenvector % 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 This example investigates eigenvalues and the similarity transformation used to diagonalize matrices. We seek the eigenvalues of the matrix $A$ below. Afterward, we can transform this matrix into a diagonal matrix which has many useful applications.\\ $$A=\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right )$$ Here, we need to solve the corresponding matrix equation;\\ $$\left( \begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array} \right ) \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right )=\lambda \left( \begin{array}{c} x_1 \\ x_2 \end{array} \right )$$ or\\ $$AX=\lambda X$$ rearranging gives $$AX-\lambda X=0$$ or\\ $$(A-\lambda I) X=0$$\\ We seek the values for $\lambda$ and $X$. First, we need to solve the characteristic equation of $A$. We do this by finding $det(A-\lambda I)$. First, calculating $A-\lambda I$ gives;\\ $$A -\lambda I= \left( \begin{array}{cc} 2-\lambda & 1 \\ 1 & 2-\lambda \end{array} \right )$$ Next, calculating $det(A-\lambda I)$ yields\\ $$det(A-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=(\lambda-1)(\lambda-3)=0$$\\ Substituting $\lambda=1$ into $(A-\lambda I)X$ gives...\\ $$\left\{ \begin{array}{c} x_1+x_2=0 \\ x_1+x_2=0 \end{array} \right.$$ so that $x_2=-x_1$ and the corresponding eigenvector is $$\left( \begin{array}{c} t \\ -t \end{array} \right )=t \left( \begin{array}{c} 1 \\ -1 \end{array} \right )$$ where $t\ne 0.$\\ Substituting $\lambda=3$ gives...\\ $$\left\{ \begin{array}{c} -x_1+x_2=0 \\ x_1-x_2=0 \end{array} \right.$$ so that $x_2=x_1$ and the corresponding eigenvector is $$\left( \begin{array}{c} t \\ t \end{array} \right )=t \left( \begin{array}{c} 1 \\ 1 \end{array} \right )$$ where $t\ne 0.$\\ Finally, to diagonalize $A$ we let the eigenvectors be the columns of a new matrix\\ $$P=\left( \begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array} \right )$$ and then since our eigenvectors are linearly independent we can also find;\\ $$P^{-1}=\frac{1}{2} \left( \begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array} \right )$$ then we create a diagonal matrix as follows...\\ $$D=P^{-1}AP=\left( \begin{array}{cc} 1 & 0 \\ 0 & 3 \end{array} \right )$$ Computing powers of $A$ is a very useful application of $D$. Solving for $A$ lets us compute powers of $A$\\ $$A=PDP^{-1}$$\\ so that\\ $$A^n=PD^nP^{-1}$$\\ or $$A^n=P\left( \begin{array}{cc} 1^n & 0 \\ 0 & 3^n \end{array} \right )P^{-1}$$\\