\(\newcommand{\bx}{\textbf{x}} \newcommand{\bo}{\textbf{0}} \newcommand{\bv}{\textbf{v}} \newcommand{\bu}{\textbf{u}} \newcommand{\bq}{\textbf{q}} \newcommand{\by}{\textbf{y}} \newcommand{\bb}{\textbf{b}} \newcommand{\ba}{\textbf{a}} \newcommand{\grad}{\boldsymbol{\nabla}} \newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\pdd}[2]{\frac{\partial^2 #1}{\partial #2^2}} \newcommand{\pddm}[3]{\frac{\partial^2 #1}{\partial #2 \partial #3}} \newcommand{\deriv}[2]{\frac{d #1}{d #2}} \newcommand{\lt}{ < } \newcommand{\gt}{ > } \newcommand{\amp}{ & } \)

Section5.4The Eigenvalues and Eigenvector Problem

Recall that the eigenvectors, \(\bx\), and the eigenvalues, \(\lambda\) of a square matrix satisfy the equation \(A \bx = \lambda \bx\). Geometrically, the eign-problem is the task of finding the special vectors \(\bx\) such that multiplication by the matrix \(A\) only produces a scalar multiple of \(\bx\). Thinking about matrix multiplication, this is rather peculiar since matrix-vector multiplication usually results in a scaling and a rotation of the vector. Therefore, in some sense the eigenvectors are the only special vectors which avoid geometric rotation under matrix multiplication.

Recall that to solve the eigen-problem for a square matrix \(A\) we complete the following steps:

  1. First rearrange the definition of the eigenvalue-eigenvector pair to \((A\bx - \lambda \bx)= \bo\).

  2. Next, factor the \(\bx\) on the right to get \((A - \lambda I) \bx = \bo\).

  3. Now observe that since \(\bx \neq \bo\) the matrix \(A - \lambda I\) must NOT have an inverse. Therefore, \(\det(A-\lambda I) = 0\).

  4. Solve the equation \(\det(A - \lambda I) = 0\) for all of the values of \(\lambda\).

  5. For each \(\lambda\), find a solution to the equation \((A - \lambda I) \bx = \bo\). Note that there will be infinitely many solutions so you will need to make wise choices for the free variables.