# Linear Algebra II

## October 27, 2014

### Lecture 18

Filed under: 2014 Fall — Y.K. Lau @ 7:00 PM

We introduce the concept of change matrix. (See Lect18.pdf) Next lecture we shall see how it is used to answer the question: what is the relation between ${M_{FE}(T)}$ and ${M_{DB}(T)}$?

Some preparation: In the coming lecture we shall (re-)visit eigenvalues and eigenvectors that you are supposed to know their definitions and the method to find them. To refresh your memory, please read the following.

Given an ${n\times n}$ matrix ${A}$. If ${\lambda\in{\mathbb R}}$ and ${\underline{0}\neq \underline{x}\in {\mathbb R}^n}$ satisfy ${A\underline{x}=\lambda \underline{x}}$, then ${\lambda}$ is called the eigenvalue of ${A}$ and ${\underline{x}}$ is said to be an eigenvector of ${A}$ corresponding to ${\lambda}$. Note that by definition ${\underline{0}}$ is NOT an eigenvector of ${A}$.

Exercise. Let ${\lambda}$ be an eigenvalue of ${A}$. Show that

$\displaystyle E_\lambda(A):=\{\underline{x}\in{\mathbb R}^n: \ A\underline{x}=\lambda \underline{x}\}$

is a subspace of ${{\mathbb R}^n}$. We call ${E_\lambda(A)}$ the eigenspace of ${A}$ corresponding to ${\lambda}$, thus ${E_\lambda(A)}$ is the set of all eigenvectors corresponding to ${\lambda}$ and the zero vector.

Method:

• The eigenvalues of ${A}$ are found by solving the characteristic equation

$\displaystyle \det (xI-A)=0$

which is in fact a polynomial in ${x}$ of degree ${n}$. (${I=}$ the ${n\times n}$ identity matrix.)

• The eigenvectors corresponding to ${\lambda}$ are found by solving the matrix equation in ${\underline{x}}$:

$\displaystyle (\lambda I- A)\underline{x}=\underline{0}$

All nonzero solutions of ${\underline{x}}$ are the eigenvectors for ${\lambda}$.

(See Textbook p.151-153 for examples.)

After Class Exercises: Ex 9.2 Qn 1(b), 4(b), 5(b), 7(b) (See textbook’s solution and L18-ace.pdf)

Revision: Ex 9.1 Qn 16, 22. (See L18-ace.pdf)