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.


  • 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)


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