notes/uni/mmme/1026_maths_for_engineering/eigenvalues.md

author	date	title	tags	uuid
Akbar Rahman	\today	MMME1026 // Eigenvalues	mmme1026 maths eigenvalues uni	f2220395-bc97-432e-a1d2-74085f16991d

An eigenvalue problem takes the form:

Find all the values of \lambda for which the equation

A\pmb{x} = \lambda \pmb{x}

has a nonzero solution \pmb x, where A is an n\times n matrix and \pmb x is a column vector.

The equation may be written as

\begin{align*} A\pmb x &= \lambda I \pmb x \ \Leftrightarrow A \pmb x - \lambda I \pmb x & = 0 \ \Leftrightarrow (A-\lambda I)\pmb x &= 0 \end{align*}

(\Leftrightarrow means "if and only if")

Non-zero solutions will exist if

\det(A-\lambda I) = 0

There are infinitely many eigenvectors for a given eigenvalue. This is because if \pmb x is an eigenvector of A corresponding to the eigenvalue \lambda and c is a non-zero scalar, then c\pmb x is also an eigenvector of A:

A(c\pmb x) = cA\pmb x = c\lambda \pmb x = \lambda(c\pmb x)

In general, if A is an n\times n matrix, then |A-\lambda I| is a polynomial of degree n in \lambda, called the characteristic polynomial. The characteristic equation is:

\lambda^n + c_{n-1}\lambda^{n-1} + c_{n-2}\lambda^{n-2} + \cdots + c_0 = 0

Example 1 (2\times2 example)

If A is the matrix

A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}

then

|A - \lambda I| = \lambda^2 - (a+d)\lambda + (ad-bc)

And the standard method for solving a quadratic can be used to find \lambda.