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| author | date | title | tags | uuid | ||||
|---|---|---|---|---|---|---|---|---|
| Akbar Rahman | \today | MMME1026 // Eigenvalues |
|
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)
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.