---
author: Akbar Rahman
date: \today
title: MMME1026 // Eigenvalues
tags: [ mmme1026, maths, eigenvalues, uni ]
uuid: 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:

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

<details>
<summary>

#### Example 1 ($2\times2$ example)

</summary>

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 $\lambda$.

</details>