65 lines
1.5 KiB
Markdown
65 lines
1.5 KiB
Markdown
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---
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author: Akbar Rahman
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date: \today
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title: Eigenvalues
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tags: [ mmme1026, maths, eigenvalues, uni ]
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uuid: f2220395-bc97-432e-a1d2-74085f16991d
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---
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An eigenvalue problem takes the form:
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Find all the values of $\lambda$ for which the equation
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$$A\pmb{x} = \lambda \pmb{x}$$
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has a nonzero solution $\pmb x$, where $A$ is an $n\times n$ matrix and
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$\pmb x$ is a column vector.
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The equation may be written as
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\begin{align*}
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A\pmb x &= \lambda I \pmb x \\
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\Leftrightarrow A \pmb x - \lambda I \pmb x & = 0 \\
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\Leftrightarrow (A-\lambda I)\pmb x &= 0
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\end{align*}
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($\Leftrightarrow$ means "if and only if")
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Non-zero solutions will exist if
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$\det(A-\lambda I) = 0$
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There are infinitely many eigenvectors for a given eigenvalue.
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This is because if $\pmb x$ is an eigenvector of $A$ corresponding to the
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eigenvalue $\lambda$ and $c$ is a non-zero scalar, then $c\pmb x$ is also
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an eigenvector of $A$:
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$$A(c\pmb x) = cA\pmb x = c\lambda \pmb x = \lambda(c\pmb x)$$
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In general, if $A$ is an $n\times n$ matrix, then $|A-\lambda I|$ is a
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polynomial of degree $n$ in $\lambda$, called the characteristic polynomial.
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The characteristic equation is:
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$$\lambda^n + c_{n-1}\lambda^{n-1} + c_{n-2}\lambda^{n-2} + \cdots + c_0 = 0$$
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<details>
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<summary>
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#### Example 1 ($2\times2$ example)
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</summary>
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If $A$ is the matrix
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$$A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
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then
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$$|A - \lambda I| = \lambda^2 - (a+d)\lambda + (ad-bc)$$
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And the standard method for solving a quadratic can be used to find $\lambda$.
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</details>
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