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MATH
[1251]

Examples from Eigenvalues and eigenvectors

All examples from Eigenvalues and eigenvectors.

An eigenvector of differentiation

Let V=C∞(R,R)V=C^\infty(\mathbb R,\mathbb R), the vector space of smooth real-valued functions and let T:V→VT:V\to V be differentiation, T(f)=f′T(f)=f'. For any real λ\lambda, the function

f(x)=eλxf(x)=e^{\lambda x}

satisfies

T(f)=f′=λeλx=λf.T(f)=f'=\lambda e^{\lambda x}=\lambda f.

Thus eλxe^{\lambda x} is an eigenvector of differentiation for the eigenvalue λ\lambda. For a matrix, the same test reads Av=λvAv=\lambda v, which is the form used throughout diagonalisation.

[routine]

  1. [1]:Eigenvalue
  2. [2]:Eigenvector

The matrix with all entries equal to one

Let

A=(1111).A= \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.

The characteristic polynomial is

pA(λ)=det⁡(1−λ111−λ)=(1−λ)2−1=λ(λ−2).p_A(\lambda) =\det\begin{pmatrix}1-\lambda&1\\1&1-\lambda\end{pmatrix} =(1-\lambda)^2-1 =\lambda(\lambda-2).

Thus the eigenvalues are 00 and 22. For λ=0\lambda=0,

Av=0⟹v=α(1−1),A v=0 \quad\Longrightarrow\quad v=\alpha\begin{pmatrix}1\\-1\end{pmatrix},

so

E0=span⁡{(1−1)}.E_0=\operatorname{span}\left\{\begin{pmatrix}1\\-1\end{pmatrix}\right\}.

For λ=2\lambda=2,

(A−2I)v=0⟹v=α(11),(A-2I)v=0 \quad\Longrightarrow\quad v=\alpha\begin{pmatrix}1\\1\end{pmatrix},

so

E2=span⁡{(11)}.E_2=\operatorname{span}\left\{\begin{pmatrix}1\\1\end{pmatrix}\right\}.

The two eigenspaces are perpendicular lines: AA doubles the component parallel to (1,1)T(1,1)^{\mathsf T} and removes the component parallel to (1,−1)T(1,-1)^{\mathsf T}. The roots 00 and 22 alone do not show this. The eigenspaces are what describe the action of the matrix and supply an eigenbasis.

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