Type a command or search the notes. Tab cycles completions. Up and down recall history when blank. Escape dismisses, then leaves the prompt.
MATH
[1251]

Examples from Spectral applications

All examples from Spectral applications.

Three repeated-eigenvalue cases

Consider the three matrices

A1=(320032003),A2=(300032003),A3=3I.A_1= \begin{pmatrix} 3&2&0\\ 0&3&2\\ 0&0&3 \end{pmatrix}, \qquad A_2= \begin{pmatrix} 3&0&0\\ 0&3&2\\ 0&0&3 \end{pmatrix}, \qquad A_3=3I.

Each has characteristic polynomial (3−λ)3(3-\lambda)^3, so the algebraic multiplicity of λ=3\lambda=3 is 33.

For A1A_1,

A1−3I=(020002000),A_1-3I= \begin{pmatrix} 0&2&0\\ 0&0&2\\ 0&0&0 \end{pmatrix},

so x2=x3=0x_2=x_3=0 and

E3=span⁡{(100)}.E_3=\operatorname{span}\left\{ \begin{pmatrix}1\\0\\0\end{pmatrix} \right\}.

The geometric multiplicity is 11.

For A2A_2,

A2−3I=(000002000),A_2-3I= \begin{pmatrix} 0&0&0\\ 0&0&2\\ 0&0&0 \end{pmatrix},

so x3=0x_3=0 and

E3=span⁡{(100),(010)}.E_3=\operatorname{span}\left\{ \begin{pmatrix}1\\0\\0\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix} \right\}.

The geometric multiplicity is 22. Finally, A3−3I=0A_3-3I=0, so the eigenspace is R3\mathbb R^3 and the geometric multiplicity is 33. Only the last case has three independent eigenvectors and hence an eigenbasis.

Powers of the all-ones matrix

For

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

use the eigenvectors computed for the all-ones matrix:

P=(111−1),D=(2000),P−1=12(111−1).P= \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}, \qquad D= \begin{pmatrix} 2&0\\ 0&0 \end{pmatrix}, \qquad P^{-1} =\frac12 \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}.

Therefore, for n≥1n\geq1,

An=PDnP−1=12(111−1)(2n000)(111−1)=2n−1(1111).\begin{aligned} A^n &=PD^nP^{-1}\\ &=\frac12 \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix} \begin{pmatrix} 2^n&0\\ 0&0 \end{pmatrix} \begin{pmatrix} 1&1\\ 1&-1 \end{pmatrix}\\ &=2^{n-1} \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}. \end{aligned}

For x0=(a,b)Tx_0=(a,b)^{\mathsf T},

Anx0=2n−1(a+b)(11).A^nx_0 =2^{n-1}(a+b) \begin{pmatrix}1\\1\end{pmatrix}.

The component in the 00-eigenspace disappears after one step, while the component in the 22-eigenspace grows like 2n2^n.

A three-by-three power

Let

A=(311020002),A= \begin{pmatrix} 3&1&1\\ 0&2&0\\ 0&0&2 \end{pmatrix},

and take the eigenvectors

v1=(100),v2=(−101),v3=(−110),v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad v_2=\begin{pmatrix}-1\\0\\1\end{pmatrix}, \qquad v_3=\begin{pmatrix}-1\\1\\0\end{pmatrix},

for eigenvalues 3,2,23,2,2, respectively. Then

P=(1−1−1001010),D=(300020002),P−1=(111001010).P= \begin{pmatrix} 1&-1&-1\\ 0&0&1\\ 0&1&0 \end{pmatrix}, \qquad D= \begin{pmatrix} 3&0&0\\ 0&2&0\\ 0&0&2 \end{pmatrix}, \qquad P^{-1}= \begin{pmatrix} 1&1&1\\ 0&0&1\\ 0&1&0 \end{pmatrix}.

The multiplication check gives A=PDP−1A=PDP^{-1} and hence

An=P(3n0002n0002n)P−1=(3n3n−2n3n−2n02n0002n).A^n=P \begin{pmatrix} 3^n&0&0\\ 0&2^n&0\\ 0&0&2^n \end{pmatrix} P^{-1} = \begin{pmatrix} 3^n&3^n-2^n&3^n-2^n\\ 0&2^n&0\\ 0&0&2^n \end{pmatrix}.

The eigenvalue 22 is repeated but has two independent eigenvectors, v2v_2 and v3v_3. Together with v1v_1 they form an eigenbasis, so the diagonal power calculation is valid.

On this page