Examples from Spectral applications All examples from Spectral applications.
[ EXAMPLE ] Consider the three matrices
A 1 = ( 3 2 0 0 3 2 0 0 3 ) , A 2 = ( 3 0 0 0 3 2 0 0 3 ) , A 3 = 3 I . 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. A 1 = 3 0 0 2 3 0 0 2 3 , A 2 = 3 0 0 0 3 0 0 2 3 , A 3 = 3 I . Each has characteristic polynomial ( 3 − λ ) 3 (3-\lambda)^3 ( 3 − λ ) 3 , so the algebraic multiplicity of λ = 3 \lambda=3 λ = 3 is 3 3 3 .
For A 1 A_1 A 1 ,
A 1 − 3 I = ( 0 2 0 0 0 2 0 0 0 ) , A_1-3I=
\begin{pmatrix}
0&2&0\\
0&0&2\\
0&0&0
\end{pmatrix}, A 1 − 3 I = 0 0 0 2 0 0 0 2 0 , so x 2 = x 3 = 0 x_2=x_3=0 x 2 = x 3 = 0 and
E 3 = span { ( 1 0 0 ) } . E_3=\operatorname{span}\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix}
\right\}. E 3 = span ⎩ ⎨ ⎧ 1 0 0 ⎭ ⎬ ⎫ . The geometric multiplicity is 1 1 1 .
For A 2 A_2 A 2 ,
A 2 − 3 I = ( 0 0 0 0 0 2 0 0 0 ) , A_2-3I=
\begin{pmatrix}
0&0&0\\
0&0&2\\
0&0&0
\end{pmatrix}, A 2 − 3 I = 0 0 0 0 0 0 0 2 0 , so x 3 = 0 x_3=0 x 3 = 0 and
E 3 = span { ( 1 0 0 ) , ( 0 1 0 ) } . E_3=\operatorname{span}\left\{
\begin{pmatrix}1\\0\\0\end{pmatrix},
\begin{pmatrix}0\\1\\0\end{pmatrix}
\right\}. E 3 = span ⎩ ⎨ ⎧ 1 0 0 , 0 1 0 ⎭ ⎬ ⎫ . The geometric multiplicity is 2 2 2 . Finally, A 3 − 3 I = 0 A_3-3I=0 A 3 − 3 I = 0 , so the eigenspace is R 3 \mathbb R^3 R 3 and the geometric multiplicity is 3 3 3 . Only the last case has three independent eigenvectors and hence an eigenbasis.
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[ EXAMPLE ] For
A = ( 1 1 1 1 ) , A=
\begin{pmatrix}
1&1\\
1&1
\end{pmatrix}, A = ( 1 1 1 1 ) , use the eigenvectors computed for the all-ones matrix :
P = ( 1 1 1 − 1 ) , D = ( 2 0 0 0 ) , P − 1 = 1 2 ( 1 1 1 − 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}. P = ( 1 1 1 − 1 ) , D = ( 2 0 0 0 ) , P − 1 = 2 1 ( 1 1 1 − 1 ) . Therefore, for n ≥ 1 n\geq1 n ≥ 1 ,
A n = P D n P − 1 = 1 2 ( 1 1 1 − 1 ) ( 2 n 0 0 0 ) ( 1 1 1 − 1 ) = 2 n − 1 ( 1 1 1 1 ) . \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} A n = P D n P − 1 = 2 1 ( 1 1 1 − 1 ) ( 2 n 0 0 0 ) ( 1 1 1 − 1 ) = 2 n − 1 ( 1 1 1 1 ) . For x 0 = ( a , b ) T x_0=(a,b)^{\mathsf T} x 0 = ( a , b ) T ,
A n x 0 = 2 n − 1 ( a + b ) ( 1 1 ) . A^nx_0
=2^{n-1}(a+b)
\begin{pmatrix}1\\1\end{pmatrix}. A n x 0 = 2 n − 1 ( a + b ) ( 1 1 ) . The component in the 0 0 0 -eigenspace disappears after one step, while the component in the 2 2 2 -eigenspace grows like 2 n 2^n 2 n .
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[ EXAMPLE ] Let
A = ( 3 1 1 0 2 0 0 0 2 ) , A=
\begin{pmatrix}
3&1&1\\
0&2&0\\
0&0&2
\end{pmatrix}, A = 3 0 0 1 2 0 1 0 2 , and take the eigenvectors
v 1 = ( 1 0 0 ) , v 2 = ( − 1 0 1 ) , v 3 = ( − 1 1 0 ) , 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}, v 1 = 1 0 0 , v 2 = − 1 0 1 , v 3 = − 1 1 0 , for eigenvalues 3 , 2 , 2 3,2,2 3 , 2 , 2 , respectively. Then
P = ( 1 − 1 − 1 0 0 1 0 1 0 ) , D = ( 3 0 0 0 2 0 0 0 2 ) , P − 1 = ( 1 1 1 0 0 1 0 1 0 ) . 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}. P = 1 0 0 − 1 0 1 − 1 1 0 , D = 3 0 0 0 2 0 0 0 2 , P − 1 = 1 0 0 1 0 1 1 1 0 . The multiplication check gives A = P D P − 1 A=PDP^{-1} A = P D P − 1 and hence
A n = P ( 3 n 0 0 0 2 n 0 0 0 2 n ) P − 1 = ( 3 n 3 n − 2 n 3 n − 2 n 0 2 n 0 0 0 2 n ) . 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}. A n = P 3 n 0 0 0 2 n 0 0 0 2 n P − 1 = 3 n 0 0 3 n − 2 n 2 n 0 3 n − 2 n 0 2 n . The eigenvalue 2 2 2 is repeated but has two independent eigenvectors, v 2 v_2 v 2
and v 3 v_3 v 3 . Together with v 1 v_1 v 1 they form an eigenbasis, so the diagonal power
calculation is valid.
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Spectral applications Builds on Diagonalisation and 2 more.