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30.2: Diagonalization

Last updated

Sep 17, 2022
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- 30.1: Pre-class Assignment Review
- 30.3: The Power of a Matrix

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%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
from urllib.request import urlretrieve

sym.init_printing()
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
            'answercheck.py');

%matplotlib inline
import matplotlib.pylab as plt
import numpy as np
import sympy as sym
from urllib.request import urlretrieve

sym.init_printing()
urlretrieve('https://raw.githubusercontent.com/colbrydi/jupytercheck/master/answercheck.py', 
            'answercheck.py');

Reminder

The eigenvalues of triangular (upper and lower) and diagonal matrices are easy:

The eigenvalues for triangular matrices are the diagonal elements.
The eigenvalues for the diagonal matrices are the diagonal elements.

Diagonalization

Definition

A square matrix $A$ is said to be diagonalizable if there exist a matrix $C$ such that $D=C^{-1}AC$ is a diagonal matrix.

Definition

$B$ is a similar matrix of $A$ if we can find $C$ such that $B=C^{-1}AC$ .

Given an $n \times n$ matrix $A$ , can we find another $n \times n$ invertable matrix $C$ such that when $D=C^{-1}AC$ is diagonal, i.e., $A$ is diagonalizable?

Because $C$ is invertible, we have

$C^{-1}AC=D \nonumber$

$CC^{-1}AC = CD \nonumber$

$AC = CD \nonumber$

Generate $C$ as the columns of $n$ linearly independent vectors $(x_1 \ldots x_n)$ We can compute $AC=CD$ as follows:

$A\begin{bmatrix} \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots \\ { x }_{ 1 } & { x }_{ 2 } & \dots & { x }_{ n } \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}=AC=CD=\begin{bmatrix} \vdots & \vdots & \vdots & \vdots \\ \vdots & \vdots & \vdots & \vdots \\ { x }_{ 1 } & { x }_{ 2 } & \dots & { x }_{ n } \\ \vdots & \vdots & \vdots & \vdots \end{bmatrix}\begin{bmatrix} { \lambda }_{ 1 } & 0 & 0 & 0 \\ 0 & { \lambda }_{ 2 } & 0 & 0 \\ \vdots & \vdots & { \dots } & \vdots \\ 0 & 0 & 0 & { \lambda }_{ n } \end{bmatrix} \nonumber$

Then we check the corresponding columns of the both sides. We have

$Ax_1 = \lambda_1x_1 \\ \vdots \\ Ax_n=\lambda x_n \nonumber$

$A$ has $n$ linear independent eigenvectors.
$A$ is saied to be similar to the diagonal matrix $D$ , and the transformation of $A$ into $D$ is called a similarity transformation.

A simple example

Consider the following:

$A = \begin{bmatrix}7& -10\\3& -4\end{bmatrix},\quad C = \begin{bmatrix}2& 5\\1& 3\end{bmatrix} \nonumber$

Do This

Find the similar matrix $D=C^{-1}AC$ of $A$ .

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#Put your answer to the above question here.

#Put your answer to the above question here.

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from answercheck import checkanswer

checkanswer.matrix(D, '8313fe0f529090d6a8cdb36248cfdd6c');

from answercheck import checkanswer

checkanswer.matrix(D, '8313fe0f529090d6a8cdb36248cfdd6c');

Do This

Find the eigenvalues and eigenvectors of $A$ . Set variables e1 and vec1 to be the smallest eigenvalue and it’s associated eigenvector and e2, vec2 to represent the largest.

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#Put your answer to the above question here.

#Put your answer to the above question here.

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from answercheck import checkanswer
checkanswer.float(e1, "e4c2e8edac362acab7123654b9e73432");

from answercheck import checkanswer
checkanswer.float(e1, "e4c2e8edac362acab7123654b9e73432");

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from answercheck import checkanswer
checkanswer.float(e2, "d1bd83a33f1a841ab7fda32449746cc4");

from answercheck import checkanswer
checkanswer.float(e2, "d1bd83a33f1a841ab7fda32449746cc4");

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from answercheck import checkanswer
checkanswer.eq_vector(vec1, "d28f0a721eedb3d5a4c714744883932e", decimal_accuracy = 4)

from answercheck import checkanswer
checkanswer.eq_vector(vec1, "d28f0a721eedb3d5a4c714744883932e", decimal_accuracy = 4)

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from answercheck import checkanswer
checkanswer.eq_vector(vec2, "09d9df5806bc8ef975074779da1f1023", decimal_accuracy = 4)

from answercheck import checkanswer
checkanswer.eq_vector(vec2, "09d9df5806bc8ef975074779da1f1023", decimal_accuracy = 4)

Theorem

Similar matrices have the same eigenvalues.

Proof: Assume $B=C^{-1}AC$ is a similar matrix of $A$ , and $\lambda$ is an eigenvalue of $A$ with corresponding eigenvector $x$ . That is, $Ax=\lambda x$ . Then we have $B(C^{-1}x) = C^{-1}AC(C^{-1}x) = C^{-1}Ax = C^{-1}(\lambda x) = \lambda (C^{-1}x)$ . That is $C^{-1}x$ is an eigenvector of $B$ with eigenvalue $\lambda$ .

A second example

Do This

Consider $A = \begin{bmatrix}-4& -6\\3& 5\end{bmatrix}$ .

Find a matrix $C$ such that $C^{-1}AC$ is diagonal.

Hint: use the function diagonalize in sympy.

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#Put your answer to the above question here.

#Put your answer to the above question here.

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#Check the output type
assert(type(C)==sym.Matrix)

#Check the output type
assert(type(C)==sym.Matrix)

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from answercheck import checkanswer
checkanswer.matrix(C,'ba963b7fef354b4a7ddd880ca4bac071')

from answercheck import checkanswer
checkanswer.matrix(C,'ba963b7fef354b4a7ddd880ca4bac071')

The third example

Do This

Consider $A = \begin{bmatrix}5& -3\\3& -1\end{bmatrix}$ .

Can we find a matrix $C$ such that $C^{-1}AC$ is diagonal.

Hint: find eigenvalues and eigenvectors using sympy.

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#Put your answer to the above question here.

#Put your answer to the above question here.

Dimensions of eigenspaces and diagonalization

Definition

The set of all eigenvectors of a $n \times n$ matrix corresponding to a eigenvalue $\lambda$ , together with the zero vector, is a subspace of $R^n$ . This subspace spaces is called eigenspace.

For the third example, we have that the characteristic equation $(\lambda - 2)^2 = 0$ .
Eigenvalue $\lambda = 2$ has multiplicity 2, but the eigenspace has dimension 1, since we can not find two lineare independent eigenvector for $\lambda = 2$ .

The dimension of an eigenspace of a matrix is less than or equal to the multiplicity of the corresponding eigenvalue as a root of the characteristic equation.

A matrix is diagonalizable if and only if the dimension of every eigenspace is equal to the multiplicity of the corresponding eigenvalue as a root of the characteristic equation.

The fourth example

Do This

Consider $A = \begin{bmatrix}2& -1\\1& 2\end{bmatrix}$ .

Can we find a matrix $C$ such that $C^{-1}AC$ is diagonal.

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#Put your answer to the above question here.

#Put your answer to the above question here.

Reminder

Diagonalization

Definition

Definition

A simple example

Do This

Do This

Theorem

A second example

Do This

The third example

Do This

Dimensions of eigenspaces and diagonalization

Definition

The fourth example

Do This

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