29.2: Diagonalizable Matrix

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In class we will be using matrix diagonalization to solve some problems.

Matrix \(A\) is diagonalizable if there exists a diagonal matrix \(D\) that is similar similar to \(A\):

\[ D = C^{-1}AC \nonumber \]

If matrix \(A\) has linearly independent eigenvectors (\(v_1, \ldots, v_n\)) then \(A\) is diagonalizable with the following solution:

\[C = \left[ v_1^T, \ldots, v_n^T \right] \nonumber \]

In other words, each column of \(C\) is a linearly independent eigenvector of \(A\). The diagonal matrix \(D\) is

\[\begin{split} D =
\left[
\begin{matrix}
\lambda_1 & 0 & 0 \\
0 & \ddots & 0 \\
0 & 0 & \lambda_n
\end{matrix}
\right]
\end{split} \nonumber \]

In other-other words, \(D\) consists of the corresponding eigenvalues.

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

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

Do This

Using numpy, Diagonalize (i.e. calculate \(C\) and \(D\)) the following matrix:

A = np.matrix([[5, -2, 2], [4, -3, 4], [4,-6,7]])
sym.Matrix(A)

# Put your answer here

from answercheck import checkanswer

checkanswer.matrix(D,'56821475223b52e0b6e751da444a1441');

Do This

Verify that \(A\) is in fact Diagonalizable by calculating \(D2 = C^{-1} AC\) and comparing it to your original \(D\) using np.allclose.

#Put your verificaiton code here.

np.allclose(D,D2)

Diagonalization of Symmetric Matrices

One special case is Symmetric Matrices. It can be shown that symmetric Matrices are Diagonalizable and the resulting eigenvectors are not only linearly independent but also orthogonal. Since this is true, the equation changes to:

\[ D = C^{T}AC \nonumber \]

Question

Why do we care if \(C\) is orthogonal? What advantages does the above equation give us?