DLA Diagonal form

Section B.3 Diagonal form

In this example we apply the diagonalization procedure to a

$3 \times 3$

$\begin{equation*} A = \begin{bmatrix} 7 \amp -12 \amp -4 \\ 4 \amp -9 \amp -4 \\ -4 \amp 12 \amp 7 \end{bmatrix} \text{.} \end{equation*}$

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A = matrix( [ [7,-12,-4], [4,-9,-4], [-4,12,7] ] )
print(A)

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Eigenvalues.

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We can calculate the characteristic polynomial ourselves.

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xA = x*identity_matrix(3) - A
print(xA)
print()
xA.det().factor()

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Or we can have Sage do it.

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A.charpoly().factor()

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So the eignevalues are

$\lambda = -1$ and

$\lambda = 3\text{.}$ But we could have had Sage do this for us.

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A.eigenvalues()

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Notice that the output is a list of eigenvalues, with each eigenvalue repeated in the list according to its algebraic multiplicity.

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Eigenvectors.

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Let's analyze

$\lambda = 3\text{,}$ since this eigenvalue had algebraic multiplicity

$2\text{.}$ We also need geometric multiplicity

$2$ in order for matrix

$A$ to be diagonalizable; if this geometric multiplicity is only

$1\text{,}$ then analyzing

$\lambda = -1$ would be a waste of time.

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C = 3 * identity_matrix(3) - A
print(C)
print()
C.rref()

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As desired, two parameters are required, so we may proceed. Assigning

$x_2 = s$ and

$x_3 = t$ leads to independent eigenvectors

$\begin{align*} \uvec{p}_1 \amp = (3,1,0) \text{,} \amp \uvec{p}_2 \amp = (1,0,1) \text{.} \end{align*}$

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p1 = vector((3,1,0))
p2 = vector((1,0,1))
print(p1,p2)
print(A*p1,A*p2)

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Notice that

$A \uvec{p}_j = 3 \uvec{p}_j$ for each of these two vectors, as expected.

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Now let's analyze

$\lambda = -1\text{.}$

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C = -1 * identity_matrix(3) - A
print(C)
print()
C.rref()

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We need one parameter, as expected from the algebraic multiplicity. Assigning

$x_3 = t$ leads to eigenvector

$\uvec{p}_3 = (-1,-1,1)\text{.}$

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p3 = vector((-1,-1,1))
print(p3)
A*p3

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Notice that

$A \uvec{p}_3 = - \uvec{p}_3\text{,}$ as expected.

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The transition matrix.

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We now have our basis of independent eigenvectors to form the transition matrix

$P\text{.}$

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P = column_matrix( [ p1, p2, p3 ] )
print(P)
print()
print("rank is", P.rank())

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The rank computation verifies that the columns of

$P$ are linearly independent (but our theory also tells us this: eigenvectors from different eigenspaces are always independent).

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The diagonal matrix.

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Let's compute

$\inv{P} A P$ in two different ways. First, directly.

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P.inverse() * A * P

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Notice the order of the eigenvalues down the diagonal — this is due to the order we placed the eigenvectors as columns in

$P\text{.}$ If we create a different transition matrix, we may get a different diagonal matrix.

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Q = column_matrix([p3, p1, p2])
Q.inverse() * A * Q

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Now let's use the idea in Subsection 26.4.2 to affirm that

$\inv{P} A P$ can also be computed by row reduction: we augment

$P$ with the result of

$A P\text{,}$ and then reducing the

$P$ part on the left to identity simultaneously applies

$\inv{P}$ to the

$A P$ part on the right.

$\begin{equation*} \left[\begin{array}{c|c} P \amp AP \end{array}\right] \qquad \rowredarrow \qquad \left[\begin{array}{c|c} I \amp \inv{P}(AP) \end{array}\right] \end{equation*}$

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P_AP = P.augment(A*P)
print(P_AP)
print()
I_iPAP = P_AP.rref()
print(I_iPAP)

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As expected, there's our diagonal matrix on the right. We can use Python-ic list comprehension to extract it, if we like.

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D = I_iPAP[:,3:6]
print(D)

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The first list-range specifier : says “take all rows.” The second list-range specifier 3:6 says “take columns with index starting at 3 and ending before 6.” Remember that Python uses 0-based indexing, so the column with index 3 is actually the fourth column. And the last column has index 5, so we want to stop before 6 to include it.