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potential rational roots:1,-1,3,-3,9,-9. synthetic division: next ... To find an eigenvector belonging to the repeated root 1, ... – PowerPoint PPT presentation

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Title: examples:


1
examples eigenvalues, eigenvectors
and diagonability
Pamela Leutwyler
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Find the eigenvalues and eigenvectors
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characteristic polynomial
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characteristic polynomial
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potential rational roots1,-1,3,-3,9,-9
synthetic division
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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 7 15 9


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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9


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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9

1
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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9
1
1 8
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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9
1 8
1 8 23
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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9
1 8 23
1 8 23 31
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potential rational roots1,-1,3,-3,9,-9
synthetic division
1 1 7 15 9
1 8 23
1 8 23 31
This is not zero. 1 is not a root.
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9


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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9

1
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3
1 4
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3 -12
1 4 3
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3 -12 -9
1 4 3 0
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3 -12 -9
1 4 3 0
This is zero. -3 is a root.
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3 -12 -9
1 4 3 0
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potential rational roots1,-1,3,-3,9,-9
synthetic division
-3 1 7 15 9
-3 -12 -9
1 4 3 0
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The eigenvalues are -3, -3, -1
synthetic division
-3 1 7 15 9
-3 -12 -9
1 4 3 0
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The eigenvalues are -3, -3, -1
To find an eigenvector belonging to the repeated
root 3, consider the null space of the matrix
3I - A
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The eigenvalues are -3, -3, -1
To find an eigenvector belonging to the repeated
root 3, consider the null space of the matrix
3I - A
The 2 dimensional null space of this matrix
has basis
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The eigenvalues are -3, -3, -1
To find an eigenvector belonging to the repeated
root 1, consider the null space of the matrix
1I - A
The null space of this matrix has basis
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The eigenvalues are -3, -3, -1
The eigenvectors are
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The eigenvalues are -3, -3, -1
The eigenvectors are
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The eigenvalues are -3, -3, -1
The eigenvectors are
A
P 1
P
diagonal matrix that is similar to A
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