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How to find the rank and nullity of (a - ai) when dealing with repeated eigenvalues and the need for generalized eigenvectors. It also covers the null space of a matrix and provides examples for understanding the concepts.
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. (^) find the Rank and Nullity of (A - AI). . (^) find linearly independent vectors when there are repeatea eigenvalues and the nulli!y of (A - XI) is less than the mult~plicity of the eigenvalue. These will be caned generalized eigenvectors.
Ref: Sections 3.8, 3.9. Fall 2004
Summary
For solutions to (A - Ail) Xi = 0, i^ -^ 1,2,...n
n linearly independent vectors, either eigenvectors or generalized eigenvectors (defined on the next pages).
If the eigenvalues are distinct, then n eigenvectors can be found and A can be diagonalized.
Repeated Eigenvalues:
.
. (^) For each repeated eigenvalue find the nullity
1 1
. (^) If the nullity is less than the multiplicity of
0 0 I 1
Zx = 10 = (xl +x3) 1 + (X2+X3)
01 2 10
The Null Space
independent ~_quafionsare there? How many unkriowns? What is the rank of Z? dimension of Z? How are these related?
. (^) All solutions to Zx = 0 are a rnultiple of
Observe from this example:
.
.
.
there are n elements in every vector in N(Z).
there are p (Z) equations that t11ese n elements must satisfy.
n - p(Z) elements can be arbitrarily chosen which means there are n - p(Z) linearly independent vectors in N(Z).
Corollary
\
Generalized Eigenvectors
(A-AI) Xl = 0
(A -AI)2 X2 = 0
(A -AI)3 X3 = 0
Del:. Xk is a generalized eigenvector of rank k of Alff
(A -AI)k-1 Xk ;/=
once you find Xk you can also find
Xk-2 = (A-AI) Xk-l
Justify the chain,
Example
i (t) = 0 1 1 x (t) 001
Solution
A) = A2 = A3 = 1. Check the rank and nullity of (A - I),
0 1 1
(A-I) = 0 0 1 000
example (cont):
Now the modal matrix, M, is
M = 10 1 0
and 1 -1 -
M-1 = 0 1 0
0 0 1
and
M-1AM =
Here the nullity is less than the mull A cannot be diagonalized.
\
Example
This example has
. (^) repeated eigenvalues of multiplicity 2,.
. (^) no generalized eigenvectors needed.
Investigate the eigenvalue-eigenvector problem for. r 120
110 Ans. M = 10001 1 0
Eigenvalues are at Al == 1, A2==3, A3==
\ 110
M = 0 1 0
001 and
1 -1 0
M-1 = 0 1 0
001
and
M-1AM =
or check
AM = MA (don't need to compute M-1)
Example
Given A, find the modal matrix M. Can A be diagonalized?.
1 0 0 Ans. M=IO 4 - Solution
The eigenvalues are Al = A2 = A3 = 0.
0 -5 4
Example (continued),
')n-I ')n-I ')n-I "'1 "'1 ... "'n
and Al' A2'..., An are the eigenvalues of A.
Solution Strategy: Show that the i-th column of M, m. , is the eIgenvector of A associated with Ai. 1
Let m. 1 = [1 m' 2 1 m' 13 ... m.In ]T.
1 1. .. 1
Al A2. ..^ An 2
