Example Let -- [1 -4 4 A = 2 -6 5. -5 4. The characteristic polynomial of A is CA(A) = -(A + 1)² (A – 1), so – 1 is an eigenvalue with algebraic multiplicity , and 1 is an eigenvalue with algebraic multiplicity Without further calculations, we may determine from the inequalities 1 < geo(A) < alg(A) that the geometric multiplicity of 1 is For the other eigenvalue, -1, all that the inequalities tell us is that 1 < geo(-1) < However, after some row operations, we find that the
Example Let -- [1 -4 4 A = 2 -6 5. -5 4. The characteristic polynomial of A is CA(A) = -(A + 1)² (A – 1), so – 1 is an eigenvalue with algebraic multiplicity , and 1 is an eigenvalue with algebraic multiplicity Without further calculations, we may determine from the inequalities 1 < geo(A) < alg(A) that the geometric multiplicity of 1 is For the other eigenvalue, -1, all that the inequalities tell us is that 1 < geo(-1) < However, after some row operations, we find that the
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![Example
Let
1
-4
4
A =
-6 5
-5
4.
The characteristic polynomial of A is CA(^) =
-(A + 1)² (A – 1), so –1 is an eigenvalue with algebraic multiplicity
and 1 is an eigenvalue with
algebraic multiplicity
Without further calculations, we may determine from the inequalities 1 < geo(A) < alg(A) that the geometric multiplicity of 1
is
For the other eigenvalue, –1, all that the inequalities tell us is that 1 < geo(-1) <
However, after some row operations, we find that the
matrix A + I has rank 2, so the dimension of its null space is
showing that the geometric multiplicity of –1 is
By computing reduced row-echelon forms, we find that the eigenspace associated to 1 is spanned by
1
and the eigenspace associated to -1 is spanned by
1|.
This example illustrates the following important case of the theorem above.
Important special case: If the algebraic multiplicity of an eigenvalue A is 1, so alg(A) = 1, then the inequality in the theorem above reduces to
1< geo(A) < 1. This can only be true when geo(A) = 1.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F001eda53-19f3-49af-8959-1aa5b0b6fb9b%2F8d76b0bf-eefa-4106-b4c6-7091cd9e5958%2Fta151uq_processed.png&w=3840&q=75)
Transcribed Image Text:Example
Let
1
-4
4
A =
-6 5
-5
4.
The characteristic polynomial of A is CA(^) =
-(A + 1)² (A – 1), so –1 is an eigenvalue with algebraic multiplicity
and 1 is an eigenvalue with
algebraic multiplicity
Without further calculations, we may determine from the inequalities 1 < geo(A) < alg(A) that the geometric multiplicity of 1
is
For the other eigenvalue, –1, all that the inequalities tell us is that 1 < geo(-1) <
However, after some row operations, we find that the
matrix A + I has rank 2, so the dimension of its null space is
showing that the geometric multiplicity of –1 is
By computing reduced row-echelon forms, we find that the eigenspace associated to 1 is spanned by
1
and the eigenspace associated to -1 is spanned by
1|.
This example illustrates the following important case of the theorem above.
Important special case: If the algebraic multiplicity of an eigenvalue A is 1, so alg(A) = 1, then the inequality in the theorem above reduces to
1< geo(A) < 1. This can only be true when geo(A) = 1.
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