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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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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.
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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