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Math Help - problem with invertible matrices

  1. #1
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    problem with invertible matrices

    Question:
    If A and B are similar matrices
    the geometric multiplicity of an eigenvalue \lambda of A is the same as its geometric multiplicity as an eigenvalue of B.

    Proof:
    {\lambda} I - B = Q({\lambda} I -A)Q^{-1}
    since Q is invertible, it follows that nullity({\lambda}I - B) = nullity({\lambda}I - A)

    I dont get the bit about "since Q is invertible" I do not see how this is relevant
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  2. #2
    Super Member girdav's Avatar
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    We can see that x\in \ker (\lambda I-B) if and only if Q^{-1}x\in\ker(\lambda I-A). We deduce that we have an isomorphism between these two space.
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  3. #3
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    An invertible matrix is one-to-one. Consequently, the null spaces of A and QA, where Q is invertible, are the same. QAv is 0 if and only if Av is 0- v is in the null space of A. Similarly, AQv= 0 only if Qv is in the null space of A. Since Q is invertible, the dimension of "all v such that Qv is in the null space of A" is the same as the dimension of the null space of A.

    Personally, I wouldn't prove this statement in that way. Instead (longer but more "fundamental"):
    Suppose \lambda is an eigenvalue of A with geometric multiplicity n- that is, there exist n independent vectors, \{v_1, v_2, \cdot\cdot\cdot, v_n\} vectors such that Av_i= \lambda v_i for i= 1, 2, ..., n. Suppose B is similar to A: A= Q^{-1}BQ for some invertible matrix Q.

    Then, for every i from 1 to n, B(Qv_i)= Q(Q^{-1}BQ)v_i= QAv_i= Q(\lambda v_i)= \lambda (Qv_i)
    so that every Qv_i is an eigenvector of B corresponding to eigenvalue \lambda. To show that they are independent, write a_1(Qv_1)+ a_2(Qv_2)+ \cdot\cdot\cdot+ a_n(Qv_n)= Q(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)= 0. Since Q is invertible, it is one-to-one: we have a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n= 0. Since the v_i vectors are independent, we have a_1= a_2= \cdot\cdot\cdot= a_n= 0 so the \{Qv_1, Qv_2, \cdot\cdot\cdot, Qv_n\} are n independent eigenvectors of B corresponding to eigenvalue \lambda.
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