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If a matrix commutes with all nxn matrices, then A must be scalar.

  1. Feb 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove: If a matrix A commutes with all matrices B \in M_{nxn}(F), then A must be scalar - i.e., A=diag.(λ,...,λ), for some λ \in F.


    2. Relevant equations
    If two nxn matrices A and B commute, then AB=BA.


    3. The attempt at a solution
    I understand that if A is scalar, it will definitely commute with all nxn matrices. But I don't get the intuition behind why commuting with more than one matrix implies that A must be scalar. The way I tried to solve it was by comparing an individual entry in the product, (AB)_{ij} = (BA)_{ij} = (AC)_{ij} = (CA)_{ji}, etc. This implies that
    Ʃa_{ik}b_{kj} = Ʃb_{ik}a_{kj} = ...
    But I'm not sure how that implies that A is scalar.
     
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  3. Feb 8, 2012 #2

    Dick

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    Think about eigenvectors. Pick special matrices B. Given any vector [itex]v[/itex], extend it to a basis [itex]{v,b_2,b_3,...b_n}[/itex] and define the matrix [itex]B[/itex] by [itex]Bv=v, Bb_k=0[/itex]. Can you show that A commuting with B means that v must also be an eigenvector of A?
     
    Last edited: Feb 8, 2012
  4. Feb 8, 2012 #3
    I'm trying to figure out what v, b2, b3,..., bn is a basis for. Is it for all nxn matrices?

    If Bv=v, then v is an eigenvector of B corresponding to eigenvalue λ=1, and B is the identity operator on the one-dimensional subspace spanned by v.

    I know that det(B-I) = 0, so maybe something with determinants?

    AB=BA

    -> det(AB) = det(BA)

    and det(B-I) = 0
    det(A)det(B-I)=0
    det(A(B-I))=0
    det(AB - BI) = 0
    det(AB - B) = 0

    I'm sorry, that's as far as I've gotten with that. Please let me know if I'm on the right track. Thanks!
     
  5. Feb 8, 2012 #4

    Dick

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    The vectors are just supposed to be a basis for F^n, the vector space your matrices act on. But, yes, the point is that the eigenvectors of B with eigenvalue 1 are a one dimensional subspace of F^n spanned by v! Now forget about the det's. BAv=ABv put together with Bv=v tells you B(Av)=(Av). So Av is an eigenvector of B with eigenvalue 1. It must lie in the same one dimensional subspace as v. So?
     
    Last edited: Feb 8, 2012
  6. Feb 8, 2012 #5
    So since Av is in the same one-dimensional subspace as v, we know that Av is a scalar multiple of v, and so A is a scalar nxn matrix!

    But does this apply to any nxn matrix B? Or does it have something to do with the specific B that we defined, such that we have to generalize it further to prove for all cases?
     
  7. Feb 8, 2012 #6

    Dick

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    No. You don't have to show anything for all matrices B. You can pick any specific ones you want. A has to commute with all of them. What you have so far is that Av is a multiple of v for ANY v. So ANY vector v is an eigenvector of A. So A is a diagonal matrix in any basis. You haven't shown it's a scalar matrix yet. To do that you have to show all of the eigenvectors of A have the same eigenvalue. Keep going.
     
  8. Feb 9, 2012 #7
    So if A is a diagonal matrix in any bases β and γ, then

    [itex][A]_β = diag(a_1,..., a_n)[/itex]
    and
    [itex][A]_γ = diag(b_1,..., b_n) [/itex]

    And for the eigenvectors in any basis,

    [itex] [A]_βe_i = a_ie_i [/itex]

    But I'm stuck there. How do I show that

    [itex] a_1 = a_2 = ... = a_n [/itex]?
     
  9. Feb 9, 2012 #8

    Dick

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    Suppose A has two linearly independent eigenvectors with two different eigenvalues. We know every vector is an eigenvector of A. See if you can find a contradiction.
     
    Last edited: Feb 9, 2012
  10. Feb 9, 2012 #9
    So if I have A = [itex]diag(a_1,...,a_n)[/itex], then

    [itex]A\vec{e_1} = a_1\vec{e_1} [/itex]
    [itex]A\vec{e_2} = a_2\vec{e_2} [/itex]
    ...
    [itex]A\vec{e_n} = a_n\vec{e_n} [/itex]

    But a vector of all 1's should also be an eigenvector of A.

    [itex]A * (1,1,...,1)^T = (a_1, a_2, ..., a_n)^T [/itex]

    And therefore this can only be an eigenvector if all the diagonal elements of A are equal! Is that right?
     
  11. Feb 9, 2012 #10

    Dick

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    That's right. Put a little more simply, if u and v are eigenvectors with different eigenvalues then u+v can't be an eigenvector.
     
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