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What can commute with a diagonal matrix?

  1. Oct 31, 2011 #1
    I have two matrices which commute, one of which is definitely diagonal:

    [itex]\textbf{B}diag\{\underline{\lambda}\} = diag\{\underline{\lambda}\}\textbf{B}[/itex]

    and I want to know what I can say about [itex]\textbf{B}[/itex] and/or [itex]\underline{\lambda}[/itex]. Specifically, I feel that either one or both of the following must be correct:

    (1) [itex]diag\{\underline{\lambda}\}[/itex] is proportional to identity.
    (2) [itex]\textbf{B}[/itex] is diagonal.
    [ignoring the trivial cases where one or both matrices equal the zero matrix]

    But are there other cases when these two matrices can commute? i.e. Is it possible for both [itex]\textbf{B}[/itex] to be non-diagonal and the elements of [itex]\underline{\lambda}[/itex] to not all be identical?
     
  2. jcsd
  3. Oct 31, 2011 #2

    I like Serena

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    Hi weetabixharry! :smile:

    What about if B is a symmetric matrix?
    Or what if some of the lambda's are identical, forming a submatrix that is proportional to identity?
    Or what if a lambda is zero?
     
  4. Oct 31, 2011 #3
    If [itex]\textbf{B}[/itex] is symmetric, then:

    [itex]\textbf{B}diag\{\underline{\lambda}\} = [diag\{\underline{\lambda}\}\textbf{B}]^T[/itex]

    Therefore the two matrices only commute if [itex]diag\{\underline{\lambda}\}\textbf{B}[/itex] is symmetric. I feel like this can only happen in the 2 cases I stated above, because each element of [itex]\underline{\lambda}[/itex] multiplies across an entire row of [itex]\textbf{B}[/itex].

    I'm not sure how to approach the other two cases you mentioned.
     
  5. Oct 31, 2011 #4

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    Yes, you are right.
    B being symmetric doesn't help.

    I just checked a 2x2 matrix with a zero on the diagonal.
    Still yields that B must be diagonal, if all lambda's are different.

    If we have a set of equal lambda's, we can split B into sub blocks matrices, and multiply the matrices as sub blocks.
    A sub block of B on the diagonal that corresponds to a block with equal lambdas always commutes.
    A sub block of B that is not on the diagonal has to be zero.
     
  6. Nov 1, 2011 #5
    To prove this it is useful to write the commutator in components:
    [itex]\sum_j B_{ij}\lambda_j \delta_{jl}=\sum_j \lambda_i \delta_{ij}B_{jl}[/itex]
    [itex]B_{il}\lambda_l=\lambda_i B_{il}[/itex]
    [itex]B_{il}(\lambda_l-\lambda_i)=0[/itex]
     
  7. Nov 1, 2011 #6
    Ah yes, this is an excellent way of seeing it. Many thanks for that! (Though, I feel the RHS of the first line should have [itex]\lambda_j[/itex] instead of [itex]\lambda_i[/itex]... even though the result will be the same).

    Quick example of a non-diagonal matrix commuting with a non-proportional-to-identity diagonal matrix:

    [itex]

    \left[\begin{array}{lll}
    7&2&0 \\
    0&1&0 \\
    0&0&4
    \end{array}\right]

    \left[\begin{array}{lll}
    3&0&0 \\
    0&3&0 \\
    0&0&2
    \end{array}\right]

    =

    \left[\begin{array}{lll}
    3&0&0 \\
    0&3&0 \\
    0&0&2
    \end{array}\right]

    \left[\begin{array}{lll}
    7&2&0 \\
    0&1&0 \\
    0&0&4
    \end{array}\right]

    =

    \left[\begin{array}{lll}
    21&6&0 \\
    0&3&0 \\
    0&0&8
    \end{array}\right]

    [/itex]

    Unfortunately, this ruins the proof I was writing... back to the drawing board I guess...
     
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