What can commute with a diagonal matrix?

[weetabixharry]

I have two matrices which commute, one of which is definitely diagonal:

\textbf{B}diag\{\underline{\lambda}\} = diag\{\underline{\lambda}\}\textbf{B}

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

(1) diag\{\underline{\lambda}\} is proportional to identity.
(2) \textbf{B} 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 \textbf{B} to be non-diagonal and the elements of \underline{\lambda} to not all be identical?

 

[I like Serena]

Hi weetabixharry!

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?

 

[weetabixharry]

Hi weetabixharry!

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?

If \textbf{B} is symmetric, then:

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

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

I'm not sure how to approach the other two cases you mentioned.

 

[I like Serena]

If \textbf{B} is symmetric, then:

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

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

I'm not sure how to approach the other two cases you mentioned.

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.

 

[aesir]

...
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.

To prove this it is useful to write the commutator in components:
\sum_j B_{ij}\lambda_j \delta_{jl}=\sum_j \lambda_i \delta_{ij}B_{jl}
B_{il}\lambda_l=\lambda_i B_{il}
B_{il}(\lambda_l-\lambda_i)=0

 

[weetabixharry]

To prove this it is useful to write the commutator in components:
\sum_j B_{ij}\lambda_j \delta_{jl}=\sum_j \lambda_i \delta_{ij}B_{jl}
B_{il}\lambda_l=\lambda_i B_{il}
B_{il}(\lambda_l-\lambda_i)=0

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 \lambda_j instead of \lambda_i... even though the result will be the same).

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

<br /> <br /> \left[\begin{array}{lll}<br /> 7&amp;2&amp;0 \\<br /> 0&amp;1&amp;0 \\<br /> 0&amp;0&amp;4<br /> \end{array}\right]<br /> <br /> \left[\begin{array}{lll}<br /> 3&amp;0&amp;0 \\<br /> 0&amp;3&amp;0 \\<br /> 0&amp;0&amp;2<br /> \end{array}\right]<br /> <br /> =<br /> <br /> \left[\begin{array}{lll}<br /> 3&amp;0&amp;0 \\<br /> 0&amp;3&amp;0 \\<br /> 0&amp;0&amp;2<br /> \end{array}\right]<br /> <br /> \left[\begin{array}{lll}<br /> 7&amp;2&amp;0 \\<br /> 0&amp;1&amp;0 \\<br /> 0&amp;0&amp;4<br /> \end{array}\right]<br /> <br /> =<br /> <br /> \left[\begin{array}{lll}<br /> 21&amp;6&amp;0 \\<br /> 0&amp;3&amp;0 \\<br /> 0&amp;0&amp;8<br /> \end{array}\right]<br /> <br />

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