What can commute with a diagonal matrix?

  • #1

Main Question or Discussion Point

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?
 

Answers and Replies

  • #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?
 
  • #3
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?
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.
 
  • #4
I like Serena
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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.
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.
 
  • #5
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...
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:
[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]
 
  • #6
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]
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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