- #1

Little Devil

Ben

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Little Devil

Ben

- #2

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Diagonal matrices commute.

- #3

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- #4

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- #5

HallsofIvy

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It follows then that A= M

Then AB= M

= M

= M

= BA.

- #6

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That is the proof I couldn't remember.

- #7

Hurkyl

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- #8

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http://hverrill.net/courses/linalg/linalg8.html

I don't have anything better to add to it.

- #9

Hurkyl

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Here's a sketch of what I have so far:

Assume A is not defective and finite dimensional. Then choose a basis that diagonalizes A.

The i,k-th entry in AB is A

The i,k-th entry in BA is B

So AB = BA iff, for all (i, k), B

If A

By repeating this process, we can produce a basis that simultaneously diagonalizes both A and B.

I don't know what to do if A is defective or infinite dimensional. (though I admit not having taken a crack at modifying the above proof to use transfinite induction to tackle the infinite dimensional case)

- #10

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- #11

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Diagonal matrices commute.

this is not IFF is it?

A * B can B* A without A being diagonal

- #12

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Can you give me and example to find matrices commute with matrices A that we know?

- #13

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Clearly an example is all matrices for which B = kA, where k is a scalar. Then, the eigenvalues differ by a constant factor and the matrix M diagonalizes both A and B as required.Can you give me and example to find matrices commute with matrices A that we know?

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