# Find the Dimension of a Subspace of Matrices

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Euge
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Given a complex matrix ##A\in M_n(\mathbb{C})##, let ##X_A## be the subspace of ##M_n(\mathbb{C})## consisting of all the complex matrices ##M## commuting with ##A## (i.e., ##MA = AM##). Suppose ##A## has ##n## distinct eigenvalues. Find the dimension of ##X_A##.

topsquark and jbergman
Suppose that you found the basis where ##A## is diagonal and it looks as
##A=\operatorname{diag}\{\lambda_1,...,\lambda_n\}##

Then the commutativity condition in index notation turns into very simple one
##(\lambda_k-\lambda_i)M^i_k=0##
Here ##i## and ##k## run from ##1## to ##n##. For ##i=k##, ##M^i_i## can be any, however all other values must zero. Hence you have the space of diagonal matrices, which dimension is ##n##.

topsquark
Korybut said:
Suppose that you found the basis where ##A## is diagonal and it looks as
##A=\operatorname{diag}\{\lambda_1,...,\lambda_n\}##

Then the commutativity condition in index notation turns into very simple one
##(\lambda_k-\lambda_i)M^i_k=0##
Here ##i## and ##k## run from ##1## to ##n##. For ##i=k##, ##M^i_i## can be any, however all other values must zero. Hence you have the space of diagonal matrices, which dimension is ##n##.

I've always felt like it's a flaw of linear algebra that more matrices don't commute with diagonal matrices

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