If $f$ is diagonalizable, but its spectrum is not simple, prove that the vector space dimension of the operators that commute with $f$ is equal to the sum of the squares of the algebraic multiplicities of each eigenvalue.
2026-03-29 17:42:54.1774806174
Algebraic multiplicities and dimension
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Hint: First show that any operator A that commutes with f has to preserve its eigenspaces (the eigenspaces of f are invariant under A). Then show that any operator that preserves the eigenspaces of f commutes with f. This tells you about the block form of the matrix of A in an eigenbasis for f, and tells you that any matrix with that block form (in the chosen eigenbasis) commutes with f.