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Let $A, B$ be two n by n matrices. Suppose that $A, B$ are diagonalizable and $AB=BA$. Do we have the following: eigenvalues of $AB$ are the products of eigenvalues of $A, B$?

Edit: Do we have the following: eigenvalues of $AB$ are products of eigenvalues of $A, B$?

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    $\begingroup$ Maybe you want to say that the eigenvalues of $AB$ are products of eigenvalues of $A$ and $ B$, without the article " the ". In this case this is true because they are simultanously diagonalizable. $\endgroup$
    – Felix Klein
    Commented Jun 27, 2017 at 11:57
  • $\begingroup$ Consider for example $A=({}^1_0\,{}^0_2)$ and $B=({}^3_0\,{}^0_5)$. "The" products of eigenvalues would be $\{3,5,6,10\}$, but the only ones of these products that are actually eigenvalues of $AB$ are $3$ and $10$. $\endgroup$
    – hmakholm left over Monica
    Commented Jun 27, 2017 at 12:06
  • $\begingroup$ @AmrAhmad, thank you very much. Yes, I mean "products of eigenvalues of $A$ and $B$". $\endgroup$
    – LJR
    Commented Jun 27, 2017 at 12:11
  • $\begingroup$ @HenningMakholm, thank you very much. $\endgroup$
    – LJR
    Commented Jun 27, 2017 at 12:12

2 Answers 2

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$A,B$ do not need to be diagonalizable.

Proposition. Let $A,B\in M_n(\mathbb{C})$ and $spectrum(A)=(\lambda_i)_i$.

If $AB=BA$, then, there is an ordering of $spectrum(B)$: $(\mu_i)_i$, s.t. $spectrum(A+B)=(\lambda_i+\mu_i)_i$ and $spectrum(AB)=(\lambda_i\mu_i)_i$.

Proof. $A,B$ are simultaneously triangularizable.

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Since $A$ and $B$ commute, they can be diagonalized simultaneously, i.e. there is an invertible matrix $S$ such that $$ A = S^{-1} \mathrm{diag}(\lambda_1, \dots, \lambda_n) S, \quad B = S^{-1} \mathrm{diag}(\mu_1, \dots, \mu_n) S.$$ where $\mathrm{diag}$ is a diagonal matrix with the specified diagonal entries (the eigenvalues of $A$ resp. $B$). Then $$ AB = S^{-1} \mathrm{diag}(\lambda_1, \dots, \lambda_n) SS^{-1} \mathrm{diag}(\mu_1, \dots, \mu_n) S = S^{-1}\mathrm{diag}(\lambda_1, \dots, \lambda_n)\mathrm{diag}(\mu_1, \dots, \mu_n) S \\ = S^{-1}\mathrm{diag}(\lambda_1 \mu_1, \dots, \lambda_n \mu_n) S $$ and the eigenvalues are indeed, as you suspected, the product of the eigenvalues. Observe that the order in which the eigenvalues are multiplied depend on the matrix $S$, i.e. is not arbitrary.

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