Suppose that $A$, $B$ and $AB$ are Hermitian matrices. show that every eigenvalue of $AB$ is the product of an eigenvalue of $A$ and eigenvalue of $B$.

I'm stuck on this problem. I could show, any eigenvector of $A$ is an eigenvector of $B$ and $AB$ as well, but I don't know why any eigenvalue of $AB$ is the product of an eigenvalue of $A$ and eigenvalue of $B$. Here is a link, which I was told that's a answer for my question, but the assumptions seems slightly different to me and honestly I didn't get it. I appreciate any hint to start this problem.

First $(AB)^H=B^HA^H=BA=AB$, which implies $A$ and $B$ commute.

Let $\lambda \in spec(B)$ and let $v$ be an eigenvecotr of $B$ associated with $\lambda$ so $Bv=\lambda v$. Hence,

$$Bv=\lambda v$$ $$B(Av)=A(Bv)=\lambda Av,$$ so $Av\in W_\lambda$, the eigenspace of $B$ associated with $\lambda$, Since, both $v$ and $Av$ are in $W_\lambda$, there exists a scalar $k$ satisfying $Av=kv$, this implies $k\in spec(A)$ and it's associated with eigenvector $v$ of $A.$ Then, $AB(v)=\lambda kv.$

Let $k\in Spec(A)$, then there exists an eigenvector $v$ of $A$ associated with $k$, it's easy to show $v$ is an eigenvector of $A$, so there exists a $\lambda\in Spec(B)$ associated with $v$. Thus $v$ is an eigenvector of $AB$ associated with the eigenvalue $\lambda k\in spec(AB).$

  • $\begingroup$ The product of two Hermitian matrices is not necessarily Hermitian $\endgroup$ – polfosol Jan 16 '17 at 6:28
  • 1
    $\begingroup$ @polfosol I didn't say the product of two Hermitian matrices is Hermitian. I said suppose that $A$, $B$ and $AB$ are hermtian matrices. $\endgroup$ – Parisina Jan 16 '17 at 7:15
  • $\begingroup$ Possible duplicate of Eigenvalues of product of two hermitian matrices $\endgroup$ – polfosol Jan 16 '17 at 7:28

Since $A$ and $B$ commute and they are diagonalizable, they can be simultaneously diagonalized. More explicitly, let us denote the distinct eigenvalues of $A$ by $\lambda_1, \dots, \lambda_k$ and write

$$ \mathbb{C}^n = \oplus_{i=1}^k V_{\lambda_i}(A) $$

where $V_{\lambda_i}(A) = \ker(A - \lambda_i I)$ is the eigenspace of $A$ associated to the eigenvalue $\lambda_i$. Your argument shows that each $V_{\lambda_i}(A)$ is $B$-invariant. Restricting $B$ to each $V_{\lambda_i}(A)$ we get a diagonalizable operator (for example, because $B|_{V_{\lambda_i}}$ is Hermitian with respect to the induced inner product on the subspace $V_{\lambda_i}$) and so each $V_{\lambda_i}(A)$ splits into a direct sum of eigenspaces of $B$. Thus, by choosing our vectors appropriately, we can obtain a basis $v_1,\dots,v_n$ of $\mathbb{C}^n$ such that $Av_i = \lambda_i v_i$ and $Bv_i = \mu_i v_i$ for some $\lambda_i,\mu_i \in \mathbb{R}$ (the indices here are not necessarily the same as before and there can be repetitions). But then

$$ (AB)(v_i) = A(\mu_iv_i) = \lambda_i \mu_i v_i $$

so the eigenvalues of $AB$ are products of the eigenvalues of $A$ and $B$.


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