I have an operator $a$ in Euclidean space. How can i show that $a^{*}a$ and $a a^{*}$ have the same characteristic polynomials.
I've understood that these operators are self-adjoint and that their eigenvalues coincide.
Proof
$(\cdot,\cdot)$ is the inner product in the Hilbert space.
$e_1,...,e_n$ is a basis consisting of eigenvectors of $a^*a$. We know that $(e_i, e_j) = 0$ for $i \ne j$. It's true for self'adjoint operators. Now, we can take basis $ae_1,...,ae_n$ and $(ae_i, ae_j)=(a^*ae_i, e_j) = \lambda_i(e_i, e_j) = 0 $ for $i \ne j$
$aa^*(ae_i) = a(a^*ae_i) = a(\lambda_ie_i) = \lambda_i(ae_i)$. It means that every eigenvalue of $a^*a$ is an eigenvalue of $aa^*$. In the same way we can show that every eigenvalue of $aa^*$ is an eigenvalue of $a^*a$. So, $a^*a$ and $aa^*$ have the same eigenvalues