Do T and T* have the same eigenvalues with the same algebraic multiplicity?

Question

I know that the eigenvalues of T* are the conjugates of T's eigenvalues , but how can I see each eigenvalue of T and it's conjugate , the eigenvalue of T*, have the same algebraic multiplicity?

Answer 1

Using the Schur decomposition: If $T=QUQ^*$ is the Schur decomposition of $T$, then $T^*=QU^*Q^*$. Since both $U$ and $U^*$ are triangular (upper/lower), their diagonal entries equal to the eigenvalues of $T$ and $T^*$, respectively. Then note that the diagonal entries of $U^*$ are complex conjugate entries of the diagonal of $U$.

Using the characteristic polynomial: If $\lambda\in\mathbb{C}$ is an eigenvalue of $T$, then $\det(\lambda I-T)=0$. But since $\det(A)=\overline{\det(A^*)}$, we have that $0=\overline{\det(\lambda I-T)}=\det(\overline{\lambda}I-T^*)$. Hence $\overline{\lambda}$ is the root of $\det(tI-T^*)$, the characteristic polynomial of $T^*$.

You can see that the algebraic multiplicities of eigenvalues of $T$ and the complex conjugate eigenvalues of $T^*$ are equal from both approaches. It is easy to see this from the Schur form. From the characteristic polynomial, it is easy to get this result as well. Note that $$ p_T(t)=\det(tI-T)=\prod_{i=1}^k(t-\lambda_i)^{\mu_i}, $$ where $\{\lambda_i\}_{i=1}^k$ are the eigenvalues of $T$ and $\{\mu_i\}_{i=1}^k$ are their algebraic multiplicities ($\sum_{i=1}^k\mu_i=n$). Since $p_{T^*}(\overline{t})=\det(\overline{t}I-T^*)=\overline{\det(tI-T)}=\overline{p_T(t)}$ (and thus $p_{T^*}(t)=\overline{p_T(\overline{t})}$), you have that $$ p_{T^*}(t)=\overline{\prod_{i=1}^k(\overline{t}-\lambda_i)^{\mu_i}}=\prod_{i=1}^k(t-\overline{\lambda_i})^{\mu_i}. $$