Jim answered the finite-dimensional case very nicely; let me address the infinite-dimensional case.
If you don't specify some conditions on your operator, the answer is no. Consider $H=\ell^2(\mathbb{N})$, and let $T$ be the "reverse shift" operator, given by $ T(a_1,a_2,\ldots)=(a_2,a_3,\ldots). $ Then it is easy to see that every $\lambda\in\mathbb{C}$ with $|\lambda|<1$ is an eigenvalue with eigenvector $(\lambda,\lambda^2,\lambda^3,\ldots)$. More properly, one is free to choose the first coordinate, but that is irrelevant here.
Now, $T^*$ is the usual shift $ T^*(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots) $ and it has no eigenvalues (and so no eigenvectors).