Suppose I have some linear transform $T: V \rightarrow W$ where $V \in \mathbb{C}^{N}$ and $W \in \mathbb{C}^{M}$ representing some Linear Time Invariant (LTI) system. Since the space is discrete, I can represent this transform as a matrix, ie:
$$ T = \begin{bmatrix} T_{11} & T_{12} & T_{13} & \dots & T_{1N} \\ T_{21} & T_{22} & T_{23} & \dots & T_{2N} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ T_{M1} & T_{M2} & T_{M3} & \dots & T_{MN} \end{bmatrix} $$
Now I know that the poles of this LTI system are given by the roots of $(\lambda I - T)$ In other words, the poles of the system are just the eigenvalues of $T$.
My question: What are the zeros of the system?
My thought is that the zeros of the system are simply the roots of $(\mathbf{v}I - T^{H})$ where $T^{H}$ means conjugate transpose. In other words, the zeros of the system are just the eigenvalues of the adjoint of $T$.
Am I correct? Is there a simple proof?