is it true, for any matrix $T \in \mathbb{C}^{n \times n}$ (with scalar product $\langle .,.\rangle$ and the norm $\left\Vert v\right\Vert = \langle v,v\rangle^{\frac{1}{2}}$), that, if $T$ is a contraction ($\left\Vert T \right\Vert = \sup\limits_{\left\Vert v \right\Vert = 1} \left\Vert Tv\right\Vert \leq 1$), the conjugate transpose $T^*$ also is a contraction? thanks in advance for your help
//Edit: $\langle .,.\rangle$ is supposed to be an arbitrary inner product
Yes. I prefer to denote by $A^H$ the conjugate transpose of a complex matrix $A$. Note that
$$ \left< Tv, Tv \right> = (Tv)^{H} Tv = v^{H} T^{H} T v. $$
The matrix $T^H T$ is Hermitian and positive-semidefinite. From the equation above, we see that $T$ is contraction if and only if $T^H T$ has no eigenvalues whose norm is greater than one. Similarly, $T^H$ is a contraction if and only if $T T^H$ has no eigenvalues whose norm is greater than one. Since $T^H T$ and $T T^H$ have the same eigenvalues, the result follows. The fact that $T^H T$ and $T T^H$ have the same eigenvalues follows from example from the existence of a SVD decomposition of a matrix.