So let us define the generalized disc of degree $n$ as
$$
\mathbb{D}\_n:=\{w\in M\_{n\times n}(\mathbb{C}):w=w^t, I\_n-w\overline{w}>0\}.
$$
For a Hermitian matrix $A$, the notation $A>0$ means that it is positive definite.


**Q**: So how do you prove cleanly that $\mathbb{D}\_n$ is bounded?



Looking for a simple proof that the generalized disc is bounded
The diagonal entries of a positive definite matrix are real and non-negative. If we let the rows of $w$ be $v\_1,\dots,v\_n$, then the diagonal entries of $I\_n-w\overline{w}$ are $1-v\_i\overline{v\_i}=1-\sum\_{j=1}^n|w\_{ij}|^2$. So in particular $|w\_{ij}|\le 1$. This implies that $\mathbb D\_n$ is bounded.


