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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?

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$||w||^2 = ||w^*w|| = ||\overline{w}w|| \leq ||I_n||$, unless I misunderstand the question. – Michael Sep 2 '11 at 21:47
    
(The third term should be $w$ times the complex conjugate of $w$; there seems to be a MathJax rendering bug.) – Michael Sep 2 '11 at 21:48
up vote 2 down vote accepted

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.

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thanks @Gjergji! – Hugo Chapdelaine Sep 2 '11 at 22:08

Your condition implies that for any $\vec{v} \in \mathbb{C}^n$, we have $\lVert \vec{v} \rVert > \lVert w\vec{v} \rVert$, except for $\vec{v} = 0$. Take $\vec{v}$ to be any of the standard basis vectors and conclude that the columns of $w$ have length less than 1, so its entries are at the least bounded. It seems that you don't need to assume $w = w^t$, if you write $I_n - ww^* > 0$ instead, and this seems more natural if you are talking about complex Hermitian matrices.

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