On the localization of eigenvalues : Gershgorin and Hadamard

Let $\lambda\in\operatorname{Sp}(U)$. By definition of the spectrum, there exists an eigenvector

$$Z= \begin{pmatrix} z_1\\ \vdots\\ z_n \end{pmatrix} \in\mathbb{C}^n\setminus\{0\}$$

such that

$$UZ=\lambda Z.$$

Since $Z\neq0$, the set

$$\left\{|z_x|\;\middle|\;x\in\llbracket 1,n\rrbracket\right\}$$

is a non-empty finite subset of $\mathbb{R}$. It therefore has a maximum. Choose

$$i\in\llbracket 1,n\rrbracket$$

such that

$$|z_i| = \max_{x\in\llbracket 1,n\rrbracket}|z_x|.$$

Since $Z\neq0$, this maximum is strictly positive, and consequently

$$z_i\neq0.$$

The equality $UZ=\lambda Z$ gives, by considering its $i$-th component,

$$\sum_{j=1}^n u_{ij}z_j=\lambda z_i.$$

Isolating the term corresponding to $j=i$, we obtain

$$(\lambda-u_{ii})z_i = \sum_{\substack{1\leqslant j\leqslant n\\j\neq i}}u_{ij}z_j.$$

Since $z_i\neq0$, we can divide by $z_i$, whence

$$\lambda-u_{ii} = \sum_{\substack{1\leqslant j\leqslant n\\j\neq i}} u_{ij}\frac{z_j}{z_i}.$$

Taking absolute values and using the triangle inequality, we obtain

$$|\lambda-u_{ii}| \leq \sum_{\substack{1\leqslant j\leqslant n\\j\neq i}} |u_{ij}| \frac{|z_j|}{|z_i|}.$$

By the definition of $i$,

$$\forall j\in\llbracket 1,n\rrbracket,\qquad |z_j|\leqslant|z_i|,$$

and hence

$$\forall j\in\llbracket 1,n\rrbracket,\qquad \frac{|z_j|}{|z_i|}\leqslant1.$$

It follows that

$$|\lambda-u_{ii}| \leqslant \sum_{\substack{1\leqslant j\leqslant n\\j\neq i}}|u_{ij}| = \Lambda_i.$$

By the definition of $D_i$, we therefore have

$$\lambda\in D_i.$$

Thus, every eigenvalue of $U$ belongs to one of the disks $D_i$. Consequently,

$$\operatorname{Sp}(U) \subset \bigcup_{i=1}^nD_i \; .$$

Suppose that

$$\forall i\in\llbracket1,n\rrbracket,\qquad |u_{ii}|> \sum_{\substack{1\leqslant j\leqslant n\\j\neq i}}|u_{ij}|.$$

For every $i\in\llbracket1,n\rrbracket$, we then have

$$|u_{ii}|>\Lambda_i.$$

Consequently,

$$0\notin D_i.$$

Indeed, if $0\in D_i$, then, by the definition of $D_i$,

$$|u_{ii}|\leqslant \Lambda_i,$$

which would contradict the hypothesis.

Thus,

$$0\notin\bigcup_{i=1}^nD_i.$$

Now, by the first question,

$$\operatorname{Sp}(U) \subset \bigcup_{i=1}^nD_i.$$

We deduce that

$$0\notin\operatorname{Sp}(U).$$

By definition of the spectrum,

$$0\notin\operatorname{Sp}(U) \quad\Longleftrightarrow\quad \ker(U)=\{0\}.$$

The linear map associated with $U$ is therefore injective. Since it is an endomorphism of the finite-dimensional vector space $\mathbb C^n$, it is then bijective. Consequently, $U$ is invertible.

Thus,

$$U\in GL_n(\mathbb C)\; .$$