Let `M` be a complex square matrix of size `n \in \mathbb{N}`.
We define for each `k\in{1,\ldots,n}`:
`D_k = {x\in\mathbb{C} ; |x-a_{kk}|\leq sum_(i=1, i\ne k)^n |a_{ki}|}.`
Then, the spectrum `Sp(M)` of `M` verifies the localization condition:
`Sp(M) \subset \bigcup_{k=1}^n D_k.`
Proof:
Let `M` be a complex square matrix of size `n \in \mathbb{N}`, and `\lambda\in\mathbb{C}` one of its eigenvalues
associated to the eigenvector `x=(x_k)_{1\leq k\leq n}\in\mathbb{C}^n`.
Let `k_0\in{1,\ldots,n}` be the index suh that `\forall k\in{1,\ldots,n}`, ` x_k \leq x_{k_0}`.
We have then :
`sum_(i=1, i\ne k_0)^n a_{k_0i}x_i = (\lambda - a_{k_0k_0}) x_{k_0k_0}.`
Hence
`|\lambda - a_{k_0k_0}| \leq sum_(i=1, i\ne k_0)^n |a_{k_0i}x_i/x_{k_0}|.`
Finally
`|\lambda - a_{k_0k_0}| \leq sum_(i=1, i\ne k_0)^n |a_{k_0i}|.`
If we define for each `k\in\{1,\ldots,n\}`:
`D_k = {x\in\mathbb{C} ; |x-a_{kk}|\leq sum_(i=1, i\ne k)^n |a_{ki}|}`
the theorem follows.
`\square`