Levy-Desplanques theorem

A strictly diagonally dominant matrix is non-singular. In other words, let $A\in {𝐂}^{n,n}$ be a matrix satisfying the property

$$\left|a_{ii}\right|>\sum _{j\ne i}\left|a_{ij}\right|\mathit{\hspace{1em}\hspace{1em}}\forall i;$$

then $det(A)\ne 0$.

Proof:Let $det(A)=0$; then a non-zero vector $𝐱$ exists such that $A𝐱=\mathrm{𝟎}$; let $M$ be the index such that $\left|x_M\right|=\max (\left|x_1\right|,\left|x_2\right|,\mathrm{\cdots },\left|x_n\right|)$, so that $\left|x_j\right|\le \left|x_M\right|\mathit{\hspace{1em}}\forall j$; we have

$a_{M1}{x}_1+a_{M2}{x}_2+\mathrm{\cdots }+a_{MM}{x}_M+\mathrm{\cdots }+a_{Mn}{x}_n=0$

which implies:

$\left|a_{MM}\right|\left|x_M\right|=\left|a_{MM}{x}_M\right|=\left|{\sum }_{j\ne M}{a}_{Mj}{x}_j\right|\le {\sum }_{j\ne M}\left|a_{Mj}\right|\left|x_j\right|\le \left|x_M\right|{\sum }_{j\ne M}\left|a_{Mj}\right|$

that is

$\left|a_{MM}\right|\le {\sum }_{j\ne M}\left|a_{Mj}\right|,$

in contrast with strictly diagonally dominance definition.$\mathrm{\square }$

Remark:the Levy-Desplanques theorem is equivalent to the well-known Gerschgorin circle theorem. In fact, let’s assume Levy-Desplanques theorem is true, and let $A$ a $n\times n$ complex-valued matrix, with an eigenvalue $\lambda$; let’s apply Levy-Desplanques theorem to the matrix $B=A-\lambda I$, which is singular by definition of eigenvalue: an index $i$ must exist for which $\left|a_{ii}-\lambda \right|=\left|b_{ii}\right|\le {\sum }_{j\ne i}^n\left|b_{ij}\right|={\sum }_{j\ne i}^n\left|a_{ij}\right|$, which is Gerschgorin circle theorem.On the other hand, let’s assume Gerschgorin circle theorem is true, and let $A$ be a strictly diagonally dominant $n\times n$ complex matrix. Then, since the absolute value of each disc center $\left|a_{ii}\right|$ is strictly greater than the same disc radius ${\sum }_{j\ne i}^n\left|a_{ij}\right|$, the point $\lambda =0$ can’t belong to any circle, so it doesn’t belong to the spectrum of $A$, which therefore can’t be singular.