Invertible matrix if the sum of the absolute values of a column is lower than the absolute of the diagonal value

Question

I've been trying to prove that if I have $A=[a_{ij}]_{n \times n}$ therefore, if;

Then the matrix is invertible.

Can anyone help me and give me some vocabulary (If I'm not wrong, there is a name for this type of matrix, isn't there?)

Answer 1

We will prove it by contradiction. Assume we have a vector $x \neq 0$ such that $Ax = 0$. This means that $A$ is singular, i.e. non-invertible. This means that for all $k$, we have \begin{equation} \sum_{j=1}^n a_{kj} x_j = 0 \qquad \forall k = 1 \ldots n \end{equation} Let's choose $k=m$, where $\vert x_m \vert$ is the largest amongst all and discuss \begin{equation} -a_{mm}x_m = \sum_{j \neq m} a_{mj} x_j \end{equation}

i.e. \begin{equation} -a_{mm} = \sum_{j \neq m} a_{mj} \frac{x_j}{x_m} \end{equation} Take the absolute value on both sides, we get \begin{equation} \vert a_{mm} \vert= \Big\vert \sum_{j \neq m} a_{mj} \frac{x_j}{x_m} \Big\vert \leq \sum_{j \neq m} \vert a_{mj}\vert . \Big\vert \frac{x_j}{x_m} \Big\vert \end{equation} But $ \Big\vert \frac{x_j}{x_m} \Big\vert \leq 1$, so \begin{equation} \vert a_{mm} \vert \leq \sum_{j \neq k} \vert a_{mj} \vert \end{equation} But this is a contradiction of the assumption. Hence, $A$ has to be invertible.