We are given the following theorem of which we need only know the result.
Theorem
Suppose an $n\times n$ matrix $A= (a_{ij})$ is tri-diagonal with $a_{i,i-1}a_{i,i+1} \neq 0$, for each $i=2,\dots,n-1$. If $|a_{1,1}| > |a_{1,2}| , |a_{i,i}| > |a_{i,i-1}| + |a_{i,i+1}|$ for $i=2, \dots, n-1$ and $|a_{n,n}| > |a_{n,n-1}|$ then $A^{-1}$ exists
I would, however, still like to see the proof of this theorem, as I would like to have some deeper knowledge of why this is true, rather than simply learning a result.
Can someone please provide me with a proof? Or send me a link to a site which proves it?
Let $A$ be a square $n\times n$ matrix and $A=D+B$, where $D$ is the diagonal part of $A$ and let $A$ be strictly diagonally dominant (necessarily, $D$ is nonsingular), that is, $\|D^{-1}B\|_\infty<1$. Since $A=D(I-D^{-1}B)$, $A$ is nonsingular if and only if $I-D^{-1}B$ is nonsingular. Assume that $I-D^{-1}B$ is singular, then $x=D^{-1}Bx$ for some nonzero $x$ and hence $\|x\|_\infty=\|D^{-1}Bx\|_\infty\leq\|D^{-1}B\|_\infty\|x\|_\infty$ which implies $\|D^{-1}B\|_\infty\geq 1$. This contradicts $\|D^{-1}B\|_\infty<1$ and hence $I-D^{-1}B$ is nonsingular and $A$ is as well.