# Theorem for Strictly Diagonally Dominant Matrices

I'm reading Matrix Analysis (Horn and Johnson). They state (but do not prove) the following theorem:

Theorem 6.1.10: Let $$A=[a_{ij}]\in M_n$$ be strictly diagonally dominant. Then

a) A is nonsingular.

b) If $$a_{ii}>0$$ for all $$i=1,\dots, n$$, then every eigenvalue of $$A$$ has positive real part.

I think I can show part a), but I'm confused with part b). From Gershgorin, I know that every eigenvalue of $$A$$ satisfies $$|\lambda - a_{ii}| \leq \sum_{j\neq i}|a_{ij}|$$, $$i=1,\dots,n$$. Since $$A$$ is strictly diagonally dominant, $$|a_{ii}| > \sum_{j\neq i} |a_{ij}|$$ for all $$i=1,\dots, n$$. I'm not sure how to use these facts to show the result.

• Draw a disk in the complex plane, of radius $\sum_{j\ne i}|a_{ij}$, centered at $a_{ii}$. Where does it intersect the imaginary axis? – kimchi lover Mar 16 at 22:26
• So the disc won't intersect the imaginary axis since $|a_{ii}| > \sum_{j\neq i} |a_{ij}|$, which implies that the disc has $x$ coordinates that are strictly positive; hence, the real part of $\lambda$ (which corresponds to the $x$ coordinate of the disc) must be positive? So the entire proof is geometric then. Is what I just said sufficient? – Dan P. Mar 16 at 23:06
• It depends on how picky your audience is. This level of detail is fine for informal discussions, or if you are lecturing people who understand what's going on. But for the persnickety kind of grader, or for the really ignorant lecture attendees, maybe more detail might help. – kimchi lover Mar 16 at 23:09
• If you want a "non-geometric proof", then here's a way to go: any $\lambda$ with non-positive real part will satisfy $$|a_{ii} - \lambda|^2 = (a_{ii} - \operatorname{Re}(\lambda))^2 + \operatorname{Im}(\lambda)^2 \geq (a_{ii} - \operatorname{Re}(\lambda))^2 \geq a_{ii}^2$$ which means that $\lambda$ must fall outside the $i$th disk. – Omnomnomnom Mar 17 at 3:25