Prove that all the eigenvalues of $A$ are real.(Gershgorin 's Theorem)

Question

The question and its answer are given below:

And this is the Greshgorin theorem:

My questions are:

1- I do not understand why a consequence of Greshgorin's theorem is that if a circle is disjoint then it contains 1 eigenvalue. could anyone explain this for me, please?

2- Why in a characteristic polynomial with real coefficients the complex roots occur in pairs?

3- Why in the last paragraph the $\lambda $ and $\bar{\lambda}$ inside the disk should be the same? and why these leads to all the eigenvalues are real? could anyone help me in answering those questions, please?

Answer 1

That's a very 'legitimate question', but you could only post a question and work from that. I 'll try

$1$. That's the theorem the proof uses continuity argument.

$2$. If a polynomial $P(x)$ has real coefficients. If $P(z)=0$ then taking conjugate $\overline{P(z)}=P(\bar{z})=0$ so both $z$ and $\bar{z}$ are roots.

$3$. Imagine the disk $C$ with center $a\in x'Ox$ on the real line it is easy to see that for any complex number $z\in C$ $\implies$ $\bar{z}\in C$.

Answer 2

The disjointness of the circles is an additional constraint imposed by your question. It is not a general feature of real matrices.