Interval of eigenvalues using Gershgorin circles

[mathmari]

Hey!

We have the matrix $$A=\begin{pmatrix}2 & 0.4 & -0.1 & 0.3 \\ 0.3 & 3 & -0.1 & 0.2 \\ 0 & 0.7 & 3 & 1 \\ 0.2 & 0.1 & 0 & 4\end{pmatrix}$$ We get the row Gershgorin circles: $$K_1=\{z\in \mathbb{C} : |z-2|\leq 0.8 \} \\ K_2=\{z\in \mathbb{C} : |z-3|\leq 0.6 \} \\ K_3=\{z\in \mathbb{C} : |z-3|\leq 1.7 \} \\ K_4=\{z\in \mathbb{C} : |z-4|\leq 0.3 \} $$ and the column Gershgorin circles: $$K_1'=\{z\in \mathbb{C} : |z-2|\leq 0.5 \} \\ K_2'=\{z\in \mathbb{C} : |z-3|\leq 1.2 \} \\ K_3'=\{z\in \mathbb{C} : |z-3|\leq 0.2 \} \\ K_4'=\{z\in \mathbb{C} : |z-4|\leq 1.5 \} $$

To get the intervals of the eigenvalues, we have to look at the union of all row circles and the union of all column circles, or not?

Then at the result we have to take the intersection, or not?

Till now I have seen only examples where $A=A^T$ and so the row and column circles are the same, and so I am confused here.

(Wondering)

 

[Ackbach]

According to the wiki article, the way I see it, ALL the eigenvalues of $A$ must lie in
$$\displaystyle\cup_{i=1}^4K_i,$$
as you've defined them, AND ALL the eigenvalues of $A$ must lie in
$$\displaystyle\cup_{i=1}^4K_i'.$$
Therefore, all the eigenvalues of $A$ must lie in
$$\displaystyle\left(\cup_{i=1}^4K_i\right)\cap\left(\cup_{i=1}^4K_i'\right).$$
I don't think you can say that the eigenvalues must lie in
$$\cup_{i=1}^4(K_i\cap K_i'),$$
which is what I would be tempted to think, unless you can show those are equivalent.

 

[I like Serena]

Hey mathmari!

I have just read the wiki article about Gershgorin circles.
As I understand it, and this is verifiable from the proofs that are given in the article:

• All eigenvalues are in the union of all row circles.
• All eigenvalues are in the union of all column circles.
• If the row circles can be divided into disjoint unions, then each disjoint union contains exactly as many eigenvalues as there are circles in it.
  Same for the column circles.

(Cool)

In this case there is a row circle that overlaps all other row circles.
And there is also a column circle that overlaps all others.
So I think the only thing we can say, is that all eigenvalues are in the intersection of the row-union and the column-union. (Thinking)

 

[I like Serena]

So this is what it looks like:
\begin{tikzpicture}
\newcommand{\coordinates}{
\draw[help lines] (0,-2) grid (6,2);
\draw[-stealth] (0,0) -- (6.4,0) node[label=Re]{};
\draw[stealth-stealth] (0,-2.3) -- (0,2.3) node[label=right:Im]{};
\draw[fill] foreach \i in {1,...,6} { (\i,0) node[yshift=-2cm,label=below:\i]{} };
\draw foreach \i in {-2,...,2} { (0,\i) node[label=left:\i]{} };
}

\newcommand{\rowcircles}{
(2,0) circle (0.8)
(3,0) circle (0.6)
circle (1.7)
(4,0) circle (0.3)
}
\newcommand{\colcircles}{
(2,0) circle (0.5)
(3,0) circle (1.2)
circle (0.2)
(4,0) circle (1.5)
}

\begin{scope}
\coordinates
\filldraw[red, fill opacity=.3] \rowcircles node[above left] at (2.3,1.4) {row circles};
\filldraw[blue, fill opacity=.3] \colcircles node[above right] at (4,1.4) {column circles};
\fill
(2,0) circle (1pt)
(3,0) circle (1pt)
(4,0) circle (1pt);
\end{scope}

\begin{scope}[xshift=8cm]
\coordinates
\clip\rowcircles;
\path[fill=red!50!blue, fill opacity=.6] \colcircles;
\draw (1.8617,0) circle (2pt)
(4.0368,0) circle (2pt)
(3.0508,0.2407) circle (2pt)
(3.0508,-0.2407) circle (2pt);
\end{scope}
\end{tikzpicture}
The small circles on the right are the actual eigenvalues. (Malthe)