# Interval of eigenvalues using Gershgorin circles

• MHB
• mathmari
In summary, all the eigenvalues of $A$ must lie in the intersection of the row-union and the column-union.
mathmari
Gold Member
MHB
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)

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.

Hey mathmari!

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)

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)

## 1. What are Gershgorin circles?

Gershgorin circles are geometric representations of the eigenvalues of a square matrix. They are formed by drawing circles on the complex plane, with the center of each circle at the diagonal element of the matrix and the radius equal to the sum of the absolute values of the remaining elements in the same row or column.

## 2. How are Gershgorin circles used to find the interval of eigenvalues?

The interval of eigenvalues can be determined by looking at the intersection of the Gershgorin circles on the complex plane. The eigenvalues of the matrix will lie within the union of these circles, providing an upper and lower bound for the interval of eigenvalues.

## 3. Why is it important to find the interval of eigenvalues using Gershgorin circles?

Finding the interval of eigenvalues allows us to understand the behavior and stability of a matrix. This information is useful in many applications, such as solving systems of linear equations, analyzing dynamical systems, and designing control systems.

## 4. Can Gershgorin circles be used for non-square matrices?

No, Gershgorin circles can only be used for square matrices. However, there are similar methods, such as the Bauer-Fike theorem, that can be used to find the interval of eigenvalues for non-square matrices.

## 5. Are there any limitations to using Gershgorin circles to find the interval of eigenvalues?

Yes, Gershgorin circles can only provide an approximation of the interval of eigenvalues and may not always give the exact values. Additionally, they only provide information about the eigenvalues and not the corresponding eigenvectors of the matrix.

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