Gershgorin Circle Theorem
Intermediate
Published: 2026-04-09
1. Statement of the Theorem
For an $n \times n$ complex square matrix $A = (a_{ij})$, define the Gershgorin disc $D_i$ corresponding to each row $i$ as follows.
Gershgorin Disc
Center: $a_{ii}$ (the $i$-th diagonal entry)
Radius: $R_i = \displaystyle\sum_{j \neq i} |a_{ij}|$ (the sum of absolute values of the off-diagonal entries in row $i$)
$$D_i = \{ z \in \mathbb{C} \mid |z - a_{ii}| \le R_i \}$$Gershgorin Circle Theorem
Every eigenvalue of $A$ lies in the union of the $n$ Gershgorin discs.
$$\sigma(A) \subseteq \bigcup_{i=1}^{n} D_i$$Furthermore, if a group of $k$ mutually connected discs is separated from (has no intersection with) the remaining $n - k$ discs, then exactly $k$ eigenvalues (counted with multiplicity) lie within that group.
Here, "connected" refers to a connected component of the graph whose vertices are the discs and whose edges link pairs sharing at least one point (overlapping or tangent). Each separated component contains exactly as many eigenvalues (counted with multiplicity) as it has discs.
2. Proof
Let $\lambda$ be an eigenvalue of $A$ and $\mathbf{x} = (x_1, \ldots, x_n)^T$ the corresponding eigenvector. Choose the component with $|x_i| = \max_j |x_j|$ (so $|x_i| > 0$). By focusing on the component with the largest absolute value, we can use $|x_j| \le |x_i|$ to obtain a clean upper bound via the triangle inequality.
The $i$-th component of the eigenvalue equation $A\mathbf{x} = \lambda \mathbf{x}$ gives
$$\sum_{j=1}^{n} a_{ij} x_j = \lambda x_i$$Rearranging,
$$(\lambda - a_{ii}) x_i = \sum_{j \neq i} a_{ij} x_j$$Taking the absolute value of both sides and applying the triangle inequality,
$$|\lambda - a_{ii}| \cdot |x_i| = \left|\sum_{j \neq i} a_{ij} x_j\right| \le \sum_{j \neq i} |a_{ij}| \cdot |x_j|$$Since $|x_j| \le |x_i|$,
$$|\lambda - a_{ii}| \cdot |x_i| \le \sum_{j \neq i} |a_{ij}| \cdot |x_i| = R_i \cdot |x_i|$$Dividing by $|x_i| > 0$,
$$|\lambda - a_{ii}| \le R_i$$Therefore $\lambda \in D_i$. $\square$
3. Example: 3x3 Matrix
Example
Consider the following matrix.
$$A = \begin{pmatrix} 4 & 1 & 0 \\ 0.5 & 6 & 0.5 \\ 0 & 1 & 9 \end{pmatrix}$$Compute the Gershgorin disc for each row.
- Row 1: Center $a_{11} = 4$, radius $R_1 = |1| + |0| = 1$ → disc $D_1 = \{z : |z - 4| \le 1\}$, i.e., $[3, 5]$
- Row 2: Center $a_{22} = 6$, radius $R_2 = |0.5| + |0.5| = 1$ → disc $D_2 = \{z : |z - 6| \le 1\}$, i.e., $[5, 7]$
- Row 3: Center $a_{33} = 9$, radius $R_3 = |0| + |1| = 1$ → disc $D_3 = \{z : |z - 9| \le 1\}$, i.e., $[8, 10]$
$D_1$ and $D_2$ are tangent at the point $z = 5$, while $D_3$ is disjoint from them. Therefore:
- Exactly one eigenvalue lies in $D_3$ (in the range $[8, 10]$).
- Exactly two eigenvalues lie in $D_1 \cup D_2$ (in the range $[3, 7]$).
The actual eigenvalues are $\lambda_1 \approx 3.766$, $\lambda_2 \approx 6.071$, $\lambda_3 \approx 9.163$, all contained in one of the discs above. The second part of the theorem guarantees that $D_1 \cup D_2$ (a single connected component formed by the tangent discs) contains 2 eigenvalues, and the separated $D_3$ contains 1 eigenvalue. The fact that $\lambda_1$ and $\lambda_2$ individually fall into $D_1$ and $D_2$ respectively is not guaranteed by the theorem; it is merely observed numerically (at the point of tangency $z=5$ the two discs cannot be distinguished).
4. Application to Eigenvalue Localization
Invertibility of Diagonally Dominant Matrices
If a matrix $A$ is strictly diagonally dominant, i.e., for all $i$,
$$|a_{ii}| > \sum_{j \neq i} |a_{ij}|$$then no Gershgorin disc contains the origin (since $|0 - a_{ii}| = |a_{ii}| > R_i$ implies $0 \notin D_i$). Therefore $0$ is not an eigenvalue, and $A$ is invertible.
Sign of the Real Part of Eigenvalues
If all discs lie in the right half of the complex plane ($\mathrm{Re}(z) > 0$), then all eigenvalues have positive real parts. This is useful in stability analysis.
Upper Bound on the Spectral Radius
For the spectral radius $\rho(A) = \max_i |\lambda_i|$,
$$\rho(A) \le \max_{i} \left( |a_{ii}| + R_i \right)$$5. Complex Eigenvalues
Gershgorin discs are closed discs in the complex plane. Even when the matrix is real, eigenvalues can be complex. In such cases, the discs must be considered in the full complex plane.
Example: Real Matrix with Complex Eigenvalues
$$B = \begin{pmatrix} 0 & -5 \\ 5 & 0 \end{pmatrix}$$- Row 1: Center $0$, radius $5$ → $D_1 = \{z : |z| \le 5\}$
- Row 2: Center $0$, radius $5$ → $D_2 = \{z : |z| \le 5\}$
The eigenvalues are $\lambda = \pm 5i$, both lying on the boundary $|z| = 5$ of the discs. This illustrates the importance of considering the full complex plane, not just the real axis.
For real symmetric matrices, eigenvalues are always real, so it suffices to look at the intersection of the discs with the real axis. For general matrices, however, the discs should be interpreted as two-dimensional regions in the complex plane.
6. Row and Column Estimates
The Gershgorin theorem can be applied not only to $A$ but also to $A^T$. Since $A^T$ has the same eigenvalues as $A$, column-based discs also provide inclusion regions for the eigenvalues.
Column Gershgorin Discs
Center: $a_{jj}$ (the $j$-th diagonal entry)
Radius: $C_j = \displaystyle\sum_{i \neq j} |a_{ij}|$ (the sum of absolute values of the off-diagonal entries in column $j$)
$$D_j^C = \{ z \in \mathbb{C} \mid |z - a_{jj}| \le C_j \}$$Taking the intersection of the row-based and column-based estimates yields a tighter inclusion region.
$$\sigma(A) \subseteq \left(\bigcup_{i=1}^{n} D_i\right) \cap \left(\bigcup_{j=1}^{n} D_j^C\right)$$Example: Improved Estimate via Intersection
$$C = \begin{pmatrix} 10 & 0.1 & 0 \\ 2 & 5 & 0.1 \\ 0 & 2 & 1 \end{pmatrix}$$Row-based discs:
- $D_1$: center $10$, radius $0.1$ → $[9.9, 10.1]$
- $D_2$: center $5$, radius $2.1$ → $[2.9, 7.1]$
- $D_3$: center $1$, radius $2$ → $[-1, 3]$
Column-based discs:
- $D_1^C$: center $10$, radius $2$ → $[8, 12]$
- $D_2^C$: center $5$, radius $2.1$ → $[2.9, 7.1]$
- $D_3^C$: center $1$, radius $0.1$ → $[0.9, 1.1]$
The row-based disc $D_3 = [-1, 3]$ is wide, but the column-based disc $D_3^C = [0.9, 1.1]$ is very narrow. By taking the intersection, the eigenvalue near $1$ can be localized to $[0.9, 1.1]$.
7. Applications
- Pre-estimation in numerical computation: Estimating the spectral radius in advance to determine the convergence of iterative methods.
- Stability analysis: Verifying that the eigenvalues of matrices arising from discretization of differential equations lie within the stability region.
- Preconditioner design: Improving the condition number by diagonal scaling that shrinks the Gershgorin discs.
- Invertibility test: Diagonal dominance immediately implies invertibility.
- Perturbation analysis: Qualitatively assessing how much eigenvalues shift when small perturbations are applied to the matrix entries.
8. References
- Wikipedia, "Gershgorin circle theorem"
- R. S. Varga, Gershgorin and His Circles, Springer, 2004.
- G. H. Golub, C. F. Van Loan, Matrix Computations, 4th ed., Johns Hopkins University Press, 2013.
- R. A. Horn, C. R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2012.
Frequently Asked Questions
What is the Gershgorin Circle Theorem?
Every eigenvalue of an $n \times n$ matrix $A$ lies in at least one Gershgorin disc $D_i = \{ z \in \mathbb{C} : |z - a_{ii}| \le \sum_{j \neq i} |a_{ij}| \}$. This means eigenvalue locations can be estimated from the matrix entries alone, without computing eigenvalues directly.
What are the applications of the Gershgorin Circle Theorem?
It is used for preliminary eigenvalue estimation in numerical computation, invertibility verification (diagonally dominant matrices are invertible), stability analysis in control theory, and convergence assessment of iterative methods. It provides a low-cost pre-check without actually computing eigenvalues.