Before computing eigenvalues, it helps to know roughly where they are: to choose a shift, to check an answer, or to prove a matrix is invertible. Gershgorin's theorem gives such a picture straight from the entries, with no characteristic polynomial at all.
🔑 Key idea
For each row $i$ of an $n \times n$ matrix $A$, draw the disk in the complex plane with center $a_{ii}$ and radius $$r_i = \sum_{j \ne i}|a_{ij}|,$$ the sum of the sizes of the other entries in that row. Every eigenvalue of $A$ lies in at least one of these disks. Moreover, if $k$ disks form a group that does not touch the others, that group holds exactly $k$ eigenvalues (counted with multiplicity).
💡 Why it works
Let $A\mathbf{v} = \lambda\mathbf{v}$ with $\mathbf{v} \ne \mathbf{0}$, and pick $i$ so that $|v_i|$ is the largest entry of $\mathbf{v}$ in size. Row $i$ of $A\mathbf{v} = \lambda\mathbf{v}$ says $\sum_j a_{ij}v_j = \lambda v_i$, so $$(\lambda - a_{ii})v_i = \sum_{j \ne i}a_{ij}v_j.$$ Take sizes and divide by $|v_i|$: $|\lambda - a_{ii}| \le \sum_{j \ne i}|a_{ij}|\frac{|v_j|}{|v_i|} \le r_i$, since each ratio is at most $1$. So $\lambda$ is in disk $i$. (The counting statement comes from shrinking the off-diagonal entries to $0$ continuously: the eigenvalues move continuously from the centers and cannot jump between separate groups.)
Row 1: center $10$, radius $|1| + |-1| = 2$. Row 2: center $0$, radius $1 + 2 = 3$. Row 3: center $-6$, radius $0 + 1 = 1$.
On the real line these are $[8, 12]$, $[-3, 3]$ and $[-7, -5]$: three separate disks, so each holds exactly one eigenvalue.
Each eigenvalue is real: a complex one would come with its conjugate in the same disk (the disks are symmetric about the real axis), making two. The eigenvalues are about $10.09$, $0.23$ and $-6.33$.
Answer: one in each of $[8, 12]$, $[-3, 3]$, $[-7, -5]$
✏️ Worked example: Proving a matrix is invertible
Show that $B = \begin{bmatrix} 5 & 2 & -1 \\ 1 & -4 & 2 \\ 0 & 3 & 6 \end{bmatrix}$ is invertible.
Disks: center $5$, radius $3$; center $-4$, radius $3$; center $6$, radius $3$.
None contains $0$ (each $|a_{ii}|$ is bigger than its radius: the matrix is strictly diagonally dominant). So $0$ is not an eigenvalue, and $B$ is invertible.
Answer: no disk contains 0
✏️ Worked example: A disk with no eigenvalue
Draw the disks for $C = \begin{bmatrix} 0 & 1 \\ 4 & 2 \end{bmatrix}$ and find which ones hold the eigenvalues.
Row 1: center $0$, radius $1$. Row 2: center $2$, radius $4$. The disks overlap.
Eigenvalues: $\lambda^{2} - 2\lambda - 4 = 0$, so $\lambda = 1 \pm \sqrt{5} \approx 3.24$ and $-1.24$.
Both lie in the second disk, and neither is within $1$ of $0$: the first disk holds none. The theorem promises only that the union of the overlapping disks holds two.
Answer: the first disk is empty
Gershgorin disks (one per row) and the actual eigenvalues (dots). Every dot sits in some disk; a group of k separate disks holds k dots.
Because $A$ and $A^{T}$ have the same eigenvalues, column sums give a second set of disks, and the eigenvalues lie in both unions. For symmetric matrices the eigenvalues are real, so the disks become intervals on the real line.
⚠️ Common mistake
Including $|a_{ii}|$ in the radius, or expecting each disk to hold exactly one eigenvalue. The radius adds only the off-diagonal entries of the row. And when disks overlap, only the group as a whole has a guaranteed count: one disk of the group can hold none. Check: the "exactly one" conclusion needs a disk that touches no other.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.