Linear-Algebra-Done-Right · Unit 13 · Video 4 · Interactive Practice

The Gershgorin Disk Theorem: Where Eigenvalues Live

IKey Formulas

Formula	Name	What it says
$\{z \in \mathbf{F} : \|z - A_{j,j}\| \le \sum_{k \ne j} \|A_{j,k}\|\}$	Gershgorin disk of row $j$	For $T \in \mathcal{L}(V)$ with $A = \mathcal{M}(T)$ in a basis $v_1, \dots, v_n$ : centre $A_{j,j}$ , radius the total size of the other entries of row $j$ (a closed interval when $\mathbf{F} = \mathbf{R}$ )
each eigenvalue of $T$ lies in some Gershgorin disk	Gershgorin disk theorem	"Some", not "each": a disk may contain no eigenvalue, and a different basis can give different disks
$\|\lambda - A_{j,j}\| \le \sum_{k \ne j} \|A_{j,k}\| \left\|\frac{c_k}{c_j}\right\| \le \sum_{k \ne j} \|A_{j,k}\|$	The proof's key inequality	For an eigenvector $w = c_1 v_1 + \cdots + c_n v_n$ with $\|c_j\|$ the largest of $\|c_1\|, \dots, \|c_n\|$ , the eigenvalue lies in the disk of row $j$
radius $\sum_{k \ne j} \|A_{k,j}\|$	Column disks	The column disks of $A$ are the row disks of $T'$ , and $T'$ has the same eigenvalues as $T$ , so columns work too

Key Insight: A scalar $\mu$ outside every disk is not an eigenvalue, so $T - \mu I$ is invertible; with $\mu = 0$ that certifies $T$ itself is invertible.

IIThe Disk the Proof Chooses

For the video's $A = \begin{pmatrix} 6 & 1 & 0 \\ 0 & 1 & -2 \\ -1 & 3 & -3 \end{pmatrix}$ , each eigenvalue lies in the disk of its eigenvector's largest coordinate.

Disks Eigenvalue

IIISmall Off-Diagonal Entries, Small Disks

Multiplying the off-diagonal entries by $t$ multiplies every radius by $t$ , pinning the eigenvalues near the diagonal entries.

t 1

💡 Between $t = 0$ and $t = 1$ the two eigenvalues starting at $1$ and $-3$ meet on the real axis and leave it as a conjugate pair, yet at every $t$ they stay inside the union of the disks.

IVRuling a Scalar Out

A scalar $\mu$ outside every disk is not an eigenvalue, so $T - \mu I$ is invertible.

Disks

VQuiz Questions

Problem 1 · Reading the Row Disks

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ whose matrix with respect to the standard basis is $B = \begin{pmatrix} 5 & -1 & 2 \\ 1 & -2 & 0 \\ -1 & 3 & 8 \end{pmatrix}$ — find the three row Gershgorin disks, each written as (centre, radius) for rows $1, 2, 3$ .

✅ Correct! Each centre is a diagonal entry and each radius is

|A_{j,k}|

summed over the other entries of that row:

1 + 2 = 3

1 + 0 = 1

1 + 3 = 4

❌ Those are the column disks. They also contain every eigenvalue, but the row-

j

radius sums

|B_{j,k}|

along row

j

: row 1 gives

|-1| + |2| = 3

, not

|1| + |-1| = 2

from column 1.

❌ Take absolute values first. The radius is

\sum_{k \ne j} |B_{j,k}|

, not

\left|\sum_{k \ne j} B_{j,k}\right|

: row 1 gives

|-1| + |2| = 3

, while

-1 + 2 = 1

lets the signs cancel.

❌ Leave out the diagonal entry. The sum runs over

k \ne j

; the diagonal entry is the centre, not part of the radius. Row 1 gives

|-1| + |2| = 3

, not

5 + 1 + 2 = 8

Show solution

Row $j$ gives the disk centred at $B_{j,j}$ with radius $\sum_{k \ne j} |B_{j,k}|$ .

\text{row } 1: \ \text{centre } 5, \ \text{radius } |-1| + |2| = 3

\text{row } 2: \ \text{centre } -2, \ \text{radius } |1| + |0| = 1

\text{row } 3: \ \text{centre } 8, \ \text{radius } |-1| + |3| = 4

So the row disks are $(5, 3)$ , $(-2, 1)$ , $(8, 4)$ . The column disks $(5, 2)$ , $(-2, 4)$ , $(8, 2)$ are a second valid set, but they come from the columns.

Problem 2 · "Some" Disk, Not Every Disk

Given: $T \in \mathcal{L}(\mathbf{C}^2)$ with matrix $M = \begin{pmatrix} 0 & 4 \\ 1 & 0 \end{pmatrix}$ with respect to the standard basis, so $T(x, y) = (4y, x)$ — find the true statement about its row disks and its eigenvalues.

✅ Correct! The eigenvalues are

\pm 2

|\pm 2| = 2 \le 4

but

2 > 1

. The theorem promises only that each eigenvalue lies in some disk.

❌ The theorem says "some". It places each eigenvalue in at least one disk; nothing forces every disk to receive one. Here both

\pm 2

miss the disk

|z| \le 1

❌ A centre is not an eigenvalue. The disks only fence the eigenvalues in.

T(x, y) = (0, 0)

forces

y = 0

and

x = 0

, so

T

is injective and

0

is not an eigenvalue.

❌ Compute the eigenvalues.

T(x, y) = \lambda(x, y)

gives

\lambda^2 = 4

, so

\lambda = \pm 2

, and

|\pm 2| = 2 > 1

Show solution

Disks. Row 1: centre $0$ , radius $|4| = 4$ . Row 2: centre $0$ , radius $|1| = 1$ .

Eigenvalues. $T(x, y) = \lambda(x, y)$ means $4y = \lambda x$ and $x = \lambda y$ . Substituting, $4y = \lambda^2 y$ . If $y = 0$ then $x = \lambda y = 0$ , which is not an eigenvector, so $y \ne 0$ and $\lambda^2 = 4$ :

\lambda = 2 \ \text{ with eigenvector } (2, 1), \qquad \lambda = -2 \ \text{ with eigenvector } (-2, 1)

Location. $|\pm 2| = 2 \le 4$ , so both lie in the row-1 disk; $2 > 1$ , so neither lies in the row-2 disk. In each eigenvector the first coordinate is the larger one, which is why the proof selects row 1 both times.

Problem 3 · An Invertibility Certificate

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $C = \begin{pmatrix} 2 & 3 & 0 \\ 0 & 5i & 1 \\ 1 & 0 & -4 \end{pmatrix}$ with respect to the standard basis — find the conclusion that the Gershgorin disks (rows and columns) justify.

✅ Correct! The column disks have centres

2

5i

-4

and radii

1

3

1

;

0

is at distance

2 > 1

5 > 3

4 > 1

, so

0

is not an eigenvalue.

❌ Lying in a disk proves nothing. The theorem says eigenvalues lie in the disks, not that every point of a disk is an eigenvalue.

❌ Check row 1. Its disk has centre

2

and radius

|3| + |0| = 3

, and

|0 - 2| = 2 \le 3

, so

0

lies in it.

❌ Try the columns. Column disks contain every eigenvalue too, so

0

outside all of them already rules

0

out.

Show solution

Rows. Centres $2$ , $5i$ , $-4$ ; radii $|3| + |0| = 3$ , $|0| + |1| = 1$ , $|1| + |0| = 1$ . Since $|0 - 2| = 2 \le 3$ , the point $0$ lies in the row-1 disk, so the rows cannot rule $0$ out.

Columns. Same centres; radii $|0| + |1| = 1$ , $|3| + |0| = 3$ , $|0| + |1| = 1$ . Then

|0 - 2| = 2 > 1, \qquad |0 - 5i| = 5 > 3, \qquad |0 - (-4)| = 4 > 1,

so $0$ lies outside every column disk. Every eigenvalue lies in some column disk, so $0$ is not an eigenvalue of $T$ . That means $T - 0I = T$ is injective, hence invertible (an operator on a finite-dimensional space).

Problem 4 · Bounding Every Eigenvalue

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 3 & -1 & 4 \\ 0 & 1 & 2i \end{pmatrix}$ . An eigenvalue $\lambda$ in disk $j$ satisfies $|\lambda| \le |A_{j,j}| + r_j$ , where $r_j$ is that disk's radius — find the smallest bound on $|\lambda|$ , valid for every eigenvalue, obtained this way from the row disks or from the column disks.

✅ Correct! Rows give

\max(3, 8, 3) = 8

, columns give

\max(5, 3, 6) = 6

, and both bounds hold, so

|\lambda| \le 6

❌ That is the row bound alone. Every eigenvalue also lies in some column disk, and the columns give

\max(5, 3, 6) = 6

❌ The centre is not $0$ .

4

is the largest column radius, but

|\lambda| \le |\lambda - A_{j,j}| + |A_{j,j}|

adds the distance of the centre from

0

: column 3 gives

|2i| + 4 = 6

❌ Use the largest, not the smallest. You do not know which disk holds a given eigenvalue, so the bound must cover every disk: take the maximum of

|A_{j,j}| + r_j

Show solution

If $\lambda$ lies in disk $j$ , the triangle inequality gives $|\lambda| \le |\lambda - A_{j,j}| + |A_{j,j}| \le r_j + |A_{j,j}|$ . Not knowing $j$ , take the maximum over $j$ .

Rows: $|2| + (1 + 0) = 3$ , $\ |-1| + (3 + 4) = 8$ , $\ |2i| + (0 + 1) = 3$ , so $|\lambda| \le 8$ .

Columns: $|2| + (3 + 0) = 5$ , $\ |-1| + (1 + 1) = 3$ , $\ |2i| + (0 + 4) = 6$ , so $|\lambda| \le 6$ .

Both bounds hold for every eigenvalue, so the better one is $|\lambda| \le 6$ .

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