Linear-Algebra-Done-Right · Unit 25 · Video 4 · Interactive Practice

Upper-Triangular Diagonals Count Multiplicities

IKey Formulas

Formula	Name	Hypotheses
$\dim \operatorname{null} T \le d$ , $\ d$ = number of zeros on the diagonal of $\mathcal{M}(T)$	Inequality one	$\mathcal{M}(T)$ upper triangular with respect to a basis $v_1, \dots, v_n$ of $V$ , any $\mathbf{F}$
$\mathcal{M}(T^n)$ is upper triangular with diagonal $\lambda_1^n, \dots, \lambda_n^n$	Diagonal of a power	same, with diagonal $\lambda_1, \dots, \lambda_n$
$m_\lambda = d_\lambda$	Multiplicity $=$ number of times $\lambda$ appears on the diagonal	$\mathbf{F} = \mathbf{C}$ , $\mathcal{M}(T)$ upper triangular, $\lambda$ an eigenvalue of $T$

Key Insight: Each eigenvalue gives only $m_\lambda \le d_\lambda$ , but both sides add up to $n = \dim V$ over all eigenvalues, so no inequality can be strict.

IICount the Diagonal, Measure the Null Space

For every upper-triangular matrix, the number of times $\lambda$ sits on the diagonal is $\dim \operatorname{null}(T - \lambda I)^4$ .

IIIZeros on the Diagonal Bound the Null Space

The diagonal zeros of $A$ bound $\dim \operatorname{null} A$ , often strictly, but $\dim \operatorname{null} A^n$ reaches the bound.

Matrix Power k 1

IVQuiz Questions

Problem 1 · Read the Multiplicity off the Diagonal

Given: $T \in \mathcal{L}(\mathbf{C}^5)$ with matrix $\begin{pmatrix} 2 & 1 & 0 & 3 & 0 \\ 0 & -1 & 2 & 0 & 1 \\ 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}$ in the standard basis — find the multiplicity of the eigenvalue $2$ .

✅ Correct! The matrix is upper triangular and

2

appears three times on its diagonal, so

\dim G(2, T) = 3

❌ That is the geometric multiplicity.

\dim \operatorname{null}(T - 2I) = 1

here, but the multiplicity counts the generalized eigenspace, which is what the diagonal counts.

❌ The copies need not be adjacent. Count every diagonal position holding

2

: positions

1

3

and

4

❌ $-1$ is also an eigenvalue. The multiplicities of

2

and

-1

add up to

5

Show solution

The matrix is upper triangular with diagonal $2, -1, 2, 2, -1$ . Over $\mathbf{C}$ , each eigenvalue appears on the diagonal exactly as many times as its multiplicity, so

\dim G(2, T) = d_2 = 3, \qquad \dim G(-1, T) = d_{-1} = 2,

and $3 + 2 = 5 = \dim \mathbf{C}^5$ . (Directly, $T - 2I$ has rank $4$ , so the geometric multiplicity of $2$ is only $1$ .)

Problem 2 · Zeros on the Diagonal Are Not the Null Space

Given: $T \in \mathcal{L}(\mathbf{C}^3)$ with matrix $\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 5 \end{pmatrix}$ in the standard basis — find $\dim \operatorname{null} T$ and $\dim \operatorname{null} T^3$ .

$\dim \operatorname{null} T$

$\dim \operatorname{null} T^3$

✅ Correct!

\dim \operatorname{null} T = 1 < 2 = d

, a strict inequality one; but

\dim \operatorname{null} T^3 = \dim G(0, T) = d_0 = 2

❌ That is the diagonal count $d$ . Inequality one only says

\dim \operatorname{null} T \le d

; here

T

has rank

2

❌ Not quite. Find the rank of the matrix: are its three rows independent?

❌ Not quite.

\operatorname{null} T^3 = G(0, T)

, and its dimension is the number of zeros on the diagonal.

Show solution

$T(z_1, z_2, z_3) = (z_2,\ z_3,\ 5z_3)$ , which is $0$ exactly when $z_2 = z_3 = 0$ : $\operatorname{null} T = \operatorname{span}(e_1)$ has dimension $1$ .

The diagonal is $0, 0, 5$ , so $d_0 = 2$ and the multiplicity of $0$ is $2$ . Since $\dim \mathbf{C}^3 = 3$ , $\operatorname{null} T^3 = G(0, T)$ , so $\dim \operatorname{null} T^3 = 2$ . Directly: $T^3(z_1, z_2, z_3) = (5z_3, 25z_3, 125z_3)$ , which vanishes exactly when $z_3 = 0$ .

Problem 3 · A Different Upper-Triangular Matrix

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ has an upper-triangular matrix with diagonal $1, 3, 1, 1$ ; with respect to another basis $T$ has an upper-triangular matrix whose first diagonal entry is $3$ — find its other three diagonal entries.

✅ Correct!

3

has multiplicity

1

and

1

has multiplicity

3

, numbers defined without any basis; every upper-triangular matrix of

T

shows them that many times.

❌ $3$ appears only once on every upper-triangular matrix of $T$ . Its multiplicity is

1

, read off the first matrix.

❌ They are determined. The multiplicity

\dim G(\lambda, T)

does not depend on a basis, and it equals the diagonal count in every upper-triangular matrix.

Show solution

From the first matrix, $\dim G(1, T) = 3$ and $\dim G(3, T) = 1$ . These are dimensions of null spaces of $(T - \lambda I)^4$ , so they do not depend on a basis.

In the second upper-triangular matrix, $1$ must appear $3$ times and $3$ once. The $3$ is already in the first position, so the remaining diagonal entries are $1, 1, 1$ .

Problem 4 · Separated Zeros

Given: $T \in \mathcal{L}(\mathbf{C}^4)$ has an upper-triangular matrix with diagonal $0, 5, 0, 5$ and unknown entries above the diagonal — find $\dim \operatorname{null} T^4$ .

✅ Correct!

\operatorname{null} T^4 = G(0, T)

, whose dimension is the number of zeros on the diagonal,

d_0 = 2

, wherever they sit.

❌ Every diagonal zero counts. The theorem counts all positions holding

0

, adjacent or not.

❌ That is true for $\dim \operatorname{null} T$ , not for $T^4$ . The power

\dim V

removes the dependence on the entries above the diagonal.

❌ $T^4 \neq 0$ . The diagonal of

\mathcal{M}(T^4)

0, 625, 0, 625

, so

T^4

has rank at least

2

Show solution

$\dim V = 4$ , so $\operatorname{null} T^4 = G(0, T)$ , and its dimension is the multiplicity of $0$ . The multiplicity equals the number of times $0$ appears on the diagonal of any upper-triangular matrix of $T$ :

\dim \operatorname{null} T^4 = d_0 = 2.

By contrast, $\dim \operatorname{null} T$ is $1$ or $2$ depending on the entries above the diagonal (inequality one only bounds it by $2$ ).

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