Linear-Algebra-Done-Right · Unit 24 · Video 4 · Interactive Practice

Nilpotent Operators: Zero Is the Only Eigenvalue

IKey Formulas

Formula	Name	Hypotheses
$T^k = 0$ for some $k \ge 1$ $\iff$ every $v \neq 0$ is a generalized eigenvector of $T$ for $0$	Nilpotent, two ways	$V$ finite-dimensional, any $\mathbf{F}$
$T$ nilpotent $\Rightarrow T^{\dim V} = 0$	The power never exceeds the dimension	$V$ finite-dimensional, any $\mathbf{F}$
$T$ nilpotent $\Rightarrow$ $0$ is the only eigenvalue of $T$ ; conversely, $0$ the only eigenvalue $\Rightarrow$ $T$ nilpotent	Eigenvalues of a nilpotent operator	$V \neq \{0\}$ finite-dimensional; the first statement for any $\mathbf{F}$ , the converse only for $\mathbf{F} = \mathbf{C}$ (it fails over $\mathbf{R}$ )
$T$ nilpotent $\iff$ minimal polynomial $z^m$ $\iff$ $\mathcal{M}(T)$ strictly upper triangular for some basis	Three equivalent descriptions	$V \neq \{0\}$ finite-dimensional, any $\mathbf{F}$

Key Insight: If the minimal polynomial of $T$ is $z^m$ , then $T^m = 0$ but $T^{m-1} \neq 0$ , so $m$ is the smallest power that vanishes; it can be less than $\dim V$ , never more.

IIA Power That Vanishes

Each nilpotent operator from the video reaches the zero operator at a power no larger than $\dim V$ .

Operator Power k 1

IIIZeros Climb One Diagonal per Power

In each power of a strictly upper-triangular $n \times n$ matrix the zeros climb one diagonal, so $T^n = 0$ .

Size n 4 Power k 1

IVOver $\mathbf{R}$ , Only Eigenvalue $0$ Is Not Enough

On $\mathbf{R}^3$ , an operator whose only real eigenvalue is $0$ can still have no power equal to $0$ .

Operator

💡 Over $\mathbf{C}$ the operator $T$ also has the eigenvalues $i$ and $-i$ , so it does not meet the hypothesis of the converse, and there is no contradiction.

VQuiz Questions

Problem 1 · The Second Derivative on $\mathcal{P}_4(\mathbf{R})$

Given: $T \in \mathcal{L}(\mathcal{P}_4(\mathbf{R}))$ , $Tp = p''$ — find the smallest positive integer $k$ with $T^k = 0$ , and the minimal polynomial of $T$ .

Smallest $k$ with $T^k = 0$

Minimal polynomial of $T$

✅ Correct!

T^3 p = p^{(6)} = 0

for every

p

of degree at most

4

, while

T^2(x^4) = 24

, so the minimal polynomial is

z^3

; the smallest vanishing power

3

is less than

\dim \mathcal{P}_4(\mathbf{R}) = 5

❌ Not the smallest.

T^5 = 0

is the guaranteed bound

T^{\dim V} = 0

, but already

T^3 p = p^{(6)} = 0

: six derivatives kill every polynomial of degree at most

4

❌ $T^2$ is not zero.

T^2 p = p^{(4)}

, and

T^2(x^4) = 4! = 24 \neq 0

❌ $T^4 = 0$ is true, but not the first.

T^3 p

is the sixth derivative of

p

, which is already

0

when

\deg p \le 4

❌ Not minimal.

z^5

does send

T

0

, but so does the lower-degree monic polynomial

z^3

❌ $z^2$ does not send $T$ to $0$ .

T^2(x^4) = 24 \neq 0

, so

T^2 \neq 0

❌ The zeros of the minimal polynomial are the eigenvalues.

1

is not an eigenvalue of the nilpotent operator

T

; its only eigenvalue is

0

Show solution

$T^k p$ is the $(2k)$ -th derivative of $p$ . Five derivatives already send every polynomial of degree at most $4$ to $0$ , so

T^3 p = p^{(6)} = 0 \text{ for every } p \in \mathcal{P}_4(\mathbf{R}), \qquad \text{so } T^3 = 0.

No smaller power works: $T^2(x^4) = (x^4)^{(4)} = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \neq 0$ , so $T^2 \neq 0$ (and $T \neq 0$ , since $T(x^2) = 2$ ).

$T$ is nilpotent, so its minimal polynomial is $z^m$ for some $m$ . Since $z^3$ sends $T$ to $0$ , $z^m$ divides $z^3$ , so $m \le 3$ ; and $m \le 2$ would give $T^2 = 0$ . Hence the minimal polynomial is $z^3$ . The smallest vanishing power $3$ is below the bound $\dim \mathcal{P}_4(\mathbf{R}) = 5$ (basis $1, x, x^2, x^3, x^4$ ).

Problem 2 · Smaller Than the Dimension

Given: $T \in \mathcal{L}(\mathbf{F}^4)$ with matrix $N = \begin{pmatrix} 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ in the standard basis — find the smallest positive integer $k$ with $T^k = 0$ .

✅ Correct! The only entry of

N^2

that could be nonzero is the

(1,4)

entry, and it is

1 \cdot 2 + 2 \cdot (-1) = 0

❌ That is the guaranteed bound, not the smallest power.

T^{\dim V} = T^4 = 0

holds, but a smaller power already vanishes.

❌ $T^3 = 0$ is true, but not the first. Compute

N^2

: its

(1,4)

entry is

1 \cdot 2 + 2 \cdot (-1)

, and the two products cancel.

❌ $T \neq 0$ .

N

has nonzero entries, for instance

N_{1,2} = 1

Show solution

$N$ is strictly upper triangular, so an entry $(i, j)$ of $N^2$ is a sum of products $N_{i,l} N_{l,j}$ with $i < l < j$ ; only $(1,3)$ , $(1,4)$ , $(2,4)$ can be nonzero:

(N^2)_{1,3} = N_{1,2} N_{2,3} = 1 \cdot 0 = 0, \qquad (N^2)_{2,4} = N_{2,3} N_{3,4} = 0 \cdot (-1) = 0,

(N^2)_{1,4} = N_{1,2} N_{2,4} + N_{1,3} N_{3,4} = 1 \cdot 2 + 2 \cdot (-1) = 0.

So $N^2 = 0$ , while $N \neq 0$ . The smallest power is $k = 2$ , below the bound $\dim \mathbf{F}^4 = 4$ ; the minimal polynomial of $T$ is $z^2$ .

Problem 3 · Minimal Polynomial of a $5 \times 5$ Matrix

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

✅ Correct!

N^3

has the single nonzero entry

(N^3)_{1,5} = 3

and

N^4 = 0

, so the minimal polynomial is

z^4

❌ $z^5$ sends $T$ to $0$ , but it is not minimal. Already

N^4 = 0

, and

z^4

has smaller degree.

❌ $N^3 \neq 0$ . Its

(1,5)

entry is

N_{1,2} N_{2,3} N_{3,5} = 1 \cdot 1 \cdot 3 = 3

❌ $N^2 \neq 0$ . For instance

(N^2)_{1,3} = N_{1,2} N_{2,3} = 1

Show solution

$N$ is strictly upper triangular, so $T$ is nilpotent and its minimal polynomial is $z^m$ with $m$ the smallest power for which $N^m = 0$ . Compute the powers:

N^2 = \begin{pmatrix} 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}, \qquad N^3 = N^2 N = \begin{pmatrix} 0 & 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}.

For example $(N^2)_{1,5} = N_{1,2} N_{2,5} + N_{1,4} N_{4,5} = 1 \cdot (-1) + 2 \cdot 1 = 1$ and $(N^3)_{1,5} = (N^2)_{1,3} N_{3,5} = 1 \cdot 3 = 3$ . Then $N^4 = N^3 N$ has rows that are multiples of row $5$ of $N$ , which is zero, so $N^4 = 0$ .

Since $N^3 \neq 0$ and $N^4 = 0$ , the minimal polynomial of $T$ is $z^4$ — one less than $\dim \mathbf{C}^5 = 5$ .

Problem 4 · A Real Operator

Given: $T \in \mathcal{L}(\mathbf{R}^3)$ , $T(x, y, z) = (0,\ y - 2z,\ y + z)$ — decide which statement is true.

✅ Correct!

Te_1 = 0

and the

yz

-plane holds no real eigenvector, yet

T

is injective on the

yz

-plane, so

T^k(0, 1, 0) \neq 0

for every

k

❌ That converse needs $\mathbf{F} = \mathbf{C}$ . Over

\mathbf{C}

this

T

also has the eigenvalues

1 \pm \sqrt{2}\, i

, and over

\mathbf{R}

no power of

T

0

❌ $T^{\dim V} = 0$ holds only for nilpotent $T$ . Here

T^3(0, 1, 0) = (0, -5, 1) \neq 0

❌ The matrix is not upper triangular, so its diagonal says nothing about eigenvalues.

T(x, y, z) = (x, y, z)

forces

x = 0

, and then

y - 2z = y

and

y + z = z

give

z = 0

and

y = 0

, so there is no eigenvector for

1

Show solution

$0$ is an eigenvalue: $T(1, 0, 0) = (0, 0, 0)$ .

No other real eigenvalue: if $T(x, y, z) = \lambda (x, y, z)$ with $\lambda \neq 0$ , the first coordinate gives $\lambda x = 0$ , so $x = 0$ , and then

y - 2z = \lambda y, \qquad y + z = \lambda z \quad\Longrightarrow\quad y = (\lambda - 1) z, \quad (1 - \lambda)(\lambda - 1) z = 2z.

If $z \neq 0$ this says $(\lambda - 1)^2 = -2$ , impossible for real $\lambda$ ; if $z = 0$ then $y = 0$ and the vector is $0$ . So $0$ is the only real eigenvalue.

Not nilpotent: $T$ maps the $yz$ -plane into itself, and there $T(0, y, z) = 0$ forces $y - 2z = 0$ and $y + z = 0$ , so $y = z = 0$ : $T$ is injective on that plane. Every power of $T$ is then injective on it too, so $T^k(0, 1, 0) \neq 0$ for every $k$ — for instance $T^3(0, 1, 0) = (0, -5, 1)$ .

Over $\mathbf{C}$ , $(\lambda - 1)^2 = -2$ has the solutions $\lambda = 1 \pm \sqrt{2}\, i$ , so $T$ does not satisfy the hypothesis " $0$ is the only eigenvalue" of the converse, which needs $\mathbf{F} = \mathbf{C}$ .

Solved: 0 / 4

Nilpotent Operators: Zero Is the Only Eigenvalue

IKey Formulas

IIA Power That Vanishes

IIIZeros Climb One Diagonal per Power

IVOver R\mathbf{R}R, Only Eigenvalue 000 Is Not Enough

VQuiz Questions

IVOver $\mathbf{R}$ , Only Eigenvalue $0$ Is Not Enough