Linear-Algebra-Done-Right · Unit 24 · Video 1 · Interactive Practice

Null Spaces of Powers Stabilize and Split the Space

IKey Formulas

Formula	Name	Hypotheses
$\{0\} = \operatorname{null} T^0 \subseteq \operatorname{null} T^1 \subseteq \cdots \subseteq \operatorname{null} T^k \subseteq \operatorname{null} T^{k+1} \subseteq \cdots$	Increasing null spaces	$T \in \mathcal{L}(V)$ , with $T^0 = I$
$\operatorname{null} T^m = \operatorname{null} T^{m+1} \Rightarrow \operatorname{null} T^m = \operatorname{null} T^{m+k}$ for every $k \ge 1$	Equality propagates	$T \in \mathcal{L}(V)$ , $m \ge 0$ an integer
$\operatorname{null} T^n = \operatorname{null} T^{n+1} = \operatorname{null} T^{n+2} = \cdots$	Null spaces stop growing	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $n = \dim V$
$V = \operatorname{null} T^n \oplus \operatorname{range} T^n$	The space splits at power $n$	$V$ finite-dimensional, $T \in \mathcal{L}(V)$ , $n = \dim V$ ; with $T$ in place of $T^n$ this can fail

Key Insight: Every strict inclusion $\operatorname{null} T^k \subsetneq \operatorname{null} T^{k+1}$ raises the dimension by at least $1$ , and no subspace of $V$ has dimension above $n$ , so the chain has at most $n$ strict steps. At power $n$ the null space and the range meet only in $0$ , and the fundamental theorem of linear maps makes their dimensions add to $n$ .

IIThe Staircase of $\dim \operatorname{null} T^k$

Each strict inclusion raises $\dim \operatorname{null} T^k$ by at least one, and after the first flat step the staircase never climbs again.

💡 Challenge: build an operator on $\mathbf{F}^6$ whose staircase climbs at every step up to $k = 6$ ; no operator on $\mathbf{F}^6$ can climb at step $7$ , since that would need $\dim \operatorname{null} T^7 \ge 7$ .

IIIPower $1$ Can Fail, Power $\dim V$ Splits

$\dim \operatorname{null} T^k + \dim \operatorname{range} T^k = 3$ always; the sum is direct only when the two subspaces meet in $\{0\}$ .

Operator on F³ Power k 1

IVQuiz Questions

Problem 1 · Null Space of a Power

Given: $T \in \mathcal{L}(\mathbf{F}^4)$ defined by $T(z_1, z_2, z_3, z_4) = (z_2, z_3, 0, 3z_4)$ — find $\operatorname{null} T^2$ .

✅ Correct!

T^2(z_1, z_2, z_3, z_4) = (z_3, 0, 0, 9z_4)

, which is

0

exactly when

z_3 = z_4 = 0

❌ That is $\operatorname{null} T$ .

T

kills only

(z_1, 0, 0, 0)

, but

T^2

also kills

(0, 1, 0, 0)

T(0,1,0,0) = (1, 0, 0, 0)

and then

T(1,0,0,0) = 0

❌ That is $\operatorname{range} T^2$ .

T^2(z) = (z_3, 0, 0, 9z_4)

has outputs

(a, 0, 0, b)

; the null space asks which inputs give

0

, not which outputs occur.

❌ That is $\operatorname{null} T^3$ , one power too many.

T^2(0, 0, 1, 0) = (1, 0, 0, 0) \ne 0

; the third coordinate needs a third application of

T

to reach

0

Show solution

Apply $T$ twice:

T^2(z_1, z_2, z_3, z_4) = T(z_2, z_3, 0, 3z_4) = (z_3, 0, 0, 9z_4).

This is $0$ exactly when $z_3 = 0$ and $z_4 = 0$ , with $z_1, z_2$ free. So

\operatorname{null} T^2 = \{(z_1, z_2, 0, 0) : z_1, z_2 \in \mathbf{F}\}.

For comparison, $\operatorname{null} T = \{(z_1, 0, 0, 0)\}$ and $\operatorname{null} T^3 = \{(z_1, z_2, z_3, 0)\}$ , so the chain has dimensions $0, 1, 2, 3$ and then stays at $3$ : the coordinate $z_4$ is multiplied by $3$ at every step and never reaches $0$ .

Problem 2 · Dimensions Add, Yet No Direct Sum

Given: $S \in \mathcal{L}(\mathbf{F}^4)$ defined by $S(z_1, z_2, z_3, z_4) = (z_2, 0, z_4, 0)$ — which statement is true?

✅ Correct! The dimensions

2 + 2

add to

4

, but the two subspaces coincide, so their sum is only

2

-dimensional.

❌ Dimensions adding to $4$ is not enough. The fundamental theorem of linear maps guarantees that for every operator; a direct sum also needs

\operatorname{null} S \cap \operatorname{range} S = \{0\}

, and here

(1, 0, 0, 0)

lies in both.

❌ That contradicts the fundamental theorem of linear maps. For every

S \in \mathcal{L}(\mathbf{F}^4)

\dim \operatorname{null} S + \dim \operatorname{range} S = 4

exactly; here it is

2 + 2

❌ That combination is impossible. If the intersection were

\{0\}

, the sum would be direct with dimension

2 + 2 = 4

, hence all of

\mathbf{F}^4

. In fact the intersection is

\{(z_1, 0, z_3, 0)\}

Show solution

Null space: $S(z) = 0$ exactly when $z_2 = z_4 = 0$ , so $\operatorname{null} S = \{(z_1, 0, z_3, 0)\}$ , of dimension $2$ .

Range: the outputs $(z_2, 0, z_4, 0)$ have second and fourth coordinates $0$ and first and third coordinates arbitrary, so $\operatorname{range} S = \{(z_1, 0, z_3, 0)\}$ , also of dimension $2$ .

The dimensions add to $4$ , as the fundamental theorem of linear maps requires, but the two subspaces are equal. Their intersection is $2$ -dimensional, the sum is not direct, and

\operatorname{null} S + \operatorname{range} S = \{(z_1, 0, z_3, 0)\} \ne \mathbf{F}^4.

Raising the power repairs it: $S^2 = 0$ , so $\operatorname{null} S^4 = \mathbf{F}^4$ and $\operatorname{range} S^4 = \{0\}$ , and $\mathbf{F}^4 = \operatorname{null} S^4 \oplus \operatorname{range} S^4$ .

Problem 3 · Where the Chain Stops, and What Is Left

Given: $R \in \mathcal{L}(\mathbf{F}^4)$ defined by $R(z_1, z_2, z_3, z_4) = (0, z_1, 2z_3, 3z_4)$ — find the smallest $m \ge 0$ with $\operatorname{null} R^m = \operatorname{null} R^{m+1}$ , and find $\operatorname{range} R^4$ .

Smallest m with null Rᵐ = null Rᵐ⁺¹

range R⁴

✅ Correct! The chain has dimensions

0, 1, 2, 2, \dots

, and

\mathbf{F}^4 = \operatorname{null} R^4 \oplus \operatorname{range} R^4 = \{(z_1, z_2, 0, 0)\} \oplus \{(0, 0, z_3, z_4)\}

❌ The chain still grows at step 1.

\operatorname{null} R = \{(0, z_2, 0, 0)\}

but

\operatorname{null} R^2 = \{(z_1, z_2, 0, 0)\}

(1, 0, 0, 0) \mapsto (0, 1, 0, 0) \mapsto 0

❌ The chain is already flat one step earlier.

\operatorname{null} R^2 = \operatorname{null} R^3 = \{(z_1, z_2, 0, 0)\}

, because

z_3

and

z_4

are only rescaled, never sent to

0

❌ $\dim V = 4$ is an upper bound, not the answer. The theorem guarantees equality by power

4

; for this operator it happens at power

2

, since

\operatorname{null} R^2 = \operatorname{null} R^3

❌ That is $\operatorname{null} R^4$ , not the range. The range is the complementary piece:

R^4(z) = (0, 0, 16z_3, 81z_4)

❌ That is $\operatorname{range} R$ , the power $1$ . After two applications the second coordinate is gone too:

R^2(z) = (0, 0, 4z_3, 9z_4)

❌ $R^4$ is not the zero operator. Only the part of

\mathbf{F}^4

that

R

shifts toward

0

dies;

z_3

and

z_4

are multiplied by

2^4 = 16

and

3^4 = 81

, which are nonzero.

Show solution

Compute the powers:

R^2(z) = R(0, z_1, 2z_3, 3z_4) = (0, 0, 4z_3, 9z_4), \qquad R^k(z) = (0, 0, 2^k z_3, 3^k z_4) \text{ for } k \ge 2.

So $\operatorname{null} R^0 = \{0\}$ , $\operatorname{null} R = \{(0, z_2, 0, 0)\}$ , and $\operatorname{null} R^k = \{(z_1, z_2, 0, 0)\}$ for every $k \ge 2$ . The dimensions are $0, 1, 2, 2, \dots$ ; the first equality is $\operatorname{null} R^2 = \operatorname{null} R^3$ , so $m = 2$ , and by "equality propagates" every later null space is the same.

For $k = 4$ : $R^4(z) = (0, 0, 16z_3, 81z_4)$ , whose outputs are all vectors $(0, 0, a, b)$ :

\operatorname{range} R^4 = \{(0, 0, z_3, z_4)\}, \qquad \mathbf{F}^4 = \{(z_1, z_2, 0, 0)\} \oplus \{(0, 0, z_3, z_4)\}.

Problem 4 · The Power dim V Can Be Needed

Given: $Q \in \mathcal{L}(\mathbf{F}^4)$ defined by $Q(z_1, z_2, z_3, z_4) = (z_2, z_3, z_4, 0)$ — find the smallest $k \ge 1$ with $\mathbf{F}^4 = \operatorname{null} Q^k \oplus \operatorname{range} Q^k$ .

✅ Correct! For

k = 1, 2, 3

the vector

(1, 0, 0, 0)

lies in both

\operatorname{null} Q^k

and

\operatorname{range} Q^k

; at

k = 4

Q^4 = 0

and

\mathbf{F}^4 = \mathbf{F}^4 \oplus \{0\}

. The guaranteed power

\dim V

cannot be lowered for this operator.

❌ $2 + 2 = 4$ is not enough.

Q^2(z) = (z_3, z_4, 0, 0)

, so

\operatorname{null} Q^2

and

\operatorname{range} Q^2

are both

\{(z_1, z_2, 0, 0)\}

: the same plane, not a direct sum.

❌ The intersection is still nonzero at $k = 3$ .

Q^3(z) = (z_4, 0, 0, 0)

, so

\operatorname{range} Q^3 = \{(z_1, 0, 0, 0)\}

, which lies inside

\operatorname{null} Q^3 = \{(z_1, z_2, z_3, 0)\}

❌ The power $1$ fails here.

Q(0, 1, 0, 0) = (1, 0, 0, 0)

, so

(1, 0, 0, 0) \in \operatorname{range} Q

, and

Q(1, 0, 0, 0) = 0

, so it is also in

\operatorname{null} Q

Show solution

The powers shift the coordinates further left each time:

Q^2(z) = (z_3, z_4, 0, 0), \qquad Q^3(z) = (z_4, 0, 0, 0), \qquad Q^4(z) = (0, 0, 0, 0).

$k = 1$ : $\operatorname{null} Q = \{(z_1, 0, 0, 0)\}$ , $\operatorname{range} Q = \{(z_1, z_2, z_3, 0)\}$ ; both contain $(1, 0, 0, 0)$ .
$k = 2$ : $\operatorname{null} Q^2 = \operatorname{range} Q^2 = \{(z_1, z_2, 0, 0)\}$ ; the intersection is a plane.
$k = 3$ : $\operatorname{null} Q^3 = \{(z_1, z_2, z_3, 0)\}$ , $\operatorname{range} Q^3 = \{(z_1, 0, 0, 0)\}$ ; both contain $(1, 0, 0, 0)$ .
$k = 4$ : $\operatorname{null} Q^4 = \mathbf{F}^4$ , $\operatorname{range} Q^4 = \{0\}$ ; the intersection is $\{0\}$ and $\mathbf{F}^4 = \operatorname{null} Q^4 \oplus \operatorname{range} Q^4$ .

The smallest such power is $k = 4 = \dim \mathbf{F}^4$ . Its null-space chain $\{0\} \subsetneq \operatorname{null} Q \subsetneq \operatorname{null} Q^2 \subsetneq \operatorname{null} Q^3 \subsetneq \operatorname{null} Q^4$ climbs at every one of the $4$ possible steps.

Solved: 0 / 4

Null Spaces of Powers Stabilize and Split the Space

IKey Formulas

IIThe Staircase of dim⁡null⁡Tk\dim \operatorname{null} T^kdimnullTk

IIIPower 111 Can Fail, Power dim⁡V\dim VdimV Splits

IVQuiz Questions

IIThe Staircase of $\dim \operatorname{null} T^k$

IIIPower $1$ Can Fail, Power $\dim V$ Splits