Linear-Algebra-Done-Right · Unit 9 · Video 2 · Interactive Practice

The Annihilator U⁰ and dim U⁰ = dim V − dim U

IKey Formulas

Formula	Name	Hypotheses
$U^0 = \{\varphi \in V' : \varphi(u) = 0 \text{ for all } u \in U\}$	Annihilator of $U$ ; a subspace of $V'$	$U$ any subset of $V$
$\dim U^0 = \dim V - \dim U$	Dimension of the annihilator	$V$ finite-dimensional, $U$ a subspace of $V$
$\text{null}\, i' = U^0$ , $\ \text{range}\, i' = U'$	Dual of the inclusion $i(u) = u$	$V$ finite-dimensional, $U$ a subspace of $V$ , $i \in \mathcal{L}(U, V)$ ; then $i' \in \mathcal{L}(V', U')$ and $i'(\varphi) = \varphi\|_U$
$U^0 = \{0\} \iff U = V$ ; $\ U^0 = V' \iff U = \{0\}$	Smallest and largest annihilator	$V$ finite-dimensional, $U$ a subspace of $V$

Key Insight: $\varphi \in U^0$ says exactly that $U \subseteq \text{null}\,\varphi$ . A spanning list of $U$ is enough to test: if $\varphi$ is zero on each vector of the list, linearity makes it zero on every linear combination. And when $V$ is finite-dimensional, to prove a subspace $U$ is all of $V$ it suffices to show that the only functional vanishing on $U$ is $\varphi = 0$ .

IIWhich Functionals Vanish on a Line

$\varphi \in U^0$ exactly when $U \subseteq \text{null}\,\varphi$ ; for a line $U$ in $\mathbf{R}^2$ , $U^0$ is a line in $(\mathbf{R}^2)'$ .

IIIThe Dual Basis Splits Off a Basis of $U^0$

Extend a basis $u_1, \dots, u_m$ of $U$ to a basis $u_1, \dots, u_5$ of $V$ ; the dual basis entries $\varphi_{m+1}, \dots, \varphi_5$ then form a basis of $U^0$ .

dim U = m 2

IVRestriction Is the Dual of the Inclusion

For $U = \operatorname{span}\big((1, 0, 1), (0, 1, 1)\big)$ in $\mathbf{R}^3$ , $i'$ sends $\varphi$ to $\varphi|_U$ : its null space is $U^0$ , its range all of $U'$ .

a 2 b −1 c 1

dim range i′ + dim null i′ = dim U′ + dim U⁰ = 2 + 1 = 3 = dim V′

VQuiz Questions

Problem 1 · Test a Functional

Given: $U = \operatorname{span}\big((1, 0, 1, 0), (0, 1, 0, 1)\big)$ in $\mathbf{R}^4$ , with $x = (x_1, x_2, x_3, x_4)$ — find the functional that lies in $U^0$ .

✅ Correct!

\varphi(1, 0, 1, 0) = 1 - 1 = 0

and

\varphi(0, 1, 0, 1) = 0

, so by linearity

\varphi

vanishes on every vector of

U

❌ It vanishes on only one spanning vector.

\varphi(0, 1, 0, 1) = 0

, but

\varphi(1, 0, 1, 0) = 1 + 1 = 2 \neq 0

, and

(1, 0, 1, 0) \in U

❌ It vanishes on only one spanning vector.

\varphi(1, 0, 1, 0) = 0

, but

\varphi(0, 1, 0, 1) = 1 + 1 = 2 \neq 0

, and

(0, 1, 0, 1) \in U

❌ Zero on one vector of $U$ is not zero on all of $U$ . This

\varphi

sends

(1, 1, 1, 1) \in U

0

, but

\varphi(1, 0, 1, 0) = 2

and

\varphi(0, 1, 0, 1) = -2

. Test each vector of a spanning list.

❌ Not quite. Evaluate each option on both spanning vectors

(1, 0, 1, 0)

and

(0, 1, 0, 1)

; both values must be

0

Show solution

Every $u \in U$ is $u = a(1, 0, 1, 0) + b(0, 1, 0, 1)$ , so for a linear functional $\varphi$ ,

\varphi(u) = a\,\varphi(1, 0, 1, 0) + b\,\varphi(0, 1, 0, 1).

Hence $\varphi \in U^0$ exactly when $\varphi$ is zero on both spanning vectors. Writing $\varphi = c_1\varphi_1 + c_2\varphi_2 + c_3\varphi_3 + c_4\varphi_4$ in the dual basis, the two conditions are $c_1 + c_3 = 0$ and $c_2 + c_4 = 0$ .

$x_1 + x_3$ : $c_1 + c_3 = 2$ . Not in $U^0$ .
$x_2 + x_4$ : $c_2 + c_4 = 2$ . Not in $U^0$ .
$x_1 - x_3$ : $c_1 + c_3 = 0$ , $c_2 + c_4 = 0$ . In $U^0$ . ✓
$x_1 - x_2 + x_3 - x_4$ : $c_1 + c_3 = 2$ . Not in $U^0$ .

In fact $U^0 = \operatorname{span}(\varphi_1 - \varphi_3,\ \varphi_2 - \varphi_4)$ , of dimension $2 = 4 - 2$ , as the dimension formula predicts.

Problem 2 · The Annihilator of a Hyperplane

Given: $U = \{x \in \mathbf{R}^4 : x_1 + x_2 + x_3 + x_4 = 0\}$ , and $\varphi_1, \dots, \varphi_4$ the dual basis of the standard basis — find $U^0$ .

✅ Correct!

\dim U = 3

, so

\dim U^0 = 4 - 3 = 1

, and the nonzero functional

x_1 + x_2 + x_3 + x_4

vanishes on

U

by the very definition of

U

❌ That annihilates the wrong space.

\{\varphi : \varphi(1, 1, 1, 1) = 0\}

is the annihilator of

\operatorname{span}\big((1, 1, 1, 1)\big)

. For

U

itself,

\dim U^0 = 4 - \dim U = 4 - 3 = 1

; for example

\varphi_1 - \varphi_2

is in that set but sends

(1, -1, 0, 0) \in U

2

❌ $U^0 = \{0\}$ only when $U = V$ . Here

(1, 0, 0, 0) \notin U

, so

U \neq \mathbf{R}^4

and

U^0

contains a nonzero functional.

❌ $U^0 = V'$ only when $U = \{0\}$ .

\varphi_1

sends

(1, -1, 0, 0) \in U

1

, so

\varphi_1 \notin U^0

❌ Not quite. Find

\dim U

first, then use

\dim U^0 = \dim V - \dim U

Show solution

Dimension. $U$ is the null space of the nonzero functional $\psi(x) = x_1 + x_2 + x_3 + x_4$ , whose range is $\mathbf{R}$ , so by the fundamental theorem of linear maps $\dim U = 4 - 1 = 3$ . The dimension formula gives

\dim U^0 = \dim \mathbf{R}^4 - \dim U = 4 - 3 = 1.

A spanning functional. $\psi = \varphi_1 + \varphi_2 + \varphi_3 + \varphi_4$ sends every $x \in U$ to $x_1 + x_2 + x_3 + x_4 = 0$ , so $\psi \in U^0$ , and $\psi \neq 0$ since $\psi(1, 0, 0, 0) = 1$ . A nonzero vector in a $1$ -dimensional space spans it, so

U^0 = \operatorname{span}(\varphi_1 + \varphi_2 + \varphi_3 + \varphi_4).

Problem 3 · Multiples of $x - 2$ in $\mathcal{P}_3(\mathbf{R})$

Given: $U = \{(x - 2)q : q \in \mathcal{P}_2(\mathbf{R})\}$ , a subspace of $V = \mathcal{P}_3(\mathbf{R})$ — find $\dim U^0$ and a functional that spans $U^0$ .

What is $\dim U^0$ ?

Which functional spans $U^0$ ?

✅ Correct!

\dim U^0 = 4 - 3 = 1

, and evaluation at

2

is a nonzero functional sending every

(x - 2)q

0 \cdot q(2) = 0

, so it spans

U^0

❌ That is $\dim U$ .

U

has basis

x - 2

(x - 2)x

(x - 2)x^2

, so

\dim U = 3

; the annihilator has dimension

\dim V - \dim U

❌ $U^0 = \{0\}$ would force $U = V$ . But

1 \notin U

, since every element of

U

0

x = 2

❌ That is $\dim V' = \dim V$ .

U^0

is all of

V'

only when

U = \{0\}

❌ Check the dimension count.

\dim \mathcal{P}_3(\mathbf{R}) = 4

and

\dim U = 3

❌ That kills multiples of $(x - 2)^2$ , not of $x - 2$ . For

p = x - 2 \in U

p'(2) = 1 \neq 0

❌ Check where the factor vanishes.

x - 2

is zero at

x = 2

; for

p = x - 2 \in U

p(-2) = -4 \neq 0

❌ That was the functional for multiples of $x^2$ . Here

p = x - 2 \in U

has

p'(0) = 1 \neq 0

❌ Not quite. Test the option on

p = x - 2 \in U

; a functional in

U^0

must send it to

0

Show solution

Dimension of $U$ . The map $q \mapsto (x - 2)q$ from $\mathcal{P}_2(\mathbf{R})$ onto $U$ is linear and injective (a nonzero $q$ gives a nonzero product), so $\dim U = \dim \mathcal{P}_2(\mathbf{R}) = 3$ ; a basis is $x - 2$ , $(x - 2)x$ , $(x - 2)x^2$ . Then

\dim U^0 = \dim \mathcal{P}_3(\mathbf{R}) - \dim U = 4 - 3 = 1.

A spanning functional. Let $\varphi(p) = p(2)$ . For $p = (x - 2)q \in U$ , $\varphi(p) = (2 - 2)\,q(2) = 0$ , so $\varphi \in U^0$ . Also $\varphi(1) = 1$ , so $\varphi \neq 0$ , and a nonzero vector spans a $1$ -dimensional space: $U^0 = \operatorname{span}(\varphi)$ .

The other options all fail on $p = x - 2 \in U$ : $p'(2) = 1$ , $p(-2) = -4$ , $p'(0) = 1$ .

Problem 4 · Four Vectors in $\mathbf{R}^4$

Given: $U = \operatorname{span}\big((1, 0, 1, 0), (0, 1, 0, 1), (1, 1, 0, 0), (0, 0, 1, 1)\big)$ in $\mathbf{R}^4$ — use $U^0$ to decide whether $U = \mathbf{R}^4$ .

✅ Correct! A nonzero functional vanishes on all four vectors, so

U^0 \neq \{0\}

and

U \neq \mathbf{R}^4

; indeed

(1, 0, 1, 0) + (0, 1, 0, 1) = (1, 1, 0, 0) + (0, 0, 1, 1)

❌ Four vectors in $\mathbf{R}^4$ need not span it.

\varphi(x) = x_1 - x_2 - x_3 + x_4

is nonzero and sends all four spanning vectors to

0

, so

U^0 \neq \{0\}

❌ Those two vanish on the first two vectors only.

(\varphi_1 - \varphi_3)(1, 1, 0, 0) = 1 \neq 0

, so

\varphi_1 - \varphi_3 \notin U^0

; all four spanning vectors impose conditions.

❌ Check that functional on a spanning vector.

\varphi_1 + \varphi_2 + \varphi_3 + \varphi_4

sends

(1, 0, 1, 0)

2 \neq 0

❌ Not quite. Write

\varphi = c_1\varphi_1 + \dots + c_4\varphi_4

and require

\varphi

to vanish on each of the four spanning vectors.

Show solution

Let $\varphi = c_1\varphi_1 + c_2\varphi_2 + c_3\varphi_3 + c_4\varphi_4$ . Since the four vectors span $U$ , $\varphi \in U^0$ exactly when $\varphi$ is zero on each of them:

c_1 + c_3 = 0, \quad c_2 + c_4 = 0, \quad c_1 + c_2 = 0, \quad c_3 + c_4 = 0.

The first three give $c_3 = -c_1$ , $c_2 = -c_1$ , $c_4 = -c_2 = c_1$ , and then the fourth holds automatically: $c_3 + c_4 = -c_1 + c_1 = 0$ . So

U^0 = \{c_1(\varphi_1 - \varphi_2 - \varphi_3 + \varphi_4) : c_1 \in \mathbf{R}\} = \operatorname{span}(\varphi_1 - \varphi_2 - \varphi_3 + \varphi_4).

This is not $\{0\}$ , so $U \neq \mathbf{R}^4$ . Quantitatively, $\dim U^0 = 1$ and $\dim U = \dim \mathbf{R}^4 - \dim U^0 = 4 - 1 = 3$ . The dependence behind it is $(1, 0, 1, 0) + (0, 1, 0, 1) = (1, 1, 1, 1) = (1, 1, 0, 0) + (0, 0, 1, 1)$ .

Solved: 0 / 4

The Annihilator U⁰ and dim U⁰ = dim V − dim U

IKey Formulas

IIWhich Functionals Vanish on a Line

IIIThe Dual Basis Splits Off a Basis of U0U^0U0

IVRestriction Is the Dual of the Inclusion

VQuiz Questions

IIIThe Dual Basis Splits Off a Basis of $U^0$