Linear-Algebra-Done-Right · Unit 9 · Video 3 · Interactive Practice

Null T′ and Range T′ as Annihilators; M(T′) Is the Transpose

IKey Formulas

Formula	Name	What it says
$\text{null}\, T' = (\text{range}\, T)^0$ , $\ \dim \text{null}\, T' = \dim \text{null}\, T + \dim W - \dim V$	Null space of the dual map	$\psi \in W'$ is killed by $T'$ exactly when $\psi$ vanishes on $\text{range}\, T$
$\text{range}\, T' = (\text{null}\, T)^0$ , $\ \dim \text{range}\, T' = \dim \text{range}\, T$	Range of the dual map	The functionals $\psi \circ T$ are exactly those that vanish on $\text{null}\, T$
$T$ surjective $\iff T'$ injective; $\ T$ injective $\iff T'$ surjective	Duality swaps the two properties	$\text{null}\, T' = \{0\} \iff \text{range}\, T = W$ , and $\text{range}\, T' = V' \iff \text{null}\, T = \{0\}$ ( $V$ , $W$ finite-dimensional)
$\mathcal{M}(T') = (\mathcal{M}(T))^{\rm t}$	Matrix of the dual map	$\mathcal{M}(T)$ in bases $v_1, \dots, v_n$ of $V$ and $w_1, \dots, w_m$ of $W$ ; $\mathcal{M}(T')$ in the dual bases $\psi_1, \dots, \psi_m$ of $W'$ and $\varphi_1, \dots, \varphi_n$ of $V'$ . With $\dim \text{range}\, T' = \dim \text{range}\, T$ , this gives column rank $=$ row rank

Key Insight: Every fact about $T$ has a mirror fact about $T'$ : the functionals $T'$ kills are those that vanish on what $T$ reaches, and the functionals $T'$ produces are those that vanish on what $T$ kills.

II $\text{null}\, T' = (\text{range}\, T)^0$

A functional $\psi$ on $\mathbf{R}^2$ lies in $\text{null}\, T'$ exactly when it vanishes on $\text{range}\, T$ .

T ψ: coefficient of w₁ 1 coefficient of w₂ 1

III $\mathcal{M}(T') = (\mathcal{M}(T))^{\rm t}$ , Entry by Entry

Each entry $C_{k,j}$ of $\mathcal{M}(T')$ is the number $(\psi_j \circ T)(v_k)$ , which $T$ computes as $A_{j,k}$ .

Map

IVColumn Rank $=$ Row Rank

For $Tx = Ax$ with $A \in \mathbf{F}^{3,4}$ , the dual map turns the column rank of $A$ into its row rank.

Row 1 Row 2 Row 3

VQuiz Questions

Problem 1 · A Functional in null T′

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $\ T(x, y, z) = (x - y + 2z,\ -2x + 2y - 4z)$ . Find a nonzero $\psi \in (\mathbf{R}^2)'$ with $\psi \in \text{null}\, T'$ .

✅ Correct!

\text{range}\, T = \text{span}\big((1, -2)\big)

and

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

, so

\psi \in (\text{range}\, T)^0 = \text{null}\, T'

❌ Check the sign.

\psi(1, -2) = 2 + 2 = 4 \ne 0

, so this

\psi

is not zero on

Te_1 = (1, -2) \in \text{range}\, T

❌ Not quite.

\psi(1, -2) = 1 - 4 = -3 \ne 0

. A functional

aw_1 + bw_2

vanishes on

\text{range}\, T = \text{span}\big((1, -2)\big)

only when

a - 2b = 0

❌ $T$ is not surjective. Every column of

\mathcal{M}(T)

is a multiple of

(1, -2)

, so

\dim \text{range}\, T = 1

and

\dim \text{null}\, T' = 2 + 2 - 3 = 1

Show solution

By the null space theorem, $\text{null}\, T' = (\text{range}\, T)^0$ . The images of the standard basis are

Te_1 = (1, -2), \quad Te_2 = (-1, 2), \quad Te_3 = (2, -4),

all multiples of $(1, -2)$ , so $\text{range}\, T = \text{span}\big((1, -2)\big)$ . A functional $\psi(w_1, w_2) = aw_1 + bw_2$ vanishes on this span exactly when $a - 2b = 0$ ; taking $b = 1$ gives $\psi(w_1, w_2) = 2w_1 + w_2$ .

Check through $T'$ : $T'(\psi) = \psi \circ T$ sends $(x, y, z)$ to

2(x - y + 2z) + (-2x + 2y - 4z) = 0,

so $T'(\psi) = 0$ . The dimension formula agrees: $\dim \text{null}\, T = 2$ , so $\dim \text{null}\, T' = 2 + 2 - 3 = 1$ , and $\psi$ spans $\text{null}\, T'$ .

Problem 2 · Which Dimensions Match

Given: $T \in \mathcal{L}(\mathbf{R}^4, \mathbf{R}^3)$ , $\ Tx = Ax$ with $A = \begin{pmatrix} 1 & 0 & 1 & 2 \\ 0 & 1 & -1 & 1 \\ 1 & 1 & 0 & 3 \end{pmatrix}.$ Find $\dim \text{range}\, T'$ and $\dim \text{null}\, T'$ .

What is dim range T′?

What is dim null T′?

✅ Correct! The ranges match (

\dim \text{range}\, T' = \dim \text{range}\, T = 2

); the null spaces do not (

\dim \text{null}\, T = 2

but

\dim \text{null}\, T' = 1

❌ That is $\dim W' = 3$ , the dimension of the domain of

T'

, not of its range. Use

\dim \text{range}\, T' = \dim \text{range}\, T

: since row 3 of

A

is row 1 plus row 2, the rank of

A

is less than 3.

❌ Not quite. Use

\dim \text{range}\, T' = \dim \text{range}\, T

, the column rank of

A

❌ That is $\dim \text{null}\, T$ . Null spaces of

T

and

T'

need not have equal dimension:

\dim \text{null}\, T' = \dim \text{null}\, T + \dim W - \dim V = 2 + 3 - 4

❌ $T$ is not surjective.

\dim \text{range}\, T = 2 \lt 3 = \dim \mathbf{R}^3

, so

T'

is not injective.

❌ Not quite. Use

\dim \text{null}\, T' = \dim \text{null}\, T + \dim W - \dim V

Show solution

Range. Row 3 of $A$ equals row 1 plus row 2, and rows 1 and 2 are not multiples of each other, so $A$ has rank $2$ . The columns $(1, 0, 1)$ and $(0, 1, 1)$ span the column space (columns 3 and 4 are $(1, -1, 0) = c_1 - c_2$ and $(2, 1, 3) = 2c_1 + c_2$ ), so $\dim \text{range}\, T = 2$ and

\dim \text{range}\, T' = \dim \text{range}\, T = 2.

Null space. By the fundamental theorem, $\dim \text{null}\, T = 4 - 2 = 2$ . Then

\dim \text{null}\, T' = \dim \text{null}\, T + \dim W - \dim V = 2 + 3 - 4 = 1.

Check: $T' \in \mathcal{L}\big((\mathbf{R}^3)', (\mathbf{R}^4)'\big)$ , and $\dim \text{null}\, T' + \dim \text{range}\, T' = 1 + 2 = 3 = \dim (\mathbf{R}^3)'$ . Indeed $\psi(w_1, w_2, w_3) = w_1 + w_2 - w_3$ spans $\text{null}\, T'$ , since it vanishes on both columns $c_1$ and $c_2$ .

Problem 3 · Reading the Transpose

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $\ T(x, y, z) = (2x - y + 3z,\ x + 4z)$ , with the standard bases $e_1, e_2, e_3$ and $e_1, e_2$ , their dual bases $\varphi_1, \varphi_2, \varphi_3$ of $(\mathbf{R}^3)'$ and $\psi_1, \psi_2$ of $(\mathbf{R}^2)'$ . Find $T'(\psi_2)$ and the entry in row 3, column 1 of $\mathcal{M}(T')$ .

What is T′(ψ₂)?

What is the entry in row 3, column 1 of M(T′)?

✅ Correct!

\mathcal{M}(T') = A^{\rm t}

: column 2 of

A^{\rm t}

is row 2 of

A

, namely

(1, 0, 4)

, and

(A^{\rm t})_{3,1} = A_{1,3} = 3

❌ That is $T'(\psi_1)$ .

\psi_1 \circ T

reads the first coordinate

2x - y + 3z

;

\psi_2

reads the second.

❌ That reads a column of $A$ . Column 2 of

A

(-1, 0)

, the coordinates of

Te_2

\mathbf{R}^2

, not the coefficients of a functional on

\mathbf{R}^3

; the coefficients of

T'(\psi_2)

come from row 2 of

A

❌ Watch the zero.

\psi_2 \circ T

sends

(x, y, z)

x + 0y + 4z

, so the

4

belongs to

\varphi_3

, the coordinate functional of

z

❌ Not quite. Compute

T'(\psi_2) = \psi_2 \circ T

and write it in the basis

\varphi_1, \varphi_2, \varphi_3

❌ Indices swapped the wrong way.

4 = A_{2,3}

is the entry in row 3, column 2 of

\mathcal{M}(T')

; row 3, column 1 is

A_{1,3}

❌ Not quite.

C_{k,j} = A_{j,k}

: row 3, column 1 of

C

is row 1, column 3 of

A

Show solution

With the standard bases,

A = \mathcal{M}(T) = \begin{pmatrix} 2 & -1 & 3 \\ 1 & 0 & 4 \end{pmatrix}.

Directly: $T'(\psi_2) = \psi_2 \circ T$ sends $(x, y, z)$ to the second coordinate of $T(x, y, z)$ , which is $x + 4z = \varphi_1(x, y, z) + 4\varphi_3(x, y, z)$ . So $T'(\psi_2) = \varphi_1 + 4\varphi_3$ .

By the theorem: $\mathcal{M}(T') = A^{\rm t}$ , and

A^{\rm t} = \begin{pmatrix} 2 & 1 \\ -1 & 0 \\ 3 & 4 \end{pmatrix}.

Column 2 of $A^{\rm t}$ is $(1, 0, 4)$ , the coefficients of $T'(\psi_2)$ in $\varphi_1, \varphi_2, \varphi_3$ , matching the direct computation. The entry in row 3, column 1 is $(A^{\rm t})_{3,1} = A_{1,3} = 3$ ; it is also the $\varphi_3$ -coefficient of $T'(\psi_1) = 2\varphi_1 - \varphi_2 + 3\varphi_3$ .

Problem 4 · When Is T′ Injective?

Given: $T \in \mathcal{L}(\mathbf{R}^3, \mathbf{R}^2)$ , $\ Tx = Ax$ with $A = \begin{pmatrix} 1 & 2 & c \\ 2 & 4 & 6 \end{pmatrix}$ , where $c \in \mathbf{R}$ .

For which c is T′ injective?

For c = 0, is T′ surjective?

✅ Correct!

T'

is injective

\iff

T

is surjective

\iff c \ne 3

; and since

\dim \text{null}\, T \ge 3 - 2 = 1

T

is never injective, so

T'

is never surjective.

❌ Reversed. At

c = 3

every column is a multiple of

(1, 2)

, so

T

is not surjective and

T'

is not injective.

❌ Not quite. At

c = 6

the column

(6, 6)

is not a multiple of

(1, 2)

, so

\text{range}\, T = \mathbf{R}^2

and

T'

is injective. The bad value is where

(c, 6)

is a multiple of

(1, 2)

❌ Not quite.

T'

is injective exactly when

T

is surjective, that is, when the columns of

A

span

\mathbf{R}^2

❌ Injective does not give surjective here.

T' \colon (\mathbf{R}^2)' \to (\mathbf{R}^3)'

goes from dimension 2 to dimension 3, so it cannot be both.

❌ Compare with the target.

\dim \text{range}\, T' = 2

, but

T'

maps into

(\mathbf{R}^3)'

, which has dimension 3.

❌ Not quite.

T'

is surjective exactly when

T

is injective.

Show solution

Injectivity of $T'$ . By duality, $T'$ is injective $\iff$ $T$ is surjective $\iff \text{range}\, T = \mathbf{R}^2$ . The range is the span of the columns $(1, 2)$ , $(2, 4)$ , $(c, 6)$ . The second column is twice the first, so $\text{range}\, T = \mathbf{R}^2$ exactly when $(c, 6)$ is not a multiple of $(1, 2)$ . A multiple $t(1, 2) = (c, 6)$ forces $t = 3$ and $c = 3$ . So $T'$ is injective for every $c \ne 3$ .

Equivalently, $\dim \text{null}\, T' = \dim \text{null}\, T + 2 - 3$ : for $c \ne 3$ , $\dim \text{null}\, T = 3 - 2 = 1$ gives $\dim \text{null}\, T' = 0$ ; for $c = 3$ , $\dim \text{null}\, T = 2$ gives $\dim \text{null}\, T' = 1$ , spanned by $\psi(w_1, w_2) = 2w_1 - w_2$ .

Surjectivity of $T'$ . $T'$ is surjective $\iff$ $T$ is injective. But $\dim \text{null}\, T = 3 - \dim \text{range}\, T \ge 3 - 2 = 1$ , so $T$ is never injective. For $c = 0$ : $\dim \text{range}\, T' = \dim \text{range}\, T = 2 \lt 3 = \dim (\mathbf{R}^3)'$ , so $T'$ is injective but not surjective.

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Null T′ and Range T′ as Annihilators; M(T′) Is the Transpose

IKey Formulas

IInull T′=(range T)0\text{null}\, T' = (\text{range}\, T)^0nullT′=(rangeT)0

IIIM(T′)=(M(T))t\mathcal{M}(T') = (\mathcal{M}(T))^{\rm t}M(T′)=(M(T))t, Entry by Entry

IVColumn Rank === Row Rank

VQuiz Questions

II $\text{null}\, T' = (\text{range}\, T)^0$

III $\mathcal{M}(T') = (\mathcal{M}(T))^{\rm t}$ , Entry by Entry

IVColumn Rank $=$ Row Rank