Annihilators and Fundamental Subspaces, linear algebra II lesson

The map $T(x, y) = (x, y, x + y)$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ only produces outputs whose third coordinate is the sum of the first two. So the measurement $\ell(w) = w_1 + w_2 - w_3$ reads zero on every output, and its reading of $-1$ on $(1, 2, 4)$ proves that this vector is never an output. This lesson studies the measurements that vanish on a subspace, counts them, and uses them to describe the kernel and image of a dual map and the four fundamental subspaces of a matrix, all without any notion of perpendicular. Throughout, $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ , a tuple such as $(1, -1, 0)$ stands for a column vector, and a functional on $\mathbb{F}^n$ is written as a formula or as a row that acts on columns by matrix multiplication, with no conjugation.

Key terms

Annihilator

Definition Let $V$ be a vector space over $\mathbb{F}$ and $U$ a subspace of $V$ . The annihilator of $U$ is

\begin{aligned} U^{0} = \{\ell \in V^{\vee} \mid {} & \ell(u) = 0 \\[0.4em] & \text{for all } u \in U\}. \end{aligned}

It is a subspace of the dual space $V^{\vee}$ , not of $V$ . The zero functional lies in it, and if $\ell_1$ and $\ell_2$ vanish on $U$ , then so does $a\ell_1 + b\ell_2$ for all scalars $a, b$ . If $u_1, \dots, u_k$ span $U$ , then $\ell \in U^{0}$ exactly when $\ell(u_1) = \cdots = \ell(u_k) = 0$ , because a linear functional that vanishes on a spanning list vanishes on every combination of it.

In plain words The annihilator collects the measurements that cannot see $U$ at all. Each of them reads zero on every vector of $U$ , so it gives the same reading on all vectors of a coset $v + U$ .

Example In $\mathbb{R}^3$ let $U$ be spanned by $(1, 2, 0)$ and $(0, 1, 1)$ . A functional $\ell(x, y, z) = ax + by + cz$ lies in $U^{0}$ when $a + 2b = 0$ and $b + c = 0$ , so $a = -2b$ and $c = -b$ . Taking $b = -1$ shows that $U^{0}$ is spanned by

\ell(x, y, z) = 2x - y + z,

and indeed $\ell(1, 2, 0) = 0$ and $\ell(0, 1, 1) = 0$ . In $\mathbb{R}^2$ , the annihilator of the $x$ -axis consists of the multiples of the coordinate functional $(x, y) \mapsto y$ .

Dimension of an annihilator

Definition Let $V$ be a finite-dimensional vector space over $\mathbb{F}$ and $U$ a subspace of $V$ . Then

\dim U^{0} = \dim V - \dim U.

To prove it, let $\dim V = n$ and $\dim U = k$ , extend a basis $b_1, \dots, b_k$ of $U$ to a basis $b_1, \dots, b_n$ of $V$ , and let $\beta_1, \dots, \beta_n$ be the dual basis, with $\beta_i(b_j) = \delta_{ij}$ . Every $\ell \in V^{\vee}$ equals $\sum_{i} \ell(b_i)\beta_i$ , and it vanishes on $U$ exactly when $\ell(b_1) = \cdots = \ell(b_k) = 0$ . So $U^{0}$ is spanned by $\beta_{k+1}, \dots, \beta_n$ , and these $n - k$ functionals are independent because they are part of a basis. The extreme cases are $\{0\}^{0} = V^{\vee}$ and $V^{0} = \{0\}$ . The proof extends a finite basis, and the formula is stated only for finite-dimensional $V$ .

In plain words A subspace of dimension $k$ imposes $k$ independent conditions on a measurement, so $n - k$ independent measurements remain that ignore it. The bigger the subspace, the fewer independent measurements vanish on it.

Example Let $U$ be the line in $\mathbb{R}^3$ spanned by $b_1 = (1, 1, 0)$ , and extend with $b_2 = (0, 1, 0)$ and $b_3 = (0, 0, 1)$ . The dual basis is

\begin{gathered} \beta_1(x, y, z) = x, \quad \beta_2(x, y, z) = y - x, \\[0.4em] \beta_3(x, y, z) = z, \end{gathered}

as the checks $\beta_i(b_j) = \delta_{ij}$ confirm. For instance $\beta_2(b_1) = 1 - 1 = 0$ and $\beta_2(b_2) = 1$ . So $U^{0}$ is spanned by $\beta_2$ and $\beta_3$ and has dimension $3 - 1 = 2$ . Directly, $ax + by + cz$ vanishes at $(1, 1, 0)$ exactly when $a = -b$ , which gives $b(y - x) + cz$ .

Kernel of the dual map

Definition Let $V$ and $W$ be vector spaces over $\mathbb{F}$ and $T \in \operatorname{Hom}(V, W)$ , with dual map $T^{\vee}(\ell) = \ell \circ T$ . Then

\ker T^{\vee} = (\operatorname{im} T)^{0},

an equality of subspaces of $W^{\vee}$ . Indeed, $\ell \in \ker T^{\vee}$ means $\ell \circ T = 0$ , that is, $\ell(Tv) = 0$ for every $v \in V$ , and the vectors $Tv$ are exactly the vectors of $\operatorname{im} T$ . If $W$ is finite-dimensional and $T$ has rank $r$ , then $\dim \ker T^{\vee} = \dim W - r$ , so $T^{\vee}$ is injective exactly when $T$ is surjective.

In plain words An output measurement pulls back to zero exactly when it reads zero on everything $T$ can produce. A nonzero reading of such a measurement on a vector proves that $T$ never produces that vector, and when $W$ is finite-dimensional, every vector that $T$ misses gets a nonzero reading from one of them.

Example For $T(x, y) = (x, y, x + y)$ from the opening, the image is the plane $w_3 = w_1 + w_2$ and the rank is $2$ . The functional $\ell(w) = w_1 + w_2 - w_3$ pulls back to $x + y - (x + y) = 0$ , so $\ell \in \ker T^{\vee}$ . Since $\dim \ker T^{\vee} = 3 - 2 = 1$ , the kernel of $T^{\vee}$ is spanned by $\ell$ . Here $T$ is injective but $T^{\vee}$ is not, because $T$ misses outputs such as $(1, 2, 4)$ .

Image of the dual map

Definition Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{F}$ and $T \in \operatorname{Hom}(V, W)$ . Then

\operatorname{im} T^{\vee} = (\ker T)^{0},

and in particular $\operatorname{rank} T^{\vee} = \operatorname{rank} T$ . One inclusion is direct. If $\ell \in W^{\vee}$ and $v \in \ker T$ , then $(T^{\vee}\ell)(v) = \ell(Tv) = \ell(0) = 0$ , so $\operatorname{im} T^{\vee} \subseteq (\ker T)^{0}$ . For equality, let $n = \dim V$ , $m = \dim W$ and $r = \operatorname{rank} T$ . The kernel identity gives $\dim \ker T^{\vee} = m - r$ , so rank-nullity for $T^{\vee}$ on the $m$ -dimensional space $W^{\vee}$ gives $\dim \operatorname{im} T^{\vee} = r$ . The annihilator formula gives $\dim (\ker T)^{0} = n - (n - r) = r$ . A subspace contained in another of the same finite dimension equals it.

In plain words A pulled-back measurement can never see a direction that $T$ sends to zero. Conversely, every input measurement that ignores those directions comes from some output measurement.

Example Let $T(x, y, z) = (x - z, y + 2z)$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ . Its kernel is spanned by $(1, -2, 1)$ , so $(\ker T)^{0}$ consists of the functionals $ax + by + cz$ with $a - 2b + c = 0$ , a space of dimension $2$ . An output functional $pw_1 + qw_2$ pulls back to

\begin{aligned} &p(x - z) + q(y + 2z) \\[0.4em] &\quad = px + qy + (2q - p)z, \end{aligned}

and its coefficients satisfy $p - 2q + (2q - p) = 0$ . Conversely, a functional $ax + by + cz$ with $c = 2b - a$ is the pullback of $aw_1 + bw_2$ . Both $T$ and $T^{\vee}$ have rank $2$ .

Row space and left nullspace

Definition Let $A \in \mathbb{F}^{m \times n}$ have rank $r$ , viewed as the map $x \mapsto Ax$ from $\mathbb{F}^n$ to $\mathbb{F}^m$ . Functionals on $\mathbb{F}^n$ and $\mathbb{F}^m$ are rows of length $n$ and $m$ , and the dual map sends an output row $y$ to the input row $yA$ . Its image is the row space $\operatorname{Row} A$ , the span of the rows of $A$ , and its kernel is the left nullspace, the set of rows $y$ with $yA = 0$ . The two previous cards give

\begin{aligned} \operatorname{Row} A &= (\operatorname{Null} A)^{0}, \\[0.4em] \{y \mid yA = 0\} &= (\operatorname{Col} A)^{0}. \end{aligned}

So the column space in $\mathbb{F}^m$ has dimension $r$ , the nullspace in $\mathbb{F}^n$ has dimension $n - r$ , the row space, made of rows of length $n$ , has dimension $r$ , and the left nullspace, made of rows of length $m$ , has dimension $m - r$ . In particular the row rank of $A$ equals its column rank. Written as columns, the left nullspace is $\operatorname{Null} A^{\mathsf{T}}$ , since $yA = 0$ exactly when $A^{\mathsf{T}}y^{\mathsf{T}} = 0$ . Over $\mathbb{C}$ this uses the plain transpose, with no conjugation.

In plain words The rows of $A$ are measurements on inputs, and the row space contains every input measurement obtained by pulling back an output measurement. The left nullspace contains the output measurements that are blind to every output $A$ produces. Each space is described by what it annihilates, with no dot product involved.

Example The real matrix

A = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 4 & 2 \end{bmatrix}

has rank $1$ . Its column space is the line spanned by $(1, 2)$ in $\mathbb{R}^2$ , and its nullspace is the plane $x + 2y + z = 0$ in $\mathbb{R}^3$ , with basis $(-2, 1, 0)$ and $(-1, 0, 1)$ . Every $yA$ equals $(y_1 + 2y_2)\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}$ , so the row space is spanned by $\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}$ , which reads $0$ on both null basis vectors, and its dimension is $3 - 2 = 1$ . The left nullspace is spanned by $\begin{bmatrix} 2 & -1 \end{bmatrix}$ , which reads $2 - 2 = 0$ on $(1, 2)$ , and its dimension is $2 - 1 = 1$ . The same row reads $2 - 3 = -1$ on $(1, 3)$ , so $Ax = (1, 3)$ has no solution.

Annihilator versus orthogonal complement

Definition For a subspace $U$ of $V$ , the annihilator $U^{0}$ is a subspace of $V^{\vee}$ and needs no extra structure. If $V$ is an inner product space over $\mathbb{F}$ , with the inner product conjugate-linear in its first argument and linear in its second, the orthogonal complement

\begin{aligned} U^{\perp} = \{v \in V \mid {} & \langle u, v \rangle = 0 \\[0.4em] & \text{for all } u \in U\} \end{aligned}

is a subspace of $V$ itself, and it depends on the inner product. When $V$ is finite-dimensional, the inner product matches the two. Each $v \in V$ gives the functional $\ell_v(x) = \langle v, x \rangle$ , every functional on $V$ arises in this way from exactly one $v$ , and $\ell_v \in U^{0}$ exactly when $v \in U^{\perp}$ . The correspondence $v \mapsto \ell_v$ is linear over $\mathbb{R}$ and conjugate-linear over $\mathbb{C}$ . With the standard inner product $\langle x, z \rangle = x^{*}z$ on $\mathbb{F}^n$ , a row $y$ annihilates $U$ exactly when the column $y^{*}$ lies in $U^{\perp}$ .

In plain words An annihilator is made of measurements, and an orthogonal complement is made of vectors. An inner product turns each vector into a measurement, and through that identification the two sets correspond. Change the inner product and $U^{\perp}$ can move, while $U^{0}$ stays where it was.

Example In $\mathbb{C}^2$ let $U$ be spanned by $(1, i)$ . The row $\begin{bmatrix} -i & 1 \end{bmatrix}$ reads $-i + i = 0$ on it, so it spans $U^{0}$ . With $\langle x, z \rangle = x^{*}z$ the orthogonal complement is spanned by $(i, 1)$ , since $\langle (1, i), (i, 1) \rangle = i - i = 0$ , and $(i, 1)$ is the conjugate transpose of that row. The plain transpose $(-i, 1)$ is not in $U^{\perp}$ , because $\langle (1, i), (-i, 1) \rangle = -i - i = -2i$ . In $\mathbb{R}^2$ with $U$ spanned by $(1, 1)$ , the annihilator is spanned by the functional $(x, y) \mapsto x - y$ whatever the inner product. The dot product gives $U^{\perp}$ spanned by $(1, -1)$ , while the inner product $\langle v, w \rangle = 2v_1w_1 + v_2w_2$ gives $U^{\perp}$ spanned by $(1, -2)$ , since $2 \cdot 1 \cdot 1 + 1 \cdot (-2) = 0$ .

Check yourself

Find a basis of the annihilator of the line spanned by $(1, 2, 3)$ in $\mathbb{R}^3$ , and check its dimension against the formula.

Show the answer

A functional $ax + by + cz$ vanishes on the line exactly when $a + 2b + 3c = 0$ , that is, $a = -2b - 3c$ with $b$ and $c$ free. So every such functional is $b(-2x + y) + c(-3x + z)$ , and $-2x + y$ and $-3x + z$ form a basis. Both read zero at $(1, 2, 3)$ , since $-2 + 2 = 0$ and $-3 + 3 = 0$ . The dimension is $2 = 3 - 1$ , as the formula predicts.

Let $T(x, y) = (x - y, 2x - 2y)$ on $\mathbb{R}^2$ . Find $\ker T^{\vee}$ and $\operatorname{im} T^{\vee}$ directly, and compare them with $(\operatorname{im} T)^{0}$ and $(\ker T)^{0}$ .

Show the answer

The output functional $pw_1 + qw_2$ pulls back to $p(x - y) + q(2x - 2y) = (p + 2q)(x - y)$ . It is zero exactly when $p + 2q = 0$ , so $\ker T^{\vee}$ is spanned by $2w_1 - w_2$ , and $\operatorname{im} T^{\vee}$ is spanned by $x - y$ . The image of $T$ is the line spanned by $(1, 2)$ , and $cw_1 + dw_2$ vanishes on it when $c + 2d = 0$ , so $(\operatorname{im} T)^{0}$ is also spanned by $2w_1 - w_2$ . The kernel of $T$ is the line spanned by $(1, 1)$ , and $ax + by$ vanishes on it when $a + b = 0$ , so $(\ker T)^{0}$ is also spanned by $x - y$ . Both identities hold, and both $T$ and $T^{\vee}$ have rank $1$ .

Let $T(x, y, z) = (x, y)$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ . True or false? Because $T$ has a nonzero kernel, $T^{\vee}$ has a nonzero kernel.

Show the answer

False. The kernel of $T^{\vee}$ is the annihilator of the image, not of the kernel. The image of $T$ is all of $\mathbb{R}^2$ , so $\ker T^{\vee} = (\mathbb{R}^2)^{0} = \{0\}$ and $T^{\vee}$ is injective. Directly, $pw_1 + qw_2$ pulls back to $px + qy$ , which is zero only when $p = q = 0$ . The kernel of $T$ , the $z$ -axis, constrains the image of $T^{\vee}$ instead. That image is $(\ker T)^{0}$ , the functionals $ax + by$ that ignore $z$ , of dimension $3 - 1 = 2$ , so $T^{\vee}$ is not surjective onto the $3$ -dimensional space $(\mathbb{R}^3)^{\vee}$ .

Annihilators and Fundamental Subspaces

Key terms

Annihilator

Dimension of an annihilator

Kernel of the dual map

Image of the dual map

Row space and left nullspace

Annihilator versus orthogonal complement

Check yourself

How this lesson connects

Builds on