Rank-Nullity and Isomorphisms, linear algebra II lesson

Suppose a real polynomial of degree at most two takes the value $2$ at $t = 0$ and the value $5$ at $t = 1$ . The linear polynomial $2 + 3t$ fits, but so does $2 + 3t + s(t^2 - t)$ for every real $s$ , because $t^2 - t$ vanishes at both points. Two readings leave one direction of a three-dimensional space undetected, and the rank-nullity theorem says that this count always balances. This lesson proves the theorem and uses it to decide when a linear map is injective, surjective or invertible, and how much rank survives a composition.

Key terms

Rank-nullity theorem

Definition Let $V$ and $W$ be vector spaces over $\mathbb{F}$ , where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ , and let $T: V \to W$ be linear. The kernel and the image of $T$ are the subspaces

\begin{aligned} \ker T &= \{x \in V : T(x) = 0\}, \\[0.4em] \operatorname{im} T &= \{T(x) : x \in V\}. \end{aligned}

The nullity of $T$ is $\dim \ker T$ and the rank is $\operatorname{rank} T = \dim \operatorname{im} T$ . If $V$ is finite-dimensional, both are finite and

\dim V = \dim \ker T + \operatorname{rank} T.

The space $W$ need not be finite-dimensional, and its dimension plays no part. For the proof, choose a basis $k_1, \dots, k_r$ of $\ker T$ and extend it to a basis $k_1, \dots, k_r, v_1, \dots, v_s$ of $V$ . The images $T(v_1), \dots, T(v_s)$ span $\operatorname{im} T$ , because expanding any $x$ in this basis and applying $T$ removes every kernel term. They are independent. If $\sum_i a_i T(v_i) = 0$ , then $\sum_i a_i v_i$ lies in $\ker T$ , so it equals $\sum_l d_l k_l$ for some scalars $d_l$ , and independence of the extended basis forces every $a_i$ to be $0$ . Hence the rank is $s$ , the nullity is $r$ , and $\dim V = r + s$ . The cases $r = 0$ and $s = 0$ are included, with an empty list as the basis of the zero space.

In plain words The dimension of the domain splits into two counts, the directions the map cannot detect and the independent directions it actually produces. The size of the codomain does not enter.

Example Over $\mathbb{R}$ let $T: \mathcal{P}_2 \to \mathbb{R}^2$ be $T(p) = (p(0), p(1))$ , where $\mathcal{P}_2$ is the space of polynomials of degree at most two, including the zero polynomial. For $p = a + bt + ct^2$ this gives $T(p) = (a, a + b + c)$ . The kernel requires $a = 0$ and $b + c = 0$ , so $\ker T = \operatorname{span}(t^2 - t)$ has dimension $1$ . Every target $(y_1, y_2)$ is reached by $y_1 + (y_2 - y_1)t$ , so $\operatorname{im} T = \mathbb{R}^2$ has dimension $2$ , and $3 = 1 + 2$ . One solution plus the kernel gives all solutions, so the polynomials with readings $(2, 5)$ are exactly

p(t) = 2 + 3t + s(t^2 - t), \qquad s \in \mathbb{R}.

Injective map

Definition A map $T: V \to W$ is injective, or one-to-one, if $T(x) = T(y)$ implies $x = y$ for all $x, y \in V$ . A linear map is injective exactly when $\ker T = \{0\}$ . If the kernel is zero and $T(x) = T(y)$ , then $T(x - y) = 0$ , so $x - y = 0$ . If instead some $k \ne 0$ lies in the kernel, then $T(k) = T(0)$ with $k \ne 0$ . When $V$ is finite-dimensional, rank-nullity turns injectivity into the condition $\operatorname{rank} T = \dim V$ . When $W$ is finite-dimensional as well, an injective linear map from $V$ to $W$ can exist only if $\dim V \le \dim W$ .

In plain words An injective map never sends two different inputs to the same output. For a linear map it is enough to check that only the zero vector goes to zero.

Example The inclusion $\iota(x, y) = (x, y, 0)$ from $\mathbb{R}^2$ to $\mathbb{R}^3$ has kernel $\{0\}$ , so it is injective. The two-reading map $T(p) = (p(0), p(1))$ on $\mathcal{P}_2$ is not injective, since $t^2 - t$ and the zero polynomial both give $(0, 0)$ .

Surjective map

Definition A map $T: V \to W$ is surjective, or onto, if $\operatorname{im} T = W$ , that is, if every $w \in W$ equals $T(x)$ for some $x \in V$ . Surjectivity depends on the stated codomain $W$ , not only on the formula. When $W$ is finite-dimensional, a linear map $T$ is surjective exactly when $\operatorname{rank} T = \dim W$ , because a subspace of $W$ with the same finite dimension as $W$ is all of $W$ . If $V$ is also finite-dimensional, rank-nullity gives $\operatorname{rank} T \le \dim V$ , so a surjective linear map from $V$ to $W$ can exist only if $\dim W \le \dim V$ .

In plain words Every vector of the codomain is hit. For a linear map between finite-dimensional spaces this comes down to comparing the dimension of the image with the dimension of the codomain.

Example The projection $\pi(x, y, z) = (x, y)$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ is surjective, since $(x, y) = \pi(x, y, 0)$ , but it is not injective, because its kernel is the $z$ -axis. The inclusion $\iota(x, y) = (x, y, 0)$ is not surjective onto $\mathbb{R}^3$ , since $(0, 0, 1)$ is never reached. The same formula viewed as a map to the plane $z = 0$ is surjective, which is why the codomain has to be stated.

Isomorphism

Definition An isomorphism is a linear map $T: V \to W$ that is bijective, meaning both injective and surjective. Two vector spaces over $\mathbb{F}$ are isomorphic when an isomorphism between them exists. Isomorphic finite-dimensional spaces have the same dimension, since an isomorphism has nullity $0$ and rank $\dim W$ . Now let $V$ and $W$ be finite-dimensional over $\mathbb{F}$ with $\dim V = \dim W = n$ , and let $T: V \to W$ be linear. Then $T$ is injective if and only if it is surjective, and either property makes $T$ an isomorphism. Rank-nullity gives $\dim \ker T + \operatorname{rank} T = n = \dim W$ , so

\begin{aligned} &\dim \ker T = 0 \\[0.4em] &\quad \iff \operatorname{rank} T = \dim W. \end{aligned}

The equivalence can fail when the dimensions differ. The inclusion of $\mathbb{R}^2$ in $\mathbb{R}^3$ is injective and not surjective, and the projection of $\mathbb{R}^3$ onto $\mathbb{R}^2$ is surjective and not injective.

In plain words An isomorphism matches the vectors of two spaces one for one while respecting addition and scaling, so the two spaces carry the same linear structure. It does not say the objects are identical, and different isomorphisms give different matchings. In equal finite dimensions one of the two checks is enough.

Example Over $\mathbb{R}$ let $T: \mathcal{P}_2 \to \mathbb{R}^3$ be $T(p) = (p(0), p(1), p(-1))$ . If $T(p) = 0$ , then $p$ has the three distinct roots $0$ , $1$ and $-1$ , and a nonzero polynomial of degree at most two has at most two roots, so $p = 0$ . The kernel is zero and both spaces have dimension $3$ , so $T$ is an isomorphism. To find the polynomial with readings $(2, 5, 1)$ , write $p = a + bt + ct^2$ . The readings give $a = 2$ , $b + c = 3$ and $-b + c = -1$ , so $c = 1$ , $b = 2$ and $p = 2 + 2t + t^2$ .

Linear inverse

Definition Let $T: V \to W$ be a bijective linear map between vector spaces over $\mathbb{F}$ . Its inverse $T^{-1}: W \to V$ sends each $y \in W$ to the unique $x \in V$ with $T(x) = y$ , so $T^{-1} \circ T$ is the identity of $V$ and $T \circ T^{-1}$ is the identity of $W$ . The inverse is linear. Take $y, z \in W$ with $T^{-1}(y) = x$ and $T^{-1}(z) = v$ , and scalars $a, b \in \mathbb{F}$ . Linearity of $T$ gives $T(ax + bv) = ay + bz$ , and since $ax + bv$ is the only preimage of $ay + bz$ ,

T^{-1}(ay + bz) = aT^{-1}(y) + bT^{-1}(z).

No finite-dimensional hypothesis is needed. When $V$ and $W$ are finite-dimensional with bases $\mathcal{B}$ and $\mathcal{C}$ , the matrix $[T^{-1}]_{\mathcal{B} \leftarrow \mathcal{C}}$ is the inverse of the square matrix $[T]_{\mathcal{C} \leftarrow \mathcal{B}}$ .

In plain words Undoing a linear bijection never breaks linearity. Reading the map backwards gives another linear map, not merely a function.

Example For the three-reading isomorphism above, solving $a = r$ , $a + b + c = s$ and $a - b + c = u$ for arbitrary readings $(r, s, u)$ gives

\begin{aligned} T^{-1}(r, s, u) &= r + \frac{s - u}{2}\, t \\[0.4em] &\quad + \frac{s + u - 2r}{2}\, t^2. \end{aligned}

Each coefficient is a linear expression in $r$ , $s$ and $u$ , as the theorem predicts. For example $T^{-1}(1, 0, 4) = 1 - 2t + t^2$ , and indeed $(1 - t)^2$ takes the values $1$ , $0$ and $4$ at $0$ , $1$ and $-1$ .

Rank of a composition

Definition Let $T: V \to W$ and $S: W \to Z$ be linear maps between vector spaces over $\mathbb{F}$ , with $V$ and $W$ finite-dimensional, and write $ST$ for the composition $S \circ T$ , which applies $T$ first. Then $\ker T \subseteq \ker(ST)$ and $\operatorname{im}(ST) \subseteq \operatorname{im} S$ , and

\operatorname{rank}(ST) \le \min(\operatorname{rank} S, \operatorname{rank} T).

The first inclusion holds because $T(x) = 0$ gives $S(T(x)) = S(0) = 0$ . Rank-nullity on the common domain $V$ then gives

\begin{aligned} \operatorname{rank}(ST) &= \dim V - \dim \ker(ST) \\[0.4em] &\le \dim V - \dim \ker T \\[0.4em] &= \operatorname{rank} T. \end{aligned}

This half uses only that $V$ is finite-dimensional. The second inclusion holds because every output $S(T(x))$ is an output of $S$ , so $\operatorname{rank}(ST) \le \operatorname{rank} S$ . For matrices over $\mathbb{F}$ this reads $\operatorname{rank}(AB) \le \min(\operatorname{rank} A, \operatorname{rank} B)$ whenever the product $AB$ is defined.

In plain words A composition cannot produce more independent outputs than either stage. What the first map loses stays lost, and the second map cannot reach outside its own image. The bound can be strict.

Example Let $T: \mathbb{R} \to \mathbb{R}^2$ be $T(t) = (t, 0)$ and $S: \mathbb{R}^2 \to \mathbb{R}$ be $S(x, y) = y$ . Each has rank one, but $S(T(t)) = 0$ for every $t$ , so $ST$ is the zero map and has rank $0$ . The map $S$ discards exactly the direction that $T$ produces.

Check yourself

A linear map $T: \mathbb{R}^3 \to \mathbb{R}^4$ has $\ker T = \{0\}$ . Is $T$ an isomorphism?

Show the answer

No. Rank-nullity gives $\operatorname{rank} T = 3 - 0 = 3$ , while $\dim \mathbb{R}^4 = 4$ , so the image is a proper subspace of $\mathbb{R}^4$ and $T$ is not surjective. In fact no linear map from $\mathbb{R}^3$ to $\mathbb{R}^4$ is surjective. The shortcut that makes an injective map automatically surjective needs the domain and codomain to have the same finite dimension, and here they do not.

Let $D: \mathcal{P}_3 \to \mathcal{P}_3$ be differentiation over $\mathbb{R}$ . Find its kernel and image, check rank-nullity, and decide whether $D$ is injective or surjective.

Show the answer

The derivative of $p$ is zero exactly when $p$ is constant, so the kernel is the space of constant polynomials, of dimension $1$ . The derivative of $a_0 + a_1 t + a_2 t^2 + a_3 t^3$ is $a_1 + 2a_2 t + 3a_3 t^2$ , and every polynomial $q_0 + q_1 t + q_2 t^2$ arises, as the derivative of $q_0 t + \frac{q_1}{2} t^2 + \frac{q_2}{3} t^3$ . So $\operatorname{im} D = \mathcal{P}_2$ , of dimension $3$ , and $4 = 1 + 3$ . The map is not injective, since its kernel is nonzero, and not surjective, since $t^3$ is not in the image. Domain and codomain have the same dimension here, so the two properties had to fail together.

Let $A$ be a real $3 \times 2$ matrix and $B$ a real $2 \times 3$ matrix. Can $AB$ be invertible? Can $BA$ ?

Show the answer

The product $AB$ is $3 \times 3$ , and $\operatorname{rank}(AB) \le \operatorname{rank} A \le 2 < 3$ , so $AB$ is never invertible. The product $BA$ is $2 \times 2$ , and the bound gives only $\operatorname{rank}(BA) \le 2$ , which does not rule out invertibility. It does happen. For the matrices

A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}

the product $BA$ is the identity $I_2$ , while $AB$ has rank $2$ .

Rank-Nullity and Isomorphisms

Key terms

Rank-nullity theorem

Injective map

Surjective map

Isomorphism

Linear inverse

Rank of a composition

Check yourself

How this lesson connects

Builds on

Used later in