The First Isomorphism Theorem, linear algebra II lesson

The map $T(x, y, z) = (x + y, z)$ from $\mathbb{R}^3$ to $\mathbb{R}^2$ sends both $(3, 1, 2)$ and $(5, -1, 2)$ to $(4, 2)$ . It cannot see how the sum $x + y$ is split between $x$ and $y$ . After every group of inputs with a common output is merged into one object, is any further information lost? The first isomorphism theorem says that nothing more is lost, and each of its four claims has its own short proof.

Key terms

Fiber of a linear map

Definition Let $T: V \to W$ be a linear map between vector spaces over $\mathbb{F}$ , where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ . The fiber of $T$ over $y \in W$ is the set of inputs with output $y$ ,

T^{-1}(y) = \{\, v \in V \mid T(v) = y \,\}.

The notation does not assume that $T$ has an inverse. The fiber is empty when $y \notin \operatorname{im} T$ . Since $T(v) - T(u) = T(v - u)$ ,

T(v) = T(u) \iff v - u \in \ker T,

so when $T(v_0) = y$ the fiber over $y$ is the coset $v_0 + \ker T$ . The fiber over $0$ is the kernel itself, and the nonempty fibers are exactly the cosets of $\ker T$ .

In plain words A fiber collects all the inputs that the map cannot tell apart from one another. Any two of them differ by a vector of the kernel.

Example For $T(x, y, z) = (x + y, z)$ the output is zero exactly when $x + y = 0$ and $z = 0$ , so $\ker T = \operatorname{span}\{(1, -1, 0)\}$ . The fiber over $(4, 2)$ is

\begin{aligned} &(3, 1, 2) + \ker T \\[0.4em] &\quad = \{\, (3 + s, 1 - s, 2) \mid s \in \mathbb{R} \,\}, \end{aligned}

a line parallel to the kernel that contains $(5, -1, 2)$ and $(4, 0, 2)$ .

Quotient map

Definition For a subspace $U$ of a vector space $V$ over $\mathbb{F}$ , the quotient map is

q: V \to V/U, \qquad q(v) = v + U.

It is linear, because $q(av + bx) = (av + bx) + U = a(v + U) + b(x + U)$ by the definition of the quotient operations. It is onto, because every coset has a representative. Its kernel is $U$ , because $v + U$ is the zero coset exactly when $v \in U$ . None of this needs finite dimension, an inner product or a choice of complement.

In plain words The quotient map replaces each vector by the coset it belongs to. Two vectors have the same image exactly when they differ by a vector of $U$ .

Example For $U = \operatorname{span}\{(1, -1, 0)\}$ in $\mathbb{R}^3$ , the quotient map sends $(3, 1, 2)$ and $(5, -1, 2)$ to the same coset, since their difference $(-2, 2, 0)$ lies in $U$ . It sends $(1, -1, 0)$ to the zero coset $U$ . In this way every subspace is a kernel, namely the kernel of its own quotient map.

Induced map

Definition Let $T: V \to W$ be linear over $\mathbb{F}$ and let $U$ be a subspace of $V$ . The rule

\widetilde{T}(v + U) = T(v)

defines a function on $V/U$ exactly when $U \subseteq \ker T$ . If $U \subseteq \ker T$ and $v + U = v' + U$ , then $v - v' \in U$ , so $T(v) - T(v') = T(v - v') = 0$ . Conversely, if the rule is well defined, every $u \in U$ satisfies $u + U = 0 + U$ , so $T(u) = T(0) = 0$ . When $U \subseteq \ker T$ , the induced map $\widetilde{T}: V/U \to W$ is linear, because

\begin{aligned} &\widetilde{T}\big(a(v + U) + b(x + U)\big) \\[0.4em] &\quad = T(av + bx) = a\,T(v) + b\,T(x) \\[0.4em] &\quad = a\,\widetilde{T}(v + U) + b\,\widetilde{T}(x + U), \end{aligned}

and it satisfies $T = \widetilde{T} \circ q$ . It is the only map with that property, because $q$ is onto.

In plain words A map can be defined on cosets only if it ignores every direction that the quotient ignores. The quotient may throw away only what the map already throws away.

Example For $T(x, y, z) = (x + y, z)$ and $U = \ker T$ , the induced map sends the coset $(x, y, z) + U$ , which consists of the vectors $(x + s, y - s, z)$ with $s \in \mathbb{R}$ , to $(x + y, z)$ . With $U$ the $z$ -axis instead, the rule fails. The vectors $0$ and $e_3$ represent the same coset $U$ , but $T(0) = (0, 0)$ and $T(e_3) = (0, 1)$ . Here $U$ is not contained in $\ker T$ .

First isomorphism 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 with kernel $K = \ker T$ . Then the induced map

\begin{gathered} \widetilde{T}: V/K \to \operatorname{im} T, \\[0.4em] \widetilde{T}(v + K) = T(v), \end{gathered}

is a linear bijection, so $V/\ker T \cong \operatorname{im} T$ . The proof checks four things separately. The map is well defined because two representatives of one coset differ by a kernel vector. It is linear by the quotient operations and the linearity of $T$ . It is injective because $T(v) = T(u)$ gives $v - u \in K$ , so $v + K = u + K$ . It is onto $\operatorname{im} T$ because each $y \in \operatorname{im} T$ equals $T(v)$ for some $v$ , and then $\widetilde{T}(v + K) = y$ . No finite-dimensional hypothesis is needed. The original map $T$ factors as $T = \iota \circ \widetilde{T} \circ q$ , where $q: V \to V/K$ is the quotient map and $\iota$ is the inclusion of $\operatorname{im} T$ in $W$ . When $V$ is finite-dimensional, comparing dimensions gives rank-nullity, $\dim V - \dim \ker T = \dim \operatorname{im} T$ .

The theorem identifies the quotient with the image, not with the codomain, and the image equals the codomain only when $T$ is onto.

In plain words Once the inputs a map cannot tell apart are merged, nothing else is lost. Each coset of the kernel matches exactly one output in the image, and each output in the image matches exactly one coset.

Example For $T(x, y, z) = (x + y, z)$ the image is all of $\mathbb{R}^2$ . The vectors $(1, -1, 0)$ , $e_1$ and $e_3$ form a basis of $\mathbb{R}^3$ , so $e_1 + K$ and $e_3 + K$ form a quotient basis, and the induced map sends them to $(1, 0)$ and $(0, 1)$ . Because $(3, 1, 2) - (4, 0, 2) = (-1, 1, 0)$ lies in $K$ ,

\begin{aligned} (3, 1, 2) + K &= 4(e_1 + K) \\[0.4em] &\quad + 2(e_3 + K), \end{aligned}

and $\widetilde{T}$ sends this coset to $(4, 2)$ , which is $T(3, 1, 2)$ .

Polynomial evaluation map

Definition Fix a scalar $s \in \mathbb{F}$ and a whole number $d \ge 0$ , and let $\mathcal{P}_d$ be the space of polynomials over $\mathbb{F}$ of degree at most $d$ , including the zero polynomial. The evaluation map

\begin{gathered} \operatorname{ev}_s: \mathcal{P}_d \to \mathbb{F}, \\[0.4em] \operatorname{ev}_s(p) = p(s), \end{gathered}

is linear, since $(p + r)(s) = p(s) + r(s)$ and $(ap)(s) = a\,p(s)$ . It is onto, because the constant polynomial with value $a$ is sent to $a$ . Its kernel, the polynomials with $p(s) = 0$ , has dimension $d$ by rank-nullity, and for $d \ge 1$ it consists of the multiples $(t - s)\,r(t)$ with $r \in \mathcal{P}_{d-1}$ . By the first isomorphism theorem,

\begin{gathered} \mathcal{P}_d / \ker \operatorname{ev}_s \cong \mathbb{F}, \\[0.4em] p + \ker \operatorname{ev}_s \mapsto p(s). \end{gathered}

In plain words Evaluating at one point remembers a single number. Polynomials with the same value at $s$ merge into one coset, and the quotient is a copy of the scalars.

Example Take $\mathbb{F} = \mathbb{R}$ , $d = 2$ and $s = 0$ . For $p = a + bt + ct^2$ we get $\operatorname{ev}_0(p) = a$ , so the kernel is $\operatorname{span}\{t, t^2\}$ and each coset consists of all polynomials with one fixed constant term. The coset $1 + \ker \operatorname{ev}_0$ is a quotient basis, and the induced map sends $(a + bt + ct^2) + \ker \operatorname{ev}_0$ to $a$ . The rule that sends a coset to the coefficient of $t$ is not well defined. The polynomials $2 + t$ and $2 + 5t + t^2$ lie in one coset, since their difference $4t + t^2$ is in the kernel, but their coefficients of $t$ are $1$ and $5$ .

Check yourself

Let $T(x, y, z) = (x + y, z)$ . Describe the fiber over $(1, 5)$ , and decide whether $(0, 1, 5)$ and $(2, -1, 5)$ represent the same coset of $\ker T$ .

Show the answer

The vector $(1, 0, 5)$ maps to $(1, 5)$ , so the fiber is $(1, 0, 5) + \operatorname{span}\{(1, -1, 0)\}$ , the points $(1 + s, -s, 5)$ with $s \in \mathbb{R}$ . Both given vectors map to $(1, 5)$ , and their difference $(-2, 2, 0) = -2(1, -1, 0)$ lies in the kernel, so they represent the same coset.

For $P(x, y) = (x, 0)$ on $\mathbb{R}^2$ , a student concludes from the first isomorphism theorem that $\mathbb{R}^2/\ker P \cong \mathbb{R}^2$ . What is wrong, and what is the correct statement?

Show the answer

The theorem gives an isomorphism onto the image, not onto the codomain. Here $\ker P$ is the $y$ -axis and $\operatorname{im} P$ is the $x$ -axis, so the quotient has dimension $2 - 1 = 1$ and cannot be isomorphic to $\mathbb{R}^2$ . The correct statement is that the induced map $(x, y) + \ker P \mapsto (x, 0)$ is a linear bijection onto the $x$ -axis. The vector $(0, 1)$ is never reached.

Over $\mathbb{R}$ let $T: \mathcal{P}_2 \to \mathbb{R}^2$ be given by $T(p) = (p(0), p(1))$ . Find $\ker T$ , show that $T$ is onto, and state what the first isomorphism theorem gives. Is the rule $p + \ker T \mapsto p(2)$ well defined?

Show the answer

Write $p = a + bt + ct^2$ , so that $T(p) = (a, a + b + c)$ . The output is zero when $a = 0$ and $b = -c$ , so $\ker T = \operatorname{span}\{t^2 - t\}$ . Given $(u, w) \in \mathbb{R}^2$ , the polynomial $u + (w - u)t$ has values $u$ at $0$ and $w$ at $1$ , so $T$ is onto. The theorem gives

\mathcal{P}_2 / \operatorname{span}\{t^2 - t\} \cong \mathbb{R}^2

through $p + \ker T \mapsto (p(0), p(1))$ , in agreement with the count $3 - 1 = 2$ . The rule $p + \ker T \mapsto p(2)$ is not well defined. The polynomials $0$ and $t^2 - t$ lie in the same coset, but their values at $2$ are $0$ and $2$ . The kernel of $T$ is not contained in the kernel of evaluation at $2$ .

The First Isomorphism Theorem

Key terms

Fiber of a linear map

Quotient map

Induced map

First isomorphism theorem

Polynomial evaluation map

Check yourself

How this lesson connects

Builds on

Used later in