Linear Maps

Structure-Preserving Maps Between Vector Spaces

The previous tutorial established that $C(K)$, equipped with the supremum norm, is a complete normed space when $K$ is nonempty and compact. Completeness concerns limits of sequences inside a space; another important question is how functions or vectors can be mapped from one vector space to another while respecting the operations that define those spaces. Such maps are called linear maps.

Linearity is an algebraic condition: it concerns addition and scalar multiplication. It does not, by itself, assert that a map is continuous or bounded. Those additional properties will be important when linear maps between normed spaces are studied. For now, the focus is the linear structure alone, together with two sets that reveal much about a map: its kernel and its range.

Definition and Basic Consequences

Definition: Let $V$ and $W$ be real vector spaces. A map $T:V\to W$ is a linear map if, for every $x,y\in V$ and every $\alpha,\beta\in\mathbb{R}$, $$ T(\alpha x+\beta y)=\alpha T(x)+\beta T(y). $$ A linear map from $V$ to $\mathbb{R}$ is also called a linear functional.

The condition combines preservation of addition and scalar multiplication. Equivalently, one may require $T(x+y)=T(x)+T(y)$ and $T(\alpha x)=\alpha T(x)$ for every $x,y\in V$ and $\alpha\in\mathbb{R}$. The combined form is often more efficient to check, especially when an example is defined by a formula.

Several useful identities follow immediately. Taking both coefficients to be zero gives $T(0_V)=0_W$. Taking one vector to be zero gives scalar homogeneity, and taking both coefficients to be $1$ gives additivity. In particular, $T(-x)=-T(x)$. These identities can help disprove linearity: any map that does not send zero to zero cannot be linear. But sending zero to zero alone is not enough.

Definition: For a linear map $T:V\to W$, its kernel and range are $$ \ker(T)=\{x\in V:T(x)=0_W\},\qquad \operatorname{range}(T)=\{T(x):x\in V\}. $$ The kernel consists of the inputs mapped to zero; the range consists of the outputs the map actually attains.

Worked Examples of Linearity

Worked Example: A Coordinate Map with an Explicit Inverse

Define $T:\mathbb{R}^2\to\mathbb{R}^2$ by $$ T(a,b)=(a+2b,\,3a-b). $$ To verify linearity, let $u=(a,b)$, $v=(c,d)$, and $\alpha,\beta\in\mathbb{R}$. Then $$ T(\alpha u+\beta v) =T(\alpha a+\beta c,\,\alpha b+\beta d) $$ $$ =(\alpha a+\beta c+2\alpha b+2\beta d,\, 3\alpha a+3\beta c-\alpha b-\beta d) $$ $$ =\alpha(a+2b,\,3a-b)+\beta(c+2d,\,3c-d) =\alpha T(u)+\beta T(v). $$ Thus $T$ is linear.

To find its kernel, solve $a+2b=0$ and $3a-b=0$. The first equation gives $a=-2b$; substituting into the second gives $-6b-b=-7b=0$, so $b=0$ and then $a=0$. Hence $\ker(T)=\{(0,0)\}$.

In fact, every $(r,s)\in\mathbb{R}^2$ has a preimage. Solving $a+2b=r$ and $3a-b=s$ gives $$ a=\frac{r+2s}{7},\qquad b=\frac{3r-s}{7}. $$ Substitution verifies both equations: $$ a+2b=\frac{r+2s+6r-2s}{7}=r,\qquad 3a-b=\frac{3r+6s-3r+s}{7}=s. $$ Therefore the range is all of $\mathbb{R}^2$. The zero kernel and the explicit preimage formula show, respectively, that this map is injective and surjective.

Worked Example: A Linear Functional on a Function Space

Consider $L:C[0,1]\to\mathbb{R}$, defined by $$ L(f)=\int_0^1(1+x)f(x)\,dx. $$ The sum and scalar-multiple laws for the Riemann integral give, for $f,g\in C[0,1]$ and $\alpha,\beta\in\mathbb{R}$, $$ L(\alpha f+\beta g) =\int_0^1(1+x)(\alpha f(x)+\beta g(x))\,dx =\alpha L(f)+\beta L(g). $$ Thus $L$ is a linear functional.

For concrete values, $L(1)=\int_0^1(1+x)\,dx=3/2$, while $$ L(x)=\int_0^1(x+x^2)\,dx=\frac12+\frac13=\frac56. $$ Consequently, linearity predicts $L(2-3x)=2L(1)-3L(x)=3-5/2=1/2$. Direct calculation confirms it: $$ L(2-3x)=\int_0^1(1+x)(2-3x)\,dx =\int_0^1(2-x-3x^2)\,dx =2-\frac12-1=\frac12. $$ The kernel here consists of continuous functions whose weighted integral is zero; it is not limited to the zero function. For instance, $f(x)=x-\frac{5}{9}$ belongs to the kernel, since $L(x)=5/6$, $L(1)=3/2$, and $L(x-\frac59)=\frac56-\frac59\cdot\frac32=0$.

Worked Example: Differentiation Between Polynomial Spaces

Let $\mathcal{P}_3$ denote the real polynomials of degree at most $3$, and let $\mathcal{P}_2$ denote those of degree at most $2$. Differentiation defines a map $D:\mathcal{P}_3\to\mathcal{P}_2$ by $D(p)=p'$. Write $$ p(x)=a_0+a_1x+a_2x^2+a_3x^3,\qquad q(x)=b_0+b_1x+b_2x^2+b_3x^3. $$ Then $$ D(\alpha p+\beta q)(x) =(\alpha a_1+\beta b_1)+2(\alpha a_2+\beta b_2)x +3(\alpha a_3+\beta b_3)x^2 =\alpha D(p)(x)+\beta D(q)(x). $$ Thus differentiation is linear.

A polynomial has derivative zero exactly when its coefficients of $x,x^2,x^3$ are all zero. Hence $\ker(D)$ is the space of constant polynomials. Also, every $u(x)=c_0+c_1x+c_2x^2\in\mathcal{P}_2$ is a derivative, because $$ p(x)=c_0x+\frac{c_1}{2}x^2+\frac{c_2}{3}x^3 \quad\Longrightarrow\quad p'(x)=c_0+c_1x+c_2x^2=u(x). $$ Therefore $D$ is onto, but it is not one-to-one: distinct polynomials that differ by a constant have the same derivative. For example, $D(x^3-2x)=3x^2-2$, and adding $7$ to the input leaves this output unchanged.

Kernels and Ranges Are Subspaces

The kernel is a subset of the domain and the range is a subset of the codomain. Linearity ensures that both inherit the vector-space operations. The key test for a subset to be a subspace is that it contains zero and is closed under linear combinations.

Theorem (Kernel and Range Are Subspaces): If $T:V\to W$ is linear, then $\ker(T)$ is a linear subspace of $V$, and $\operatorname{range}(T)$ is a linear subspace of $W$.

Proof. Since $T(0_V)=0_W$, the zero vector $0_V$ belongs to $\ker(T)$. If $x,y\in\ker(T)$ and $\alpha,\beta\in\mathbb{R}$, then $$ T(\alpha x+\beta y)=\alpha T(x)+\beta T(y) =\alpha 0_W+\beta 0_W=0_W. $$ Thus $\alpha x+\beta y\in\ker(T)$, which proves that the kernel is a subspace.

The range contains $0_W$, since $0_W=T(0_V)$. If $u,v\in\operatorname{range}(T)$, there are $x,y\in V$ with $u=T(x)$ and $v=T(y)$. For any $\alpha,\beta\in\mathbb{R}$, linearity gives $$ \alpha u+\beta v=\alpha T(x)+\beta T(y) =T(\alpha x+\beta y). $$ The right-hand side is an output of $T$, so $\alpha u+\beta v\in\operatorname{range}(T)$. The range is therefore a subspace of $W$. $\square$

This theorem provides a useful check on proposed descriptions of a kernel or range: each must be closed under linear combinations and contain zero. For the differentiation example, constants form a subspace of $\mathcal{P}_3$, and the whole space $\mathcal{P}_2$ is a subspace of itself. The theorem does not say that every subspace of $V$ is a kernel of a particular map, or that every subspace of $W$ is the range; it asserts only that the sets produced by a linear map have the subspace property.

The Kernel Detects Injectivity

Theorem (Injectivity and the Kernel): A linear map $T:V\to W$ is injective if and only if $\ker(T)=\{0_V\}$.

Proof. First suppose $T$ is injective. Since $T(0_V)=0_W$, if $x\in\ker(T)$, then $T(x)=0_W=T(0_V)$. Injectivity implies $x=0_V$, so the kernel contains only $0_V$.

Conversely, suppose $\ker(T)=\{0_V\}$, and let $x,y\in V$ satisfy $T(x)=T(y)$. By linearity, $$ T(x-y)=T(x)-T(y)=0_W. $$ Thus $x-y\in\ker(T)$, so $x-y=0_V$, which means $x=y$. Therefore $T$ is injective. $\square$

The theorem turns an equation about pairs of inputs into a simpler equation about the zero output. To test injectivity, one solves $T(x)=0_W$. If the only solution is $x=0_V$, no two distinct inputs can share an output. The converse implication also matters: any nonzero vector in the kernel gives two distinct inputs, $x$ and $0_V$, with the same output.

Compositions Preserve Linearity

Linear maps can be chained when the codomain of one is the domain of another. The resulting map is again linear, so multi-stage transformations can be analyzed one stage at a time.

Proposition: Let $T:V\to W$ and $S:W\to Z$ be linear maps between real vector spaces. Then the composition $S\circ T:V\to Z$, defined by $(S\circ T)(x)=S(T(x))$, is linear.

Proof. For $x,y\in V$ and $\alpha,\beta\in\mathbb{R}$, first use linearity of $T$, then linearity of $S$: $$ (S\circ T)(\alpha x+\beta y) =S(T(\alpha x+\beta y)) =S(\alpha T(x)+\beta T(y)) =\alpha S(T(x))+\beta S(T(y)). $$ This is $\alpha(S\circ T)(x)+\beta(S\circ T)(y)$, as required. $\square$

For example, differentiation $D:\mathcal{P}_3\to\mathcal{P}_2$ followed by evaluation at $0$, $E:\mathcal{P}_2\to\mathbb{R}$ with $E(q)=q(0)$, gives a linear map $E\circ D:\mathcal{P}_3\to\mathbb{R}$. It sends a polynomial $p$ to $p'(0)$. If $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, then $(E\circ D)(p)=a_1$, a linear function of the polynomial’s coefficients.

Linearity Is Not the Same as Continuity

The definition uses only vector addition and scalar multiplication. In particular, it contains no norm, distance, or limiting condition. A map can be linear between normed spaces, but deciding whether it is bounded or continuous requires additional analysis. Those questions are distinct from the algebraic tests used here.

Another common pitfall is to check only scalar homogeneity or only additivity. Neither condition by itself is the definition. For example, $F:\mathbb{R}\to\mathbb{R}$, $F(x)=x^2$, satisfies $F(-x)=F(x)$, but it is not linear: $F(1+1)=4$, whereas $F(1)+F(1)=2$. Likewise, a translated map such as $G(x)=x+1$ fails the necessary identity $G(0)=0$.

The kernel-range theorem and the injectivity criterion give a compact structural picture. The kernel records exactly which directions are lost when applying a map; the range records the part of the codomain that is reachable. Linearity forces each of these sets to be a subspace, and a trivial kernel means that no nonzero direction is lost. These ideas apply whether the vectors are coordinate tuples, polynomials, or continuous functions.

Takeaway: A linear map preserves every linear combination. Its kernel and range are subspaces, and it is injective exactly when its kernel is trivial. These are algebraic facts; boundedness and continuity require further hypotheses and analysis.

Check Your Understanding

Use the definition and results above to answer the following questions.

What single identity must a map satisfy to be linear, and which two familiar properties does it combine?
Why must the kernel and range of a linear map contain their respective zero vectors?
For a linear map $T$, how does a nonzero vector in $\ker(T)$ show that $T$ is not injective?
What are the kernel and range of differentiation from $\mathcal{P}_3$ to $\mathcal{P}_2$?
Why does the composition of two linear maps remain linear?
Does linearity alone imply boundedness or continuity? Explain which kind of condition the definition actually expresses.

What You'll Learn