# Linear Map A linear map is a function between vector spaces that preserves the two operations which give vector spaces their structure: addition and scalar multiplication. Despite this deceptively simple requirement, linear maps form the backbone of nearly all of modern mathematics. Differential operators, rotations of the plane, projections onto subspaces, quantum observables, and the derivative itself are all linear maps. The theory developed in this article --- the kernel-image decomposition, the rank-nullity theorem, matrix representations, and duality --- provides the language in which these diverse phenomena are understood as instances of a single algebraic pattern. The power of linearity lies in what it makes possible. A linear map is completely determined by its values on a basis, which means that every linear map between finite-dimensional spaces can be encoded as a matrix. This passage from abstract to concrete --- and back --- is the central theme of linear algebra. ## Why Structure Preservation Matters [motivation] ### The search for well-behaved transformations Consider a function $f \colon \mathbb{R}^2 \to \mathbb{R}^2$ that transforms the plane. Among all possible such [functions](/page/Function) --- reflections, translations, dilations, polynomials, wild discontinuous maps --- which ones can we hope to understand completely? The answer begins with a structural observation: the functions that respect the algebraic structure of $\mathbb{R}^2$ are precisely the ones amenable to systematic analysis. What does "respect the structure" mean? A vector space has two operations: we can add vectors and scale them. A function $\alpha \colon \mathbb{R}^2 \to \mathbb{R}^2$ that preserves both operations satisfies \begin{align*} \alpha(u + v) &= \alpha(u) + \alpha(v), \\ \alpha(\lambda u) &= \lambda \, \alpha(u), \end{align*} for all vectors $u, v \in \mathbb{R}^2$ and all scalars $\lambda \in \mathbb{R}$. These two conditions are the definition of a linear map. ### What fails without linearity To appreciate why these conditions are so powerful, consider what goes wrong without them. Take the squaring function $g \colon \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x^2$. This map is decidedly nonlinear: $g(2 + 3) = 25$, while $g(2) + g(3) = 13$. As a consequence, $g$ has no well-defined kernel in the algebraic sense. The preimage $g^{-1}(\{0\}) = \{0\}$ is a single point, but there is no analogue of the rank-nullity theorem: knowing $\dim(\ker g)$ tells us nothing about the dimension of the image. More critically, a nonlinear function cannot be represented by a matrix. The map $h \colon \mathbb{R}^2 \to \mathbb{R}^2$ given by $h(x, y) = (x^2, xy)$ sends lines through the origin to curves, not lines. There is no hope of finding a $2 \times 2$ matrix $A$ such that $h(v) = Av$ for all $v$. Without matrix representation, we lose the entire computational apparatus of determinants, eigenvalues, and canonical forms. ### The matrix connection For a linear map $\alpha \colon U \to V$ between finite-dimensional vector spaces, choosing a basis for $U$ and a basis for $V$ produces a matrix $A$ such that $\alpha(u)$ is computed by matrix multiplication. Different choices of bases yield different matrices, but all such matrices are related by a change-of-basis transformation. This means that intrinsic properties of $\alpha$ --- its rank, nullity, determinant (when $U = V$), and eigenvalues --- are invariants that do not depend on the coordinate system. The single definition of linearity thus unifies three perspectives: the geometric (transformations of space), the computational (matrix arithmetic), and the abstract (structure-preserving homomorphisms between algebraic objects). [/motivation] ## The Definition of a Linear Map Throughout this article, let $F$ denote a field (such as $\mathbb{R}$ or $\mathbb{C}$), and let $U$ and $V$ be vector spaces over $F$. [definition:Linear Map] A **linear map** (also called a **linear transformation** or **homomorphism of vector spaces**) is a function \begin{align*} \alpha \colon U &\to V \end{align*} satisfying the following two conditions for all $u, u' \in U$ and all $\lambda \in F$: 1. **Additivity**: $\alpha(u + u') = \alpha(u) + \alpha(u')$. 2. **Homogeneity**: $\alpha(\lambda u) = \lambda \, \alpha(u)$. Equivalently, $\alpha$ is linear if and only if $\alpha(\lambda u + \mu u') = \lambda \, \alpha(u) + \mu \, \alpha(u')$ for all $\lambda, \mu \in F$ and all $u, u' \in U$. The [set](/page/Set) of all linear maps from $U$ to $V$ is denoted $\operatorname{Hom}(U, V)$ or $\mathcal{L}(U, V)$, and is itself a vector space over $F$ under pointwise addition and scalar multiplication. [/definition] An immediate consequence of the definition is that every linear map sends the zero vector to the zero vector: setting $\lambda = 0$ in the homogeneity condition gives $\alpha(\mathbf{0}_U) = \alpha(0 \cdot u) = 0 \cdot \alpha(u) = \mathbf{0}_V$. This provides a quick test for non-linearity: any function $f$ with $f(\mathbf{0}) \neq \mathbf{0}$ cannot be linear. A second fundamental property is that a linear map is completely determined by its action on a basis. If $\{e_1, \ldots, e_n\}$ is a basis for $U$, then for any $u = \lambda_1 e_1 + \cdots + \lambda_n e_n$ we have \begin{align*} \alpha(u) = \lambda_1 \, \alpha(e_1) + \cdots + \lambda_n \, \alpha(e_n). \end{align*} Conversely, given any choice of vectors $v_1, \ldots, v_n \in V$, there exists a unique linear map $\alpha \colon U \to V$ satisfying $\alpha(e_i) = v_i$ for each $i$. This is the basis of the matrix representation, which we develop in a later section. [example:Rotation] **Rotation of the plane.** Fix an angle $\theta \in \mathbb{R}$ and define $R_\theta \colon \mathbb{R}^2 \to \mathbb{R}^2$ by \begin{align*} R_\theta(x, y) = (x \cos \theta - y \sin \theta, \; x \sin \theta + y \cos \theta). \end{align*} To verify linearity, take $u = (x_1, y_1)$, $u' = (x_2, y_2)$, and $\lambda \in \mathbb{R}$. Then \begin{align*} R_\theta(u + u') &= R_\theta(x_1 + x_2, \; y_1 + y_2) \\ &= \bigl((x_1 + x_2)\cos\theta - (y_1 + y_2)\sin\theta, \; (x_1 + x_2)\sin\theta + (y_1 + y_2)\cos\theta\bigr) \\ &= (x_1 \cos\theta - y_1 \sin\theta, \; x_1 \sin\theta + y_1 \cos\theta) + (x_2 \cos\theta - y_2 \sin\theta, \; x_2 \sin\theta + y_2 \cos\theta) \\ &= R_\theta(u) + R_\theta(u'). \end{align*} Homogeneity follows by a similar computation. Therefore $R_\theta$ is a linear map for every $\theta$. [/example] [example:Differentiation] **The derivative as a linear map.** Let $P_n(\mathbb{R})$ denote the vector space of real polynomials of degree at most $n$, and define the [differentiation](/page/Derivative) map \begin{align*} D \colon P_n(\mathbb{R}) &\to P_{n-1}(\mathbb{R}), \\ p &\mapsto p'. \end{align*} Linearity of $D$ is precisely the statement that the derivative of a sum is the sum of derivatives, and the derivative of a scalar multiple is the scalar times the derivative --- both standard properties from calculus. The map $D$ is surjective: every polynomial $q$ of degree at most $n - 1$ is the derivative of some polynomial in $P_n(\mathbb{R})$. Its kernel consists of the constant polynomials, so $\ker D = P_0(\mathbb{R})$, which has dimension $1$. [/example] [example:Projection] **Projection onto a subspace.** Let $V = W \oplus W'$ be a direct sum decomposition. The projection onto $W$ along $W'$ is the map \begin{align*} \pi \colon V &\to V, \\ w + w' &\mapsto w, \end{align*} where $w \in W$ and $w' \in W'$. This map is well-defined because every vector in $V$ has a unique decomposition as a sum of an element of $W$ and an element of $W'$. Linearity follows from the uniqueness of this decomposition: if $v_1 = w_1 + w_1'$ and $v_2 = w_2 + w_2'$, then $v_1 + v_2 = (w_1 + w_2) + (w_1' + w_2')$, so $\pi(v_1 + v_2) = w_1 + w_2 = \pi(v_1) + \pi(v_2)$. A projection satisfies $\pi^2 = \pi$: applying it twice gives the same result as applying it once. [/example] ## Kernel and Image The two most important subsets associated with a linear map are its kernel and its image. Together, they measure how far $\alpha$ is from being injective and surjective, respectively. Their interplay is the subject of the rank-nullity theorem and the [first isomorphism theorem](/theorems/791). [definition:Kernel] Let $\alpha \colon U \to V$ be a linear map. The **kernel** (or **null space**) of $\alpha$ is the set \begin{align*} \ker \alpha = \{ u \in U : \alpha(u) = \mathbf{0}_V \}. \end{align*} The **nullity** of $\alpha$ is $n(\alpha) = \dim(\ker \alpha)$. [/definition] [definition:Image] Let $\alpha \colon U \to V$ be a linear map. The **image** (or **range**) of $\alpha$ is the set \begin{align*} \operatorname{im} \alpha = \{ \alpha(u) : u \in U \} = \{ v \in V : v = \alpha(u) \text{ for some } u \in U \}. \end{align*} The **rank** of $\alpha$ is $r(\alpha) = \dim(\operatorname{im} \alpha)$. [/definition] Both $\ker \alpha$ and $\operatorname{im} \alpha$ are subspaces --- the kernel is a subspace of $U$, the image a subspace of $V$. Verifying this is a direct application of linearity: if $\alpha(u_1) = \mathbf{0}$ and $\alpha(u_2) = \mathbf{0}$, then $\alpha(\lambda u_1 + \mu u_2) = \lambda \alpha(u_1) + \mu \alpha(u_2) = \mathbf{0}$, so $\ker \alpha$ is closed under linear combinations. The argument for $\operatorname{im} \alpha$ is analogous. The kernel controls injectivity: $\alpha$ is injective if and only if $\ker \alpha = \{\mathbf{0}\}$. To see this, suppose $\alpha(u_1) = \alpha(u_2)$. By linearity, $\alpha(u_1 - u_2) = \mathbf{0}$, so $u_1 - u_2 \in \ker \alpha$. If the kernel is trivial, this forces $u_1 = u_2$. The kernel also defines an equivalence relation on $U$: declare $u_1 \sim u_2$ whenever $\alpha(u_1) = \alpha(u_2)$, or equivalently, whenever $u_1 - u_2 \in \ker \alpha$. The equivalence classes are the cosets $u + \ker \alpha$, and they form the quotient space $U / \ker \alpha$. This quotient space is the domain of the first isomorphism theorem, which establishes a canonical relationship between the structure that $\alpha$ "collapses" (its kernel) and the structure it "creates" (its image). [quotetheorem:384] The first isomorphism theorem says that $U / \ker \alpha \cong \operatorname{im} \alpha$ via the map $u + \ker \alpha \mapsto \alpha(u)$. This result is the linear-algebraic analogue of the first isomorphism theorem for [groups](/page/Group): in both settings, quotienting by the kernel yields the image. The theorem tells us that no information is lost by passing to the quotient --- the "compressed" domain $U / \ker \alpha$ is a perfect copy of the image. It also provides a conceptual explanation for the rank-nullity theorem: since $U / \ker \alpha \cong \operatorname{im} \alpha$, the dimensions must agree, and by the dimension formula for quotient spaces, $\dim(U / \ker \alpha) = \dim U - \dim(\ker \alpha)$. [example:KernelImage] **Kernel and image of a concrete map.** Define $\alpha \colon \mathbb{R}^3 \to \mathbb{R}^2$ by \begin{align*} \alpha(x, y, z) = (x + y, \; y + z). \end{align*} To find $\ker \alpha$, solve $\alpha(x, y, z) = (0, 0)$: \begin{align*} x + y &= 0, \\ y + z &= 0. \end{align*} From the first equation $x = -y$, and from the second $z = -y$. Setting $y = t$ gives $\ker \alpha = \{(-t, t, -t) : t \in \mathbb{R}\} = \operatorname{span}\{(-1, 1, -1)\}$, so $n(\alpha) = 1$. For the image, note that $\alpha(1, 0, 0) = (1, 0)$ and $\alpha(0, 1, 0) = (1, 1)$. These two vectors are linearly independent in $\mathbb{R}^2$, so $\operatorname{im} \alpha = \mathbb{R}^2$ and $r(\alpha) = 2$. As expected, $n(\alpha) + r(\alpha) = 1 + 2 = 3 = \dim \mathbb{R}^3$. [/example] ## The Rank-Nullity Theorem The rank-nullity theorem is the fundamental dimension-counting result for linear maps. It asserts that no matter how a linear map distributes the dimensions of its domain between the kernel and the image, the total is always accounted for. [quotetheorem:385] The rank-nullity theorem states that for a linear map $\alpha \colon U \to V$ with $U$ finite-dimensional, \begin{align*} r(\alpha) + n(\alpha) = \dim U. \end{align*} The proof proceeds by extending a basis for $\ker \alpha$ to a basis for $U$. If $\{u_1, \ldots, u_k\}$ is a basis for $\ker \alpha$, extend it to a basis $\{u_1, \ldots, u_k, u_{k+1}, \ldots, u_n\}$ for $U$. The claim is that $\{\alpha(u_{k+1}), \ldots, \alpha(u_n)\}$ is a basis for $\operatorname{im} \alpha$. To verify spanning, take any $v \in \operatorname{im} \alpha$, say $v = \alpha(u)$. Write $u = \sum_{i=1}^n \lambda_i u_i$. Then $v = \sum_{i=1}^n \lambda_i \alpha(u_i) = \sum_{i=k+1}^n \lambda_i \alpha(u_i)$, since $\alpha(u_i) = \mathbf{0}$ for $i \leq k$. For linear independence, suppose $\sum_{i=k+1}^n \mu_i \alpha(u_i) = \mathbf{0}$. Then $\alpha\bigl(\sum_{i=k+1}^n \mu_i u_i\bigr) = \mathbf{0}$, so $\sum_{i=k+1}^n \mu_i u_i \in \ker \alpha$. This means $\sum_{i=k+1}^n \mu_i u_i = \sum_{i=1}^k \nu_i u_i$ for some scalars $\nu_i$, which implies $\sum_{i=1}^k (-\nu_i) u_i + \sum_{i=k+1}^n \mu_i u_i = \mathbf{0}$. Since $\{u_1, \ldots, u_n\}$ is a basis, all coefficients are zero, so $\mu_i = 0$ for all $i$. Therefore $\operatorname{im} \alpha$ has a basis of $n - k$ elements, giving $r(\alpha) = n - k = \dim U - n(\alpha)$. This theorem has several important consequences that recur throughout linear algebra. **Surjectivity bound.** If $\alpha \colon U \to V$ is surjective, then $r(\alpha) = \dim V$, so $\dim U = \dim V + n(\alpha) \geq \dim V$. A linear map from a smaller space to a larger one can never be surjective. **Injectivity bound.** If $\alpha$ is injective, then $n(\alpha) = 0$, so $r(\alpha) = \dim U$. A linear map from a larger space to a smaller one can never be injective. **Conservation of dimension.** The theorem explains why the dimension of the image and the dimension of the kernel are not independent: increasing one decreases the other, always summing to $\dim U$. [example:RankNullityDifferentiation] **Rank-nullity for differentiation.** The differentiation map $D \colon P_3(\mathbb{R}) \to P_2(\mathbb{R})$ sends $p(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3$ to $p'(x) = a_1 + 2a_2 x + 3a_3 x^2$. The domain $P_3(\mathbb{R})$ has dimension $4$. The kernel consists of constant polynomials $\{a_0 : a_0 \in \mathbb{R}\}$, so $n(D) = 1$. The image consists of all polynomials of the form $a_1 + 2a_2 x + 3a_3 x^2$ --- since $a_1$, $a_2$, $a_3$ range freely over $\mathbb{R}$, this is all of $P_2(\mathbb{R})$, so $r(D) = 3$. Rank-nullity gives $1 + 3 = 4 = \dim P_3(\mathbb{R})$. [/example] [example:RankNullityHomogeneous] **Dimension of solution spaces.** Consider a homogeneous linear system $Ax = \mathbf{0}$, where $A$ is an $m \times n$ matrix and $x \in \mathbb{R}^n$. The system defines a linear map $\alpha \colon \mathbb{R}^n \to \mathbb{R}^m$ by $\alpha(x) = Ax$, and the solution space is $\ker \alpha$. Rank-nullity gives \begin{align*} \dim(\text{solution space}) = n(\alpha) = n - r(\alpha) = n - \operatorname{rank}(A). \end{align*} For instance, if $A$ is a $3 \times 5$ matrix of rank $2$, then the solution space has dimension $5 - 2 = 3$. This is the reason behind the common observation that a system with more unknowns than equations tends to have infinitely many solutions: the kernel must have positive dimension whenever $n > \operatorname{rank}(A)$. [/example] ## Representing Linear Maps as Matrices A linear map between finite-dimensional vector spaces, once bases are chosen, becomes a matrix. This passage from the abstract to the concrete is what makes linear algebra computable. Let $\alpha \colon U \to V$ be a linear map, and let $\mathcal{B} = \{e_1, \ldots, e_n\}$ and $\mathcal{C} = \{f_1, \ldots, f_m\}$ be ordered bases for $U$ and $V$, respectively. For each basis vector $e_j$, the image $\alpha(e_j)$ lies in $V$ and can be expressed uniquely as a linear combination of the basis $\mathcal{C}$: \begin{align*} \alpha(e_j) = a_{1j} f_1 + a_{2j} f_2 + \cdots + a_{mj} f_m = \sum_{i=1}^{m} a_{ij} f_i. \end{align*} The $m \times n$ matrix $A = (a_{ij})$ is the **matrix of $\alpha$ with respect to $\mathcal{B}$ and $\mathcal{C}$**, often written $[\alpha]_{\mathcal{B}}^{\mathcal{C}}$ or ${}_{\mathcal{C}}[\alpha]_{\mathcal{B}}$. The $j$-th column of $A$ is the coordinate vector of $\alpha(e_j)$ in the basis $\mathcal{C}$. This representation converts function evaluation into matrix multiplication: if $u \in U$ has coordinate vector $x = (\lambda_1, \ldots, \lambda_n)^T$ with respect to $\mathcal{B}$, then the coordinate vector of $\alpha(u)$ with respect to $\mathcal{C}$ is $Ax$. [example:MatrixRepresentation] **Matrix of a rotation.** The rotation $R_\theta \colon \mathbb{R}^2 \to \mathbb{R}^2$ from the earlier example, with respect to the standard basis $\mathcal{E} = \{(1,0), (0,1)\}$, has matrix \begin{align*} [R_\theta]_{\mathcal{E}}^{\mathcal{E}} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}. \end{align*} The columns are the images of the standard basis vectors: $R_\theta(1, 0) = (\cos\theta, \sin\theta)$ and $R_\theta(0, 1) = (-\sin\theta, \cos\theta)$. [/example] **Change of basis.** What happens to the matrix when we change coordinates? Suppose $\mathcal{B}'$ is another basis for $U$ and $\mathcal{C}'$ is another basis for $V$. Let $P$ be the change-of-basis matrix from $\mathcal{B}'$ to $\mathcal{B}$ (so that $P$ converts $\mathcal{B}'$-coordinates to $\mathcal{B}$-coordinates), and let $Q$ be the change-of-basis matrix from $\mathcal{C}'$ to $\mathcal{C}$. Then \begin{align*} [\alpha]_{\mathcal{B}'}^{\mathcal{C}'} = Q^{-1} \, [\alpha]_{\mathcal{B}}^{\mathcal{C}} \, P. \end{align*} Two matrices $A$ and $B$ are **equivalent** if $B = Q^{-1} A P$ for some invertible matrices $P$ and $Q$. The key insight is that equivalent matrices represent the same linear map in different coordinate systems. When $U = V$ and we use the same basis for domain and codomain ($\mathcal{B} = \mathcal{C}$, $\mathcal{B}' = \mathcal{C}'$), the change-of-basis formula becomes $B = P^{-1} A P$, and the matrices are called **similar**. Similarity is a stricter relation than equivalence: it preserves not just the rank but also the determinant, trace, eigenvalues, and characteristic polynomial. [quotetheorem:388] The canonical form theorem for linear maps (Smith normal form) tells us that every $m \times n$ matrix of rank $r$ is equivalent to the block matrix $\begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}$, where $I_r$ is the $r \times r$ identity matrix. This means that, up to a choice of bases, the only thing that distinguishes one linear map from another (in the [equivalence relation](/page/Equivalence%20Relation) sense) is its rank. The geometry confirms this: a rank-$r$ linear map $\alpha \colon F^n \to F^m$ projects onto an $r$-dimensional subspace, collapsing an $(n - r)$-dimensional kernel to zero. [example:CanonicalForm] **Reducing a matrix to canonical form.** Consider the linear map $\alpha \colon \mathbb{R}^3 \to \mathbb{R}^2$ defined by $\alpha(x, y, z) = (x + y, \; y + z)$, which has matrix \begin{align*} A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix} \end{align*} with respect to the standard bases. Since $\operatorname{rank}(A) = 2$ (the two rows are linearly independent), the canonical form is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. To realize this explicitly, choose the basis $\mathcal{B}' = \{(1, 0, 0), (0, 0, 1), (-1, 1, -1)\}$ for $\mathbb{R}^3$ (where the third vector spans $\ker \alpha$) and keep the standard basis for $\mathbb{R}^2$. Then $\alpha(1, 0, 0) = (1, 0)$ and $\alpha(0, 0, 1) = (0, 1)$, and $\alpha(-1, 1, -1) = (0, 0)$. The matrix in these coordinates is exactly $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$. [/example] ## Invertibility and Isomorphisms When does a linear map have an inverse, and what can we say about vector spaces connected by an invertible linear map? A linear map $\alpha \colon U \to V$ is an **isomorphism** if it is bijective --- both injective and surjective. In this case, the inverse function $\alpha^{-1} \colon V \to U$ is also linear (a fact that requires proof: if $\alpha(u_1) = v_1$ and $\alpha(u_2) = v_2$, then $\alpha(\lambda u_1 + \mu u_2) = \lambda v_1 + \mu v_2$, so $\alpha^{-1}(\lambda v_1 + \mu v_2) = \lambda u_1 + \mu u_2 = \lambda \alpha^{-1}(v_1) + \mu \alpha^{-1}(v_2)$). If an isomorphism exists between $U$ and $V$, we write $U \cong V$ and say that $U$ and $V$ are **isomorphic**. In finite dimensions, a remarkable collapse occurs: for a linear map $\alpha \colon U \to V$ with $\dim U = \dim V$, injectivity, surjectivity, and bijectivity are all equivalent. [quotetheorem:386] This equivalence is special to finite-dimensional spaces and fails in infinite dimensions. For instance, the right-shift operator $S \colon \ell^2 \to \ell^2$ defined by $S(a_1, a_2, a_3, \ldots) = (0, a_1, a_2, \ldots)$ is injective but not surjective, despite $\ell^2$ being isomorphic to itself. The finite-dimensional result follows from rank-nullity: if $\dim U = \dim V = n$, then $r(\alpha) + n(\alpha) = n$. Injectivity gives $n(\alpha) = 0$, hence $r(\alpha) = n = \dim V$, which is surjectivity. Conversely, surjectivity gives $r(\alpha) = n$, hence $n(\alpha) = 0$, which is injectivity. When $U = V$ and we represent $\alpha$ by a square matrix $A$, invertibility of the linear map is equivalent to invertibility of the matrix, which in turn is equivalent to a nonzero determinant. [quotetheorem:396] The determinant invertibility criterion states that a square matrix $A$ over a field $F$ is invertible if and only if $\det A \neq 0$, which holds if and only if $\operatorname{rank}(A) = n$. This result links the algebraic condition ($\det A \neq 0$), the rank condition ($\operatorname{rank}(A) = n$), and the geometric condition (the column vectors of $A$ span $F^n$, or equivalently, they are linearly independent). For linear maps, it provides a computable criterion: to check whether $\alpha \colon F^n \to F^n$ is an isomorphism, compute the determinant of its matrix in any basis and check whether it is nonzero. [example:Isomorphism] **An isomorphism between polynomial and coordinate spaces.** Define $\phi \colon P_2(\mathbb{R}) \to \mathbb{R}^3$ by \begin{align*} \phi(a_0 + a_1 x + a_2 x^2) = (a_0, a_1, a_2). \end{align*} This map assigns to each polynomial its coefficient vector. Linearity follows from the fact that polynomial addition and scalar multiplication operate componentwise on coefficients. The map is injective: if $\phi(p) = \mathbf{0}$, then $a_0 = a_1 = a_2 = 0$, so $p = 0$. It is surjective: every vector $(a_0, a_1, a_2) \in \mathbb{R}^3$ is the image of the polynomial $a_0 + a_1 x + a_2 x^2$. Therefore $\phi$ is an isomorphism, and $P_2(\mathbb{R}) \cong \mathbb{R}^3$. More generally, every $n$-dimensional vector space over $F$ is isomorphic to $F^n$. The isomorphism is given by the coordinate map with respect to any chosen basis. This is why, in finite dimensions, the study of abstract vector spaces reduces to the study of $F^n$. [/example] ## The Dual Space and Dual Maps The dual space construction reveals a deep symmetry in the theory of linear maps. Where a vector space $V$ consists of vectors, its dual $V^*$ consists of linear maps from $V$ to the base field --- the "measurements" one can perform on vectors. This passage from objects to functions on objects is one of the most powerful ideas in mathematics. [definition:Dual Space] Let $V$ be a vector space over a field $F$. The **dual space** of $V$ is the vector space \begin{align*} V^* = \operatorname{Hom}(V, F) = \mathcal{L}(V, F), \end{align*} whose elements are called **linear functionals** (or **covectors**). A linear functional is a linear map $\varphi \colon V \to F$. If $\mathcal{B} = \{e_1, \ldots, e_n\}$ is a basis for $V$, the **dual basis** is $\mathcal{B}^* = \{e_1^*, \ldots, e_n^*\}$, where each $e_i^* \colon V \to F$ is defined by \begin{align*} e_i^*(e_j) = \delta_{ij} = \begin{cases} 1 & \text{if } i = j, \\ 0 & \text{if } i \neq j. \end{cases} \end{align*} The dual basis is indeed a basis for $V^*$, so $\dim V^* = \dim V$. [/definition] The dual basis has a natural interpretation: $e_i^*$ is the linear functional that "reads off the $i$-th coordinate." If $v = \lambda_1 e_1 + \cdots + \lambda_n e_n$, then $e_i^*(v) = \lambda_i$. In this sense, $V^*$ contains all the coordinate functions simultaneously. [definition:Dual Map] Let $\alpha \colon U \to V$ be a linear map. The **dual map** (or **transpose**) of $\alpha$ is the linear map \begin{align*} \alpha^* \colon V^* &\to U^*, \\ \varphi &\mapsto \varphi \circ \alpha. \end{align*} In other words, $\alpha^*(\varphi) = \varphi \circ \alpha$ for each linear functional $\varphi \in V^*$. [/definition] The dual map reverses direction: $\alpha$ goes from $U$ to $V$, but $\alpha^*$ goes from $V^*$ to $U^*$. If $\alpha$ has matrix $A$ with respect to bases $\mathcal{B}$ and $\mathcal{C}$, then $\alpha^*$ has matrix $A^T$ (the transpose) with respect to the dual bases $\mathcal{C}^*$ and $\mathcal{B}^*$. This is because for basis functionals $f_i^* \in V^*$ and basis vectors $e_j \in U$, we have $\alpha^*(f_i^*)(e_j) = f_i^*(\alpha(e_j)) = f_i^*\bigl(\sum_k a_{kj} f_k\bigr) = a_{ij}$, which is the $(i, j)$-entry of $A$, read as the $(j, i)$-entry of $A^T$ when we track the matrix of $\alpha^*$. The annihilator provides a dual perspective on subspaces. [definition:Annihilator] Let $W \subseteq V$ be a subspace. The **annihilator** of $W$ is \begin{align*} W^0 = \{ \varphi \in V^* : \varphi(w) = 0 \text{ for all } w \in W \}. \end{align*} The annihilator $W^0$ is a subspace of $V^*$. [/definition] [quotetheorem:420] The dimension of the annihilator satisfies $\dim W + \dim W^0 = \dim V$. This is a duality version of rank-nullity: measuring a subspace $W$ through its annihilator in $V^*$ gives complementary dimension information. If $W$ has dimension $k$ in an $n$-dimensional space, then $W^0$ has dimension $n - k$. The annihilator of $\{0\}$ is all of $V^*$ (every functional vanishes at zero), and the annihilator of $V$ is $\{0\}$ (the only functional vanishing on all vectors is the zero functional). A natural question arises: is $V$ isomorphic to $V^*$? When $V$ is finite-dimensional, the answer is yes --- choosing a basis for $V$ and pairing it with its dual basis gives an isomorphism. But this isomorphism depends on the choice of basis and is therefore not **canonical** (basis-independent). The situation is different for the double dual. [quotetheorem:423] The evaluation map $\operatorname{ev} \colon V \to V^{**}$ defined by $\operatorname{ev}(v)(\varphi) = \varphi(v)$ --- which sends each vector $v$ to the functional on $V^*$ that evaluates at $v$ --- is always injective, and when $V$ is finite-dimensional, it is an isomorphism. The key point is that this isomorphism requires no choices: it is defined purely in terms of the algebraic structure, without reference to any basis. This is what "canonical" means in linear algebra. The existence of a canonical isomorphism $V \cong V^{**}$, but no canonical isomorphism $V \cong V^*$, is a fundamental asymmetry that echoes throughout mathematics, from tensor algebra to the theory of [distributions](/page/Distribution). [example:DualMap] **The dual of a projection.** Let $V = \mathbb{R}^3$ with standard basis $\{e_1, e_2, e_3\}$, and let $\pi \colon \mathbb{R}^3 \to \mathbb{R}^3$ be the projection onto the $xy$-plane defined by $\pi(x, y, z) = (x, y, 0)$. The matrix of $\pi$ with respect to the standard basis is \begin{align*} A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \end{align*} The dual map $\pi^*$ has matrix $A^T = A$ (since $A$ is symmetric). The dual map acts on functionals: if $\varphi = a \, e_1^* + b \, e_2^* + c \, e_3^*$, then $\pi^*(\varphi) = \varphi \circ \pi$. Evaluating, $\pi^*(\varphi)(x, y, z) = \varphi(x, y, 0) = ax + by$. So $\pi^*(\varphi) = a \, e_1^* + b \, e_2^*$: the dual map "drops the $e_3^*$ component." The kernel of $\pi^*$ is $\operatorname{span}\{e_3^*\}$, which is precisely the annihilator of $\operatorname{im} \pi = \operatorname{span}\{e_1, e_2\}$. This illustrates the general identity $\ker \alpha^* = (\operatorname{im} \alpha)^0$. [/example] ## Problems [problem] Let $\alpha \colon \mathbb{R}^4 \to \mathbb{R}^3$ be the linear map defined by $\alpha(x_1, x_2, x_3, x_4) = (x_1 - x_2 + x_3, \; x_2 - x_3 + x_4, \; x_1 + x_4)$. Find bases for $\ker \alpha$ and $\operatorname{im} \alpha$, and verify the rank-nullity theorem. [/problem] [solution] Setting $\alpha(x_1, x_2, x_3, x_4) = (0, 0, 0)$ gives the system \begin{align*} x_1 - x_2 + x_3 &= 0, \\ x_2 - x_3 + x_4 &= 0, \\ x_1 + x_4 &= 0. \end{align*} From the third equation, $x_4 = -x_1$. Substituting into the second: $x_2 - x_3 - x_1 = 0$, so $x_2 = x_1 + x_3$. Substituting into the first: $x_1 - (x_1 + x_3) + x_3 = 0$, which is $0 = 0$. So the free parameters are $x_1 = s$ and $x_3 = t$, giving \begin{align*} (x_1, x_2, x_3, x_4) = s(1, 1, 0, -1) + t(0, 1, 1, 0). \end{align*} A basis for $\ker \alpha$ is $\{(1, 1, 0, -1), \; (0, 1, 1, 0)\}$, so $n(\alpha) = 2$. For the image, compute the images of the standard basis vectors: $\alpha(e_1) = (1, 0, 1)$, $\alpha(e_2) = (-1, 1, 0)$, $\alpha(e_3) = (1, -1, 0)$, $\alpha(e_4) = (0, 1, 1)$. Note $\alpha(e_3) = -\alpha(e_2)$, and $\alpha(e_4) = \alpha(e_1) + \alpha(e_2)$. So $\operatorname{im} \alpha = \operatorname{span}\{(1, 0, 1), (-1, 1, 0)\}$, which has dimension $r(\alpha) = 2$. Verification: $n(\alpha) + r(\alpha) = 2 + 2 = 4 = \dim \mathbb{R}^4$. [/solution] [problem] Let $V$ be an $n$-dimensional vector space over $F$, and let $\alpha \colon V \to V$ be a linear map satisfying $\alpha^2 = \alpha$ (an idempotent, or projection). Prove that $V = \ker \alpha \oplus \operatorname{im} \alpha$. [/problem] [solution] We verify the two conditions for a direct sum: $V = \ker \alpha + \operatorname{im} \alpha$ and $\ker \alpha \cap \operatorname{im} \alpha = \{\mathbf{0}\}$. **Spanning.** For any $v \in V$, write $v = (v - \alpha(v)) + \alpha(v)$. The second summand $\alpha(v)$ is in $\operatorname{im} \alpha$ by definition. For the first, apply $\alpha$: \begin{align*} \alpha(v - \alpha(v)) = \alpha(v) - \alpha^2(v) = \alpha(v) - \alpha(v) = \mathbf{0}, \end{align*} using $\alpha^2 = \alpha$. So $v - \alpha(v) \in \ker \alpha$, and $V = \ker \alpha + \operatorname{im} \alpha$. **Trivial intersection.** Suppose $v \in \ker \alpha \cap \operatorname{im} \alpha$. Then $\alpha(v) = \mathbf{0}$ (since $v \in \ker \alpha$) and $v = \alpha(u)$ for some $u \in V$ (since $v \in \operatorname{im} \alpha$). Therefore $v = \alpha(u)$, and applying $\alpha$: $\alpha(v) = \alpha^2(u) = \alpha(u) = v$. But $\alpha(v) = \mathbf{0}$, so $v = \mathbf{0}$. Therefore $V = \ker \alpha \oplus \operatorname{im} \alpha$. [/solution] [problem] Let $\alpha \colon U \to V$ and $\beta \colon V \to W$ be linear maps between finite-dimensional vector spaces. Prove that $\operatorname{rank}(\beta \circ \alpha) \leq \min(\operatorname{rank} \alpha, \operatorname{rank} \beta)$. [/problem] [solution] For the first inequality, note that $\operatorname{im}(\beta \circ \alpha) = \beta(\alpha(U)) = \beta(\operatorname{im} \alpha)$. Since $\operatorname{im} \alpha \subseteq V$, we have $\beta(\operatorname{im} \alpha) \subseteq \beta(V) = \operatorname{im} \beta$, so $\operatorname{im}(\beta \circ \alpha) \subseteq \operatorname{im} \beta$. Taking dimensions: $\operatorname{rank}(\beta \circ \alpha) \leq \operatorname{rank} \beta$. For the second inequality, $\operatorname{im}(\beta \circ \alpha) = \beta(\operatorname{im} \alpha)$. The restriction $\beta|_{\operatorname{im} \alpha} \colon \operatorname{im} \alpha \to W$ is a linear map from a space of dimension $\operatorname{rank} \alpha$. By rank-nullity, $\dim(\beta(\operatorname{im} \alpha)) \leq \dim(\operatorname{im} \alpha) = \operatorname{rank} \alpha$. Therefore $\operatorname{rank}(\beta \circ \alpha) \leq \operatorname{rank} \alpha$. Combining both inequalities: $\operatorname{rank}(\beta \circ \alpha) \leq \min(\operatorname{rank} \alpha, \operatorname{rank} \beta)$. [/solution]