A subspace is the way linear algebra recognises a smaller linear world inside a larger one. A subset of a [vector space](/page/Vector%20Space) may contain many vectors, but it supports the same linear methods only when it is stable under the operations that make vectors useful: addition and scalar multiplication. Once that stability is present, the subset inherits bases, dimension, [linear maps](/page/Linear%20Map), quotient constructions, kernels, images, and direct-sum decompositions from the ambient theory. The point of the definition is not merely to name certain subsets. It separates subsets on which linear equations still make sense from subsets that are only set-theoretic pieces of a vector space. Lines through the origin in $\mathbb{R}^2$ are subspaces; affine lines not passing through the origin are not. The difference is structural: a subspace contains the zero vector and is closed under every linear combination of its elements. ## Definition The ambient object is a vector space over a field. The subset deserves to be called a subspace exactly when the vector space operations do not leave the subset. This definition is the local test that tells us whether a subset can be treated as a vector space in its own right. [definition: Subspace] Let $F$ be a field, and let $V$ be an $F$-vector space. A subset $U \subset V$ is a subspace of $V$ if the following conditions hold: 1. $0_V \in U$. 2. If $u, v \in U$, then $u + v \in U$. 3. If $u \in U$ and $\lambda \in F$, then $\lambda u \in U$. [/definition] A subspace is therefore not additional structure placed on $U$; it is a condition saying that the structure already on $V$ restricts to $U$. In calculations, however, checking addition and scalar multiplication separately can obscure the real invariant: closure under linear combinations. The following definition isolates that invariant so that subspace verification can match the way vectors are actually manipulated. [definition: Linear Combination Closure] Let $F$ be a field, let $V$ be an $F$-vector space, and let $U \subset V$. The subset $U$ is closed under binary linear combinations if for all $u, v \in U$ and all $\lambda, \mu \in F$, one has \begin{align*} \lambda u + \mu v \in U. \end{align*} [/definition] Linear combination closure packages the two closure axioms into one condition, but it says nothing by itself if the set is empty. In the vector-space setting, the resulting working rule is the standard subspace test: a nonempty subset $U \subset V$ is a subspace exactly when it is closed under all binary linear combinations $\lambda u + \mu v$ with $u,v \in U$ and $\lambda,\mu \in F$. This rule is useful because it matches ordinary calculations. Instead of verifying addition and scalar multiplication as separate tasks, one checks that every linear combination of two already-admissible vectors remains admissible. The nonempty hypothesis is part of the test, not a cosmetic add-on: without it, the empty set would satisfy the closure condition vacuously while still failing to contain the zero vector required of a subspace. ## Equivalent Characterisations Subspaces are also exactly the subsets that can be described as all linear combinations of some family of vectors. This point of view explains why subspaces are the natural domains for [basis](/page/Basis) and [dimension](/page/Dimension). Before using generators to describe subspaces, we need a construction that turns any proposed family of vectors into the collection of all vectors it can produce. [definition: Span] Let $F$ be a field, let $V$ be an $F$-vector space, and let $S \subset V$. The span of $S$ is the subset \begin{align*} \operatorname{span}(S) = \left\{\sum_{i=1}^{n} \lambda_i s_i : n \in \mathbb{N},\ s_i \in S,\ \lambda_i \in F\right\}. \end{align*} If $S = \varnothing$, then $\operatorname{span}(S) = \{0_V\}$. [/definition] The span construction starts with an arbitrary set and forces closure under finite linear combinations. For it to be useful, it must be the least possible subspace containing the original set, rather than merely some large subspace built from it. Thus $\operatorname{span}(S)$ is the subspace generated by $S$: it contains $S$, it is closed under finite linear combinations, and every subspace of $V$ that contains $S$ must also contain $\operatorname{span}(S)$. This minimality is what makes spanning sets precise. It says that a vector lies in $\operatorname{span}(S)$ only because it can be built from finitely many vectors of $S$ using the vector-space operations; no extra vectors are included for unrelated reasons. In the language used later for bases, $S$ spans a subspace exactly when its finite linear combinations account for the whole subspace. Generators describe subspaces from inside, while maps produce subspaces by testing or transporting vectors. Since linear maps preserve linear combinations, the vectors sent to zero should form a closed linear system inside the domain. The following definition formalises this solution space for homogeneous linear equations. [definition: Kernel] Let $F$ be a field, and let $T: V \to W$ be a linear map of $F$-vector spaces. The kernel of $T$ is \begin{align*} \ker(T) = \{v \in V : T(v) = 0_W\}. \end{align*} [/definition] The kernel records the directions invisible to a linear map. Homogeneous linear systems behave linearly exactly because adding two solutions or rescaling one solution stays invisible to the map. Thus $\ker(T)$ is not merely a set of solutions; it is a subspace of the domain $V$, so it can be studied using bases, dimension, and coordinates. Kernels describe what a map collapses. To understand what a map reaches, we need the corresponding construction in the codomain. Because sums and scalar multiples of attainable outputs are again attainable outputs, the image is the natural target-side subspace attached to a linear map. [definition: Image of a Linear Map] Let $F$ be a field, and let $T: V \to W$ be a linear map of $F$-vector spaces. The image of $T$ is \begin{align*} \operatorname{im}(T) = \{T(v) : v \in V\}. \end{align*} [/definition] The image measures the effective range of a linear transformation, which may be smaller than the declared codomain. Because linear maps preserve addition and scalar multiplication, attainable outputs remain attainable after taking linear combinations. Thus $\operatorname{im}(T)$ is a subspace of $W$, and later dimension counts compare what $T$ collapses, recorded by $\ker(T)$, with what $T$ reaches, recorded by $\operatorname{im}(T)$. ## Standard Examples The smallest possible subspace is needed as a boundary case throughout linear algebra. It is what remains when no nonzero direction is available, and it appears naturally as the kernel of any injective linear map. [definition: Zero Subspace] Let $F$ be a field, and let $V$ be an $F$-vector space. The zero subspace of $V$ is the subset \begin{align*} \{0_V\} \subset V. \end{align*} [/definition] The ambient space itself must also count as a subspace. Otherwise statements about surjective images, full spans, and decompositions would require a separate exception whenever no restriction has actually been imposed. The largest subspace is therefore the case where every vector of the vector space is allowed. [definition: Whole Subspace] Let $F$ be a field, and let $V$ be an $F$-vector space. The whole subspace of $V$ is the subset $V \subset V$. [/definition] A first geometric example is a line through the origin. It shows how a single vector generates a complete linear object. [example: Line Through the Origin] In the real vector space $\mathbb{R}^2$, fix a nonzero vector $a=(a_1,a_2)$ and set \begin{align*} L=\{ra:r\in\mathbb{R}\}. \end{align*} We verify directly that $L$ is a subspace of $\mathbb{R}^2$. Since $0a=(0,0)$, the zero vector belongs to $L$. Let $u,v\in L$. Then there are [real numbers](/page/Real%20Numbers) $s,t\in\mathbb{R}$ such that $u=sa$ and $v=ta$. For any $\lambda,\mu\in\mathbb{R}$, \begin{align*} \lambda u+\mu v=\lambda(sa)+\mu(ta). \end{align*} By associativity of scalar multiplication, $\lambda(sa)=(\lambda s)a$ and $\mu(ta)=(\mu t)a$, so \begin{align*} \lambda u+\mu v=(\lambda s)a+(\mu t)a. \end{align*} By distributivity of scalar multiplication over addition of scalars, \begin{align*} (\lambda s)a+(\mu t)a=(\lambda s+\mu t)a. \end{align*} Because $\lambda s+\mu t\in\mathbb{R}$, the vector $(\lambda s+\mu t)a$ has the required form to belong to $L$. Thus $L$ is closed under arbitrary binary linear combinations, and in particular under addition and scalar multiplication, so $L$ is a subspace of $\mathbb{R}^2$. Geometrically, passing through the origin is exactly what makes every scalar multiple of a point on the line remain on the same line. [/example] The corresponding failure example is often more instructive. Translating a subspace usually destroys the zero vector and hence destroys linear closure. [example: Affine Line Not Passing Through the Origin] In $\mathbb{R}^2$, consider \begin{align*} A = \{(1,t) : t \in \mathbb{R}\}. \end{align*} Every vector in $A$ has first coordinate $1$. The zero vector in $\mathbb{R}^2$ is $(0,0)$, whose first coordinate is $0$, so there is no $t \in \mathbb{R}$ with \begin{align*} (0,0) = (1,t). \end{align*} Thus $(0,0) \notin A$, so $A$ fails the zero-vector condition in the definition of a subspace. The same set also fails closure under scalar multiplication. Let $u=(1,0)$. Since $0 \in \mathbb{R}$, we have $u=(1,0)\in A$. Multiplying by the scalar $2$ gives \begin{align*} 2u = 2(1,0) = (2\cdot 1,2\cdot 0) = (2,0). \end{align*} If $(2,0)$ belonged to $A$, then for some $t\in\mathbb{R}$ we would have $(2,0)=(1,t)$, forcing $2=1$ by equality of first coordinates. This is impossible in $\mathbb{R}$, so $(2,0)\notin A$. The obstruction is exactly the translation away from the origin: the set is an affine line, not a linear subspace. [/example] Subspaces also arise from equations. Homogeneous linear equations produce subspaces, while nonhomogeneous equations usually produce affine translates. [example: Solution Space of a Homogeneous Linear Equation] Let $T: \mathbb{R}^3 \to \mathbb{R}$ be the linear map \begin{align*} T(x_1,x_2,x_3) = x_1 - 2x_2 + x_3. \end{align*} For $x=(x_1,x_2,x_3)\in\mathbb{R}^3$, the condition $x\in\ker(T)$ means \begin{align*} T(x)=0. \end{align*} Using the formula for $T$, this is exactly \begin{align*} x_1-2x_2+x_3=0. \end{align*} Hence \begin{align*} \ker(T)=\{x\in\mathbb{R}^3:x_1-2x_2+x_3=0\}=U. \end{align*} We can also verify the linear structure directly. Since \begin{align*} T(0,0,0)=0-2\cdot 0+0=0, \end{align*} the zero vector belongs to $U$. If $x,y\in U$ and $\lambda,\mu\in\mathbb{R}$, then $T(x)=0$ and $T(y)=0$. By linearity of $T$, \begin{align*} T(\lambda x+\mu y)=\lambda T(x)+\mu T(y). \end{align*} Substituting $T(x)=0$ and $T(y)=0$ gives \begin{align*} T(\lambda x+\mu y)=\lambda\cdot 0+\mu\cdot 0=0. \end{align*} Thus $\lambda x+\mu y\in U$, so $U$ is closed under binary linear combinations and is a subspace of $\mathbb{R}^3$. To describe the same subspace by generators, start from the defining equation \begin{align*} x_1-2x_2+x_3=0. \end{align*} Adding $2x_2$ to both sides gives \begin{align*} x_1+x_3=2x_2. \end{align*} Subtracting $x_3$ from both sides gives \begin{align*} x_1=2x_2-x_3. \end{align*} Now set $s=x_2$ and $t=x_3$. Then every $x\in U$ has the form \begin{align*} x=(2s-t,s,t). \end{align*} This vector is the sum of two scalar multiples: \begin{align*} s(2,1,0)=(2s,s,0). \end{align*} \begin{align*} t(-1,0,1)=(-t,0,t). \end{align*} Adding the two coordinatewise gives \begin{align*} (2s,s,0)+(-t,0,t)=(2s-t,s,t). \end{align*} Therefore every element of $U$ lies in $\operatorname{span}\{(2,1,0),(-1,0,1)\}$. Conversely, for any $s,t\in\mathbb{R}$, the vector $(2s-t,s,t)$ satisfies \begin{align*} (2s-t)-2s+t=0. \end{align*} So every vector of the form $s(2,1,0)+t(-1,0,1)$ lies in $U$. Hence \begin{align*} U=\operatorname{span}\{(2,1,0),(-1,0,1)\}. \end{align*} The homogeneous equation cuts out a plane through the origin, and the two displayed vectors give spanning directions for that plane. [/example] The homogeneous hypothesis is essential. Moving the right-hand side away from zero changes the set from a subspace into a translate of one. [example: Nonhomogeneous Equation] In $\mathbb{R}^3$, consider \begin{align*} A = \{x \in \mathbb{R}^3 : x_1 - 2x_2 + x_3 = 1\}. \end{align*} We show that $A$ is not a subspace of $\mathbb{R}^3$ by checking the defining equation at the zero vector. The zero vector is $(0,0,0)$, and substituting its coordinates into the left-hand side gives \begin{align*} 0 - 2\cdot 0 + 0 = 0 - 0 + 0 = 0. \end{align*} Since $0 \ne 1$, the vector $(0,0,0)$ does not satisfy the equation $x_1-2x_2+x_3=1$. Therefore $(0,0,0)\notin A$, so $A$ fails the zero-vector condition in the definition of a subspace. The same failure appears under scalar multiplication. The vector $(1,0,0)$ belongs to $A$ because \begin{align*} 1 - 2\cdot 0 + 0 = 1 - 0 + 0 = 1. \end{align*} Multiplying by the scalar $2$ gives \begin{align*} 2(1,0,0)=(2\cdot 1,2\cdot 0,2\cdot 0)=(2,0,0). \end{align*} But $(2,0,0)$ does not belong to $A$, since \begin{align*} 2 - 2\cdot 0 + 0 = 2 - 0 + 0 = 2. \end{align*} Since $2 \ne 1$, scalar multiplication has taken a vector of $A$ outside $A$. Thus $A$ is an affine solution set rather than a linear subspace; the nonzero right-hand side shifts the homogeneous plane away from the origin. [/example] Function spaces give another important family of examples. They show that subspaces are not limited to finite-dimensional coordinate spaces. [example: Polynomial Subspaces] Let $F$ be a field, and let $F[x]$ be the vector space of polynomials in one variable with coefficients in $F$. For $n \in \mathbb{N}$, define $P_n$ to be the set consisting of the zero polynomial together with all nonzero polynomials of degree at most $n$: \begin{align*} P_n = \{0\}\cup \{p \in F[x] : p\ne 0 \text{ and } \deg p \le n\}. \end{align*} We show that $P_n$ is a subspace of $F[x]$ by checking closure under binary linear combinations. The zero polynomial belongs to $P_n$ by definition. Now let $p,q\in P_n$ and let $\lambda,\mu\in F$. Since $p$ and $q$ have degree at most $n$ unless they are zero, we can write them with coefficients padded by zeros: \begin{align*} p=a_0+a_1x+\cdots+a_nx^n. \end{align*} \begin{align*} q=b_0+b_1x+\cdots+b_nx^n. \end{align*} Multiplying by scalars gives \begin{align*} \lambda p=\lambda a_0+\lambda a_1x+\cdots+\lambda a_nx^n. \end{align*} \begin{align*} \mu q=\mu b_0+\mu b_1x+\cdots+\mu b_nx^n. \end{align*} Adding coefficientwise in $F[x]$ gives \begin{align*} \lambda p+\mu q=(\lambda a_0+\mu b_0)+(\lambda a_1+\mu b_1)x+\cdots+(\lambda a_n+\mu b_n)x^n. \end{align*} This polynomial has no terms of degree greater than $n$. If all displayed coefficients are zero, then $\lambda p+\mu q$ is the zero polynomial, so it belongs to $P_n$. Otherwise its highest nonzero term has degree at most $n$, so $\deg(\lambda p+\mu q)\le n$ and again $\lambda p+\mu q\in P_n$. Thus $P_n$ contains zero and is closed under binary linear combinations, so it is a subspace of $F[x]$. The example shows that finite-degree polynomial spaces sit as finite-dimensional linear pieces inside the infinite-dimensional polynomial space. [/example] ## Properties Many constructions in linear algebra impose several linear constraints at once. The vectors satisfying every constraint should still form a linear object, because each individual constraint is compatible with linear combinations. This is why intersections of subspaces are the natural way to combine simultaneous homogeneous conditions. Intersections work because every subspace in the family accepts the same linear combination. Combining subspaces by union behaves differently: a union may fail to be closed under addition because one summand can come from each side. [example: Union of Two Subspaces Need Not Be a Subspace] In $\mathbb{R}^2$, let \begin{align*} U = \operatorname{span}\{(1,0)\}, \qquad W = \operatorname{span}\{(0,1)\}. \end{align*} By the definition of span, these sets are \begin{align*} U=\{s(1,0):s\in\mathbb{R}\}=\{(s,0):s\in\mathbb{R}\} \end{align*} and \begin{align*} W=\{t(0,1):t\in\mathbb{R}\}=\{(0,t):t\in\mathbb{R}\}. \end{align*} Each is a subspace: if $(s,0),(s',0)\in U$ and $\lambda,\mu\in\mathbb{R}$, then \begin{align*} \lambda(s,0)+\mu(s',0)=(\lambda s,0)+(\mu s',0)=(\lambda s+\mu s',0)\in U. \end{align*} The same coordinate calculation shows that if $(0,t),(0,t')\in W$, then \begin{align*} \lambda(0,t)+\mu(0,t')=(0,\lambda t)+(0,\mu t')=(0,\lambda t+\mu t')\in W. \end{align*} The union $U\cup W$ is not a subspace. We have $(1,0)\in U\subset U\cup W$ and $(0,1)\in W\subset U\cup W$, but their sum is \begin{align*} (1,0)+(0,1)=(1+0,0+1)=(1,1). \end{align*} This vector is not in $U$, because every vector in $U$ has second coordinate $0$, while $(1,1)$ has second coordinate $1$. It is also not in $W$, because every vector in $W$ has first coordinate $0$, while $(1,1)$ has first coordinate $1$. Hence $(1,1)\notin U\cup W$, so $U\cup W$ is not closed under addition. The failure comes from mixing two directions that are separately closed but not jointly closed. [/example] A reliable replacement for union must allow vectors from the two pieces to be added. The obstruction in the example is not that either line fails to be linear, but that the union forgets to include mixed sums such as one vector from each direction. The needed construction therefore records exactly all vectors forced by adding one vector from each subspace. [definition: Sum of Subspaces] Let $F$ be a field, let $V$ be an $F$-vector space, and let $U,W \subset V$ be subspaces. The sum of $U$ and $W$ is \begin{align*} U + W = \{u+w : u \in U,\ w \in W\}. \end{align*} [/definition] The sum construction is meant to repair the failure of unions by closing under mixed addition. For it to serve that role, it must itself be a subspace and it must contain no unnecessary vectors beyond those forced by $U$ and $W$. The following theorem records both facts. [quotetheorem:9463] The minimality clause says that $U+W$ adds exactly the mixed sums needed to contain both subspaces and nothing extra. Any subspace containing both $U$ and $W$ must contain every vector $u+w$ by closure under addition, so $U+W$ is the smallest possible subspace replacement for the union. Sums can contain redundancy: the same vector may be expressible using different pairs of summands. In coordinate decompositions, projections, and normal forms, we need uniqueness of representation. This motivates the [direct sum](/page/Direct%20Sum) condition. [definition: Direct Sum of Subspaces] Let $F$ be a field, let $V$ be an $F$-vector space, and let $U,W \subset V$ be subspaces. The vector space $V$ is the internal direct sum of $U$ and $W$, written \begin{align*} V = U \oplus W, \end{align*} if every $v \in V$ has a unique representation \begin{align*} v = u + w \end{align*} with $u \in U$ and $w \in W$. [/definition] The definition asks for uniqueness for every vector, but checking uniqueness directly can be inconvenient. A simpler obstruction is overlap: if a nonzero vector lies in both subspaces, it can be moved from one summand to the other. The next theorem gives the standard criterion in terms of this overlap. [quotetheorem:9464] The condition $U\cap W=\{0_V\}$ captures uniqueness because any two decompositions $v=u+w=u'+w'$ imply $u-u'=w'-w$, a vector lying in both subspaces. If the intersection contains only zero, then $u=u'$ and $w=w'$. Conversely, a nonzero vector in the intersection gives an immediate ambiguity by shifting that vector from one summand to the other, so the criterion is exactly the algebraic test for unique splitting. For finite-dimensional spaces, dimension turns these constructions into numerical identities. When two subspaces are added, the dimensions of the parts overcount any common directions. The next formula measures exactly that overcount. [quotetheorem:9465] The subtraction term $\dim(U\cap W)$ corrects the double-counting of directions that belong to both subspaces. If two distinct lines through the origin in $\mathbb{R}^2$ are added, their intersection is $\{0\}$ and their sum is all of $\mathbb{R}^2$, so the dimensions add as $1+1-0=2$. If the two lines are the same line, the intersection has dimension $1$, and the formula gives $1+1-1=1$. The finite-dimensional hypothesis is what lets this bookkeeping be expressed as an ordinary numerical identity. In practice, the formula is a way to compute the size of $U+W$ by separately understanding $U$, $W$, and the overlap between them. The same idea prepares quotienting: instead of correcting for an overlap between two subspaces, a quotient deliberately identifies every direction in one subspace with zero. [definition: Quotient Vector Space] Let $F$ be a field, let $V$ be an $F$-vector space, and let $U \subset V$ be a subspace. For $v\in V$, the coset of $U$ represented by $v$ is \begin{align*} v+U=\{v+u:u\in U\}. \end{align*} The quotient vector space $V/U$ is the set of all such cosets: \begin{align*} V/U = \{v + U : v \in V\}, \end{align*} with addition defined by \begin{align*} (v+U) + (w+U) = (v+w)+U \end{align*} and scalar multiplication defined by \begin{align*} \lambda(v+U) = (\lambda v)+U. \end{align*} [/definition] Equivalently, two vectors $v,w\in V$ represent the same coset exactly when $v-w\in U$; quotienting identifies vectors that differ by an element of $U$. The subspace condition is what makes the quotient operations well-defined: different representatives of the same coset must give the same result after addition or scalar multiplication. Once this is secured, quotienting identifies all directions lying inside $U$ with zero. This raises the central finite-dimensional question for quotient spaces: exactly how much dimension is lost when the whole subspace $U$ is collapsed to zero? The answer is the dimension formula for quotients, which makes precise the idea that the remaining independent directions are those not already accounted for by $U$. [quotetheorem:377] The formula says that quotienting by $U$ removes exactly the independent directions already present in $U$, and no others. It applies in the finite-dimensional setting, where dimension can count this loss cleanly: representatives may look different in $V$, but their cosets only remember the part of a vector not absorbed by $U$. This is the same bookkeeping principle that later appears in rank-nullity: a linear map collapses its kernel to zero, and the image records the independent directions that survive. ## Relationship to Other Concepts Subspaces are to vector spaces what [subgroups](/page/Subgroup) are to groups and submodules are to modules: subsets closed under the operations of the ambient structure. The same pattern appears throughout algebra, but vector spaces are more rigid because scalar multiplication allows all finite linear combinations. A subspace is also a special case of a [module](/page/Module) subobject. If $F$ is a field, an $F$-vector space is the same thing as a module over $F$, and a subspace is exactly an $F$-submodule. Many results about subspaces have module-theoretic analogues, though bases and dimension behave much better over fields than over general rings. In linear algebra, subspaces are inseparable from [linear independence](/page/Linear%20Independence), bases, and dimension. A basis is a linearly independent set whose span is the entire subspace under discussion, so changing the subspace changes what it means for a list of vectors to be complete. For a linear map $T: V \to W$, the two canonical associated subspaces are $\ker(T) \subset V$ and $\operatorname{im}(T) \subset W$. Their dimensions measure the loss and reach of the map. The [rank-nullity theorem](/theorems/916) is the central finite-dimensional statement tying these two subspaces together. [quotetheorem:916] Here the rank is the dimension of the image subspace, while the nullity is the dimension of the kernel subspace. The theorem is a finite-dimensional conservation law: every independent direction in the domain either survives visibly in the image or is lost inside the kernel, and these two counts exhaust the dimension of the original space. This is why kernels and images are not just auxiliary constructions; together they organize how a linear map compresses, preserves, and records subspace information. In [inner product](/page/Inner%20Product) spaces, subspaces interact with orthogonality. To discuss vectors perpendicular to an entire subspace, we need a construction that collects all vectors annihilating that subspace under the inner product. This construction is the orthogonal complement. [definition: Orthogonal Complement] Let $H$ be an [inner product space](/page/Inner%20Product%20Space) over $F$, and let $U \subset H$ be a subspace. The orthogonal complement of $U$ is \begin{align*} U^\perp = \{h \in H : (h,u)_H = 0 \text{ for all } u \in U\}. \end{align*} [/definition] Orthogonal complements are subspaces even when $U$ is not finite-dimensional, but in infinite-dimensional spaces orthogonality alone need not recover every vector by a clean decomposition. The analytic obstruction is that a subspace may fail to contain limits of its own convergent sequences. In Hilbert spaces, closedness is the extra hypothesis that makes projection onto the subspace and its orthogonal complement behave like the finite-dimensional picture. [quotetheorem:241] This result shows how the elementary concept of subspace persists in analysis. In finite-dimensional linear algebra, every subspace is closed; in infinite-dimensional normed spaces, closedness becomes an additional condition with strong consequences. ## References Axler, *Linear Algebra Done Right* (2015). Hoffman and Kunze, *Linear Algebra* (1971). Lang, *Linear Algebra* (1987). [Vector Space](/page/Vector%20Space). [Linear Map](/page/Linear%20Map). [Basis](/page/Basis).