A vector space supplies addition and scalar multiplication, but it does not automatically supply coordinates. The central problem of finite-dimensional linear algebra is to decide when every vector can be described by a finite list of scalars, in a way that respects addition, scalar multiplication, subspaces, and linear maps. The failure appears as soon as we leave Euclidean space. Polynomial spaces and function spaces obey the axioms of a [vector space](/page/Vector%20Space), but most of them cannot be controlled by finitely many vectors. Finite-dimensional vector spaces are the part of linear algebra where abstract vectors can be replaced by finite coordinate data. [example: Polynomial Space Is Not Controlled by Finitely Many Polynomials] Let $k$ be a field and let $k[x]$ be the vector space of polynomials with coefficients in $k$. Take any finite list $p_1,\ldots,p_m \in k[x]$. If every $p_i$ is the zero polynomial, then every linear combination $a_1p_1+\cdots+a_mp_m$ is $0$, so $x$ is not in its span. Otherwise, let $d$ be the maximum degree among the nonzero polynomials in the list. For each $i$, write \begin{align*} p_i=c_{i0}+c_{i1}x+\cdots+c_{id}x^d \end{align*} where $c_{ir}=0$ when the actual degree of $p_i$ is less than $r$. Then for scalars $a_1,\ldots,a_m \in k$, \begin{align*} a_1p_1+\cdots+a_mp_m=a_1(c_{10}+c_{11}x+\cdots+c_{1d}x^d)+\cdots+a_m(c_{m0}+c_{m1}x+\cdots+c_{md}x^d). \end{align*} Collecting coefficients of equal powers of $x$ gives \begin{align*} a_1p_1+\cdots+a_mp_m=(a_1c_{10}+\cdots+a_mc_{m0})+(a_1c_{11}+\cdots+a_mc_{m1})x+\cdots+(a_1c_{1d}+\cdots+a_mc_{md})x^d. \end{align*} This polynomial has no $x^{d+1}$ term, so it cannot equal $x^{d+1}$, whose coefficient of $x^{d+1}$ is $1$. Therefore no finite list of polynomials spans $k[x]$, even though $k[x]$ is a vector space. [/example] This example explains why finite-dimensionality is not a decorative adjective. It separates vector spaces where linear algebra is governed by finite systems of scalar equations from vector spaces where additional structure is needed. ## Definition The relevant finiteness condition is not that the vector space has finitely many elements. If $k$ is infinite, even the line $k$ has infinitely many vectors. What must be finite is the amount of data needed to produce every vector by linear combination. [definition: Finite-Dimensional Vector Space] Let $k$ be a field and let $V$ be a vector space over $k$. The vector space $V$ is finite-dimensional if there exists a finite subset $S \subset V$ such that every $v \in V$ is a linear combination of elements of $S$. [/definition] The zero vector space is finite-dimensional, generated by the empty set. This convention matters because kernels, intersections, and quotient constructions can naturally produce the zero space. Finite-dimensionality depends on producing vectors from a finite list, but that phrase is meaningless until the allowed algebraic recipe is fixed. We need a precise name for the finite scalar recipes that are allowed to build new vectors from chosen ones; no limits, topology, or infinite sums are being used. [definition: Linear Combination] Let $k$ be a field, let $V$ be a vector space over $k$, and let $v_1, \ldots, v_m \in V$. A linear combination of $v_1, \ldots, v_m$ is a vector of the form $a_1v_1+\cdots+a_mv_m$, where $a_1, \ldots, a_m \in k$. [/definition] Once a set of candidate vectors has been chosen, we need to separate the vectors it actually produces from the vectors that may still be unreachable. This reachable part should itself be a subset of the ambient space, so generation can later be expressed as equality with the whole space. [definition: Span] Let $k$ be a field, let $V$ be a vector space over $k$, and let $S \subset V$. The span of $S$ is the subset $\operatorname{span}(S) \subset V$ consisting of all finite linear combinations of elements of $S$. For $S=\varnothing$, set $\operatorname{span}(S)=\{0\}$. [/definition] Span tells us what a proposed list can reach, but it does not say whether the list has repeated information. We next name lists that reach everything, because finite-dimensionality begins with the existence of such a finite reaching set. [definition: Generating Set] Let $k$ be a field and let $V$ be a vector space over $k$. A subset $S \subset V$ is a generating set for $V$ if $\operatorname{span}(S)=V$. A finite generating set is a generating set with finitely many elements. [/definition] A generating set can be too large, as happens in $k^2$ with $(1,0)$, $(0,1)$, and $(1,1)$. The defect is that one vector can be recovered from the others, so the same vector may have competing recipes. To make coordinates unique, we must forbid nontrivial scalar relations among the chosen vectors. [definition: Linearly Independent Set] Let $k$ be a field, let $V$ be a vector space over $k$, and let $S \subset V$. The subset $S$ is linearly independent if every equality $a_1v_1+\cdots+a_mv_m=0$, with $v_1, \ldots, v_m$ distinct elements of $S$ and $a_1, \ldots, a_m \in k$, forces $a_1=\cdots=a_m=0$. [/definition] Spanning alone guarantees that every vector has at least one recipe, while independence prevents hidden redundancy in those recipes. A usable coordinate system needs both properties at once: every vector must be reachable, and its scalar expansion must not depend on arbitrary choices. [definition: Basis] Let $k$ be a field and let $V$ be a vector space over $k$. A basis of $V$ is a linearly independent generating set for $V$. [/definition] A basis should let us write each vector by coordinates, but this has to be justified from the two halves of the definition. The next theorem is the reason a basis is more than a convenient spanning list: it gives both existence and uniqueness of scalar expansion. [quotetheorem:372] The theorem turns an abstract vector into a finite scalar list after a basis is chosen. It also warns us that the scalar list depends on the chosen basis. Counting coordinates should measure the space, not the arbitrary basis chosen to describe it. The obstruction is that two different bases might a priori have different lengths, so dimension would be ill-defined unless basis size is invariant. [quotetheorem:915] Once basis size is known to be invariant, that common size becomes a property of the vector space itself. This gives the numerical invariant that will be used to compare spaces, subspaces, quotients, and linear maps. [definition: Dimension] Let $k$ be a field and let $V$ be a finite-dimensional vector space over $k$. The dimension of $V$, denoted $\dim_k V$, is the number of elements in any basis of $V$. [/definition] When the field is fixed, one often writes $\dim V$. The subscript is important when the same set carries vector-space structures over different fields. [example: The Complex Numbers Over Different Fields] Over $\mathbb{C}$, the single vector $1$ spans $\mathbb{C}$ because every $z \in \mathbb{C}$ can be written as $z=z\cdot 1$ with scalar $z \in \mathbb{C}$. It is linearly independent over $\mathbb{C}$ because if $a\cdot 1=0$ with $a \in \mathbb{C}$, then $a=0$. Hence $(1)$ is a basis of $\mathbb{C}$ as a complex vector space, so $\dim_{\mathbb{C}}\mathbb{C}=1$. Over $\mathbb{R}$, the allowed scalars are [real numbers](/page/Real%20Numbers), so the same vector space has basis $(1,i)$. Indeed, every $z \in \mathbb{C}$ has real and imaginary parts $a,b \in \mathbb{R}$, and therefore \begin{align*} z=a+bi=a\cdot 1+b\cdot i. \end{align*} This proves that $(1,i)$ spans $\mathbb{C}$ over $\mathbb{R}$. To check [linear independence](/page/Linear%20Independence) over $\mathbb{R}$, suppose \begin{align*} a\cdot 1+b\cdot i=0 \end{align*} with $a,b \in \mathbb{R}$. The left side is the complex number $a+bi$, so equality with $0=0+0i$ forces equality of real and imaginary parts: \begin{align*} a=0 \text{ and } b=0. \end{align*} Thus $(1,i)$ is a basis of $\mathbb{C}$ as a real vector space, so $\dim_{\mathbb{R}}\mathbb{C}=2$. The underlying set is the same in both cases, but changing the scalar field changes which linear combinations are allowed, and therefore changes the dimension. [/example] ## Coordinates and Change of Basis The main reward for finite-dimensionality is the ability to compute with coordinates. The price is that coordinates are not intrinsic: they require a basis, and changing the basis changes the coordinate list. ### Coordinate Columns A basis must be ordered before it can produce coordinate columns. Without an order, the coefficients in a basis expansion are known, but their positions in a column have not been assigned. [definition: Ordered Basis] Let $k$ be a field and let $V$ be a finite-dimensional vector space over $k$. An ordered basis of $V$ is a finite list $\mathcal{B}=(v_1, \ldots, v_n)$ whose underlying set is a basis of $V$. [/definition] An ordered basis assigns each coefficient a definite position, but the vector itself still lives in $V$, not in $k^n$. To compute with matrices, we need a notation that packages the unique coefficients of a vector into an ordered column while remembering which basis produced them. [definition: Coordinate Column] Let $k$ be a field, let $V$ be a finite-dimensional vector space over $k$, and let $\mathcal{B}=(v_1, \ldots, v_n)$ be an ordered basis of $V$. If $v=a_1v_1+\cdots+a_nv_n$, the coordinate column of $v$ in the basis $\mathcal{B}$ is $[v]_{\mathcal{B}}=(a_1, \ldots, a_n)^\top \in k^n$. [/definition] The notation $[v]_{\mathcal{B}}$ records coordinates, not an equivalence class. Coordinate columns are useful only if passing from $V$ to $k^n$ preserves the vector-space operations. Without that compatibility, matrix calculations with columns would not reliably describe calculations in the original space. This raises the basic structural question behind coordinates: once a basis has been chosen, does the coordinate assignment merely label vectors, or does it actually identify the original vector space with the standard coordinate space in a way that preserves addition and scalar multiplication? The potential obstruction is that vectors in $V$ may be polynomials, functions, or geometric arrows, while elements of $k^n$ are columns of scalars; there is no reason a priori that addition and scaling should look identical in these two settings. The point of the next result is to certify that the coordinate column construction is not just a convenient encoding, but a linear isomorphism determined by the ordered basis. That certification is what allows later matrix computations to stand in for computations with the original vectors. [quotetheorem:9921] This theorem says that every $n$-dimensional vector space over $k$ is structurally the same as $k^n$ after a basis is chosen. The phrase "after a basis is chosen" matters because a different basis gives a different isomorphism. [example: Coordinates in a Polynomial Space] Let $P_2(k)$ be the vector space of polynomials over $k$ of degree at most $2$, and let $p(x)=3-5x+2x^2$. With ordered basis $\mathcal{B}=(1,x,x^2)$, the coefficients of $1$, $x$, and $x^2$ in $p$ are respectively $3$, $-5$, and $2$, so \begin{align*} p(x)=3\cdot 1+(-5)\cdot x+2\cdot x^2. \end{align*} Therefore $[p]_{\mathcal{B}}=(3,-5,2)^\top$. Now use the ordered basis $\mathcal{C}=(1,x-1,(x-1)^2)$. Since \begin{align*} (x-1)^2=x^2-2x+1, \end{align*} we compute \begin{align*} 0\cdot 1-1\cdot(x-1)+2\cdot(x-1)^2=-(x-1)+2(x^2-2x+1). \end{align*} Expanding the right side gives \begin{align*} -(x-1)+2(x^2-2x+1)=-x+1+2x^2-4x+2. \end{align*} Collecting equal powers of $x$ gives \begin{align*} -x+1+2x^2-4x+2=3-5x+2x^2. \end{align*} Thus \begin{align*} p(x)=0\cdot 1-1\cdot(x-1)+2\cdot(x-1)^2, \end{align*} so $[p]_{\mathcal{C}}=(0,-1,2)^\top$. The polynomial is the same vector in both computations; only the ordered basis used to record its coordinates has changed. [/example] ### Changing Coordinates If coordinates depend on a basis, computation needs a way to translate between coordinate systems. The change-of-basis matrix is the device that performs that translation without reconstructing the vector each time. [definition: Change-of-Basis Matrix] Let $k$ be a field, let $V$ be a finite-dimensional vector space over $k$, and let $\mathcal{B}=(v_1, \ldots, v_n)$ and $\mathcal{C}=(w_1, \ldots, w_n)$ be ordered bases of $V$. The change-of-basis matrix from $\mathcal{B}$-coordinates to $\mathcal{C}$-coordinates is the matrix $P_{\mathcal{C} \leftarrow \mathcal{B}} \in k^{n \times n}$ whose $j$th column is $[v_j]_{\mathcal{C}}$. [/definition] The arrow in $P_{\mathcal{C} \leftarrow \mathcal{B}}$ points toward the coordinate system being produced. This matrix records how the identity map on $V$ looks when the input coordinates are written in the basis $\mathcal{B}$ and the output coordinates are written in the basis $\mathcal{C}$. The same issue appears for an arbitrary [linear map](/page/Linear%20Map): its matrix depends on the basis used in the domain and the basis used in the codomain. Once coordinates are changed on either side, the representing matrix must be adjusted by the corresponding change-of-basis matrices. The notation in this comparison is standard: $\mathrm{GL}_m(\mathbb{F})$ denotes the group of invertible $m\times m$ matrices over a field $\mathbb{F}$, $\mathrm{Mat}_{n,m}(\mathbb{F})$ denotes the set of $n\times m$ matrices over $\mathbb{F}$, $\mathcal{L}(U,V)$ denotes the vector space of linear maps from $U$ to $V$, and $\operatorname{Hom}_k(V,k)$ denotes the $k$-linear maps from $V$ to the scalar field $k$. We also use $\delta_{ij}$ for the Kronecker delta: it equals $1$ when $i=j$ and equals $0$ otherwise. With this notation fixed, the next formal result gives the matrix-level transformation. [quotetheorem:387] [Change of basis](/page/Change%20Of%20Basis) explains why neither a coordinate column nor a matrix representation should be confused with the underlying vector or linear map. The object is basis-free; the column or matrix is its representation through chosen bases. ## Bases, Extension, and Dimension Tests Finite-dimensional spaces are manageable because rough finite data can be repaired. A spanning list can be trimmed to remove redundancy, and an independent list can be enlarged until it reaches the whole space. ### Exchange The engine behind these repairs is the exchange principle. It says that independent vectors can be inserted into a spanning list, at the cost of removing old spanning vectors, without losing the ability to generate the space. [quotetheorem:373] The inequality $n \le m$ is the first rigorous expression of the idea that an independent list cannot contain more information than a spanning list can hold. It is the counting principle underneath dimension. A generating set may contain extra vectors, and an independent set may be missing directions. To use bases in practice, we use two complementary repairs: discard vectors that add no span, or add vectors that supply missing directions, while preserving the property already under control. The extraction repair removes redundant vectors from a spanning list until a basis remains; the extension repair adds new directions to an independent list until it spans the whole space. [example: Extracting a Basis in $\mathbb{R}^3$] Consider $v_1=(1,0,0)$, $v_2=(0,1,0)$, $v_3=(1,1,0)$, and $v_4=(0,0,1)$ in $\mathbb{R}^3$. We first show that the list spans $\mathbb{R}^3$. For any $(x,y,z)\in \mathbb{R}^3$, \begin{align*} xv_1+yv_2+zv_4=x(1,0,0)+y(0,1,0)+z(0,0,1). \end{align*} Using coordinatewise scalar multiplication, this is \begin{align*} (x,0,0)+(0,y,0)+(0,0,z). \end{align*} Using coordinatewise addition, the sum is \begin{align*} (x,y,z). \end{align*} Thus every vector of $\mathbb{R}^3$ is a linear combination of $v_1,v_2,v_4$, so the larger list $v_1,v_2,v_3,v_4$ also spans $\mathbb{R}^3$. The third vector contributes no new direction, because \begin{align*} v_1+v_2=(1,0,0)+(0,1,0)=(1,1,0)=v_3. \end{align*} Now check that the remaining list is linearly independent. If \begin{align*} av_1+bv_2+cv_4=0 \end{align*} for $a,b,c\in\mathbb{R}$, then \begin{align*} a(1,0,0)+b(0,1,0)+c(0,0,1)=(0,0,0). \end{align*} The left side is $(a,b,c)$, so equality of coordinates gives $a=0$, $b=0$, and $c=0$. Therefore $(v_1,v_2,v_4)$ is linearly independent and spans $\mathbb{R}^3$, hence it is a basis. Removing $v_3$ has removed exactly the redundant vector while preserving all three coordinate directions. [/example] ### Counting Tests After dimension has been established, many basis checks become counting arguments. In an $n$-dimensional space, a list of $n$ vectors cannot be both spanning and redundant, and it cannot be independent while missing a direction. [quotetheorem:374] This theorem is one of the main computational conveniences of finite dimension. It lets a spanning calculation or an independence calculation do the work of both. [example: A Determinant as a Basis Test] In $\mathbb{R}^3$, let $v_1=(1,2,0)$, $v_2=(0,1,1)$, and $v_3=(2,0,1)$. Put these vectors as the columns of a matrix $A$, so the first column is $(1,2,0)^\top$, the second is $(0,1,1)^\top$, and the third is $(2,0,1)^\top$. Using the $3\times 3$ determinant expansion along the first row, \begin{align*} \det A=1(1\cdot 1-0\cdot 1)-0(2\cdot 1-0\cdot 0)+2(2\cdot 1-1\cdot 0)=1+4=5. \end{align*} To see the independence test explicitly, suppose \begin{align*} av_1+bv_2+cv_3=0. \end{align*} Then \begin{align*} a(1,2,0)+b(0,1,1)+c(2,0,1)=(a+2c,2a+b,b+c). \end{align*} Thus \begin{align*} (a+2c,2a+b,b+c)=(0,0,0). \end{align*} Equality of coordinates gives $a+2c=0$, $2a+b=0$, and $b+c=0$. From $b+c=0$ we get $c=-b$, and from $a+2c=0$ we get $a=-2c=2b$. Substituting into $2a+b=0$ gives \begin{align*} 2(2b)+b=5b=0. \end{align*} Since $5\ne 0$ in $\mathbb{R}$, we have $b=0$, and then $c=-b=0$ and $a=2b=0$. Therefore $v_1,v_2,v_3$ are linearly independent. There are three linearly independent vectors in the three-dimensional space $\mathbb{R}^3$, so by *Finite-Dimensional Basis Tests* they form a basis of $\mathbb{R}^3$. [/example] ## Subspaces, Quotients, and Dimension Accounting Finite-dimensionality is inherited by the standard constructions of linear algebra. The deeper point is that the dimensions of those constructions obey precise accounting laws. ### Subspaces A subspace is a smaller vector space sitting inside a larger one. In finite dimensions, a proper subspace must lose dimension, so dimension detects when a subspace has already filled the ambient space. [definition: Subspace] Let $k$ be a field and let $V$ be a vector space over $k$. A subset $W \subset V$ is a subspace of $V$ if $W$ is a vector space over $k$ under the addition and scalar multiplication inherited from $V$. [/definition] Subspaces are the natural places where linear constraints live, but a constraint can leave many vectors and still reduce the number of independent directions. In finite-dimensional spaces we need to know that subspaces remain finite-dimensional and that equality of dimension is strong enough to force equality of the subspace with the ambient space. [quotetheorem:9922] This theorem is a major contrast with infinite-dimensional spaces, where a proper subspace can have the same infinite dimension as the whole space. Finite-dimensional subspaces have stricter arithmetic. [example: A Plane in $\mathbb{R}^3$] Let $W=\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1+x_2+x_3=0\}$. If $(x_1,x_2,x_3)\in W$, then $x_1+x_2+x_3=0$, so subtracting $x_1+x_2$ from both sides gives $x_3=-x_1-x_2$. Hence \begin{align*} (x_1,x_2,x_3)=(x_1,x_2,-x_1-x_2). \end{align*} Using coordinatewise scalar multiplication and addition, \begin{align*} x_1(1,0,-1)+x_2(0,1,-1)=(x_1,0,-x_1)+(0,x_2,-x_2)=(x_1,x_2,-x_1-x_2). \end{align*} Therefore every vector in $W$ is a linear combination of $(1,0,-1)$ and $(0,1,-1)$, so these two vectors span $W$. Both vectors lie in $W$, since \begin{align*} 1+0+(-1)=0 \end{align*} and \begin{align*} 0+1+(-1)=0. \end{align*} To check linear independence, suppose \begin{align*} a(1,0,-1)+b(0,1,-1)=(0,0,0) \end{align*} for $a,b\in\mathbb{R}$. The left side is \begin{align*} (a,0,-a)+(0,b,-b)=(a,b,-a-b), \end{align*} so equality with $(0,0,0)$ gives $a=0$ and $b=0$ from the first two coordinates. Thus $(1,0,-1)$ and $(0,1,-1)$ are linearly independent and span $W$, so they form a basis of $W$. Hence $W$ is a two-dimensional subspace of $\mathbb{R}^3$. [/example] ### Sums and Intersections When two subspaces sit inside the same vector space, their dimensions do not simply add. The overlap is counted once as part of the first subspace and again as part of the second, so the intersection must be subtracted. [definition: Sum of Subspaces] Let $k$ be a field, let $V$ be a vector space over $k$, and let $U,W \subset V$ be subspaces. The sum of $U$ and $W$ is $U+W=\{u+w:u \in U, w \in W\}$. [/definition] The sum is the smallest subspace containing both inputs, and the intersection records the part shared by them. If we simply add $\dim U$ and $\dim W$, the common directions are counted twice, so a correction term is needed to measure the dimension of $U+W$ accurately. [quotetheorem:9923] The formula behaves like the inclusion-exclusion formula for finite sets. It is one of the most useful ways dimension turns geometry into arithmetic. [example: Two Planes in $\mathbb{R}^3$] Let \begin{align*} U=\{(x_1,x_2,0):x_1,x_2 \in \mathbb{R}\} \end{align*} and \begin{align*} W=\{(x_1,0,x_3):x_1,x_3 \in \mathbb{R}\}. \end{align*} The vectors $e_1=(1,0,0)$ and $e_2=(0,1,0)$ form a basis of $U$, because every $(x_1,x_2,0)\in U$ satisfies \begin{align*} (x_1,x_2,0)=x_1(1,0,0)+x_2(0,1,0), \end{align*} and if $a(1,0,0)+b(0,1,0)=(0,0,0)$, then $(a,b,0)=(0,0,0)$, so $a=0$ and $b=0$. Thus $\dim U=2$. Similarly, $e_1=(1,0,0)$ and $e_3=(0,0,1)$ form a basis of $W$, since \begin{align*} (x_1,0,x_3)=x_1(1,0,0)+x_3(0,0,1), \end{align*} and $a(1,0,0)+c(0,0,1)=(0,0,0)$ gives $(a,0,c)=(0,0,0)$, hence $a=0$ and $c=0$. Therefore $\dim W=2$. A vector lies in $U\cap W$ exactly when it has both forms \begin{align*} (x_1,x_2,0)=(y_1,0,y_3). \end{align*} Equality of coordinates gives $x_1=y_1$, $x_2=0$, and $y_3=0$, so the common vectors are precisely \begin{align*} U\cap W=\{(t,0,0):t\in\mathbb{R}\}. \end{align*} This line has basis $(1,0,0)$, so $\dim(U\cap W)=1$. By *[Dimension Formula for Two Subspaces](/theorems/9465)*, \begin{align*} \dim(U+W)=\dim U+\dim W-\dim(U\cap W). \end{align*} Substituting the dimensions just computed gives \begin{align*} \dim(U+W)=2+2-1=3. \end{align*} Moreover, every $(x,y,z)\in\mathbb{R}^3$ decomposes as \begin{align*} (x,y,z)=(x,y,0)+(0,0,z), \end{align*} with $(x,y,0)\in U$ and $(0,0,z)\in W$. Hence $U+W=\mathbb{R}^3$. The two planes each have dimension $2$, but their one-dimensional overlap is counted twice unless the intersection is subtracted. [/example] ### Quotients Sometimes the right simplification is not to choose a smaller subspace, but to declare a subspace invisible. Quotient spaces formalize the operation of collapsing chosen directions to zero. [definition: Quotient Vector Space] Let $k$ be a field, let $V$ be a vector space over $k$, and let $W \subset V$ be a subspace. The quotient vector space $V/W$ is the set of cosets $v+W$ with $v \in V$, equipped with the operations $(v+W)+(u+W)=(v+u)+W$ and $a(v+W)=av+W$. [/definition] Collapsing $W$ identifies vectors that differ by an element of $W$, so the quotient no longer remembers directions lying inside $W$. The dimension question is how many independent directions survive this collapse; finite-dimensionality should make that loss measurable. [quotetheorem:377] Quotients are the language of factor spaces, cokernels, and many classification arguments. They let us speak about a vector space after selected directions have been ignored. ## Linear Maps and Matrices A linear map between finite-dimensional vector spaces can be recorded by finitely many scalars. The reason is that a linear map is determined by the images of basis vectors. ### Maps from Basis Data Linearity forces a map to respect every linear combination. Therefore once the values on a basis are known, the values on all vectors are forced by expansion in that basis. [definition: Linear Map] Let $k$ be a field, and let $V$ and $W$ be vector spaces over $k$. A function $T:V \to W$ is a linear map if $T(u+v)=T(u)+T(v)$ and $T(av)=aT(v)$ for all $u,v \in V$ and all $a \in k$. [/definition] The definition imposes conditions on all vectors, which looks like infinitely much data even when the domain is finite-dimensional. A basis removes this obstruction: every vector is forced by its basis expansion, so arbitrary choices on basis vectors should determine exactly one linear map if the proposed rule is consistent with linearity. [quotetheorem:381] The theorem turns a basis into a set of instructions for building maps. Choosing the image of each basis vector is the same as choosing the whole linear map. A matrix is the coordinate record of such a map. To write it down, we need a basis in the domain to identify the input columns and a basis in the codomain to express the output columns. [definition: Matrix of a Linear Map] Let $k$ be a field, let $V$ and $W$ be finite-dimensional vector spaces over $k$, let $\mathcal{B}=(v_1, \ldots, v_n)$ be an ordered basis of $V$, let $\mathcal{C}$ be an ordered basis of $W$, and let $T:V \to W$ be a linear map. The matrix of $T$ with respect to $\mathcal{B}$ and $\mathcal{C}$ is the matrix $[T]_{\mathcal{C} \leftarrow \mathcal{B}}$ whose $j$th column is $[T(v_j)]_{\mathcal{C}}$. [/definition] The notation mirrors change of basis: input coordinates are in $\mathcal{B}$, output coordinates are in $\mathcal{C}$. The remaining issue is whether this matrix is merely a record of basis-vector images or actually computes the coordinates of $T(v)$ for every input vector $v$. [quotetheorem:382] The formula is the justification for doing linear algebra with matrices after bases have been chosen. It also keeps track of which basis is being used at each end of the map. [example: Differentiation on Quadratic Polynomials] Let $P_2(\mathbb{R})$ be the vector space of real polynomials of degree at most $2$, and let $P_1(\mathbb{R})$ be the vector space of real polynomials of degree at most $1$. Consider $D:P_2(\mathbb{R}) \to P_1(\mathbb{R})$ given by $D(p)=p'$. Use the ordered bases $\mathcal{B}=(1,x,x^2)$ in the domain and $\mathcal{C}=(1,x)$ in the codomain. The basis vectors of $P_2(\mathbb{R})$ differentiate as \begin{align*} D(1)=0 \end{align*} \begin{align*} D(x)=1 \end{align*} \begin{align*} D(x^2)=2x. \end{align*} Now express each image in the basis $\mathcal{C}=(1,x)$. Since \begin{align*} 0=0\cdot 1+0\cdot x, \end{align*} we have $[D(1)]_{\mathcal{C}}=(0,0)^\top$. Since \begin{align*} 1=1\cdot 1+0\cdot x, \end{align*} we have $[D(x)]_{\mathcal{C}}=(1,0)^\top$. Since \begin{align*} 2x=0\cdot 1+2\cdot x, \end{align*} we have $[D(x^2)]_{\mathcal{C}}=(0,2)^\top$. Therefore $[D]_{\mathcal{C} \leftarrow \mathcal{B}}$ is the $2\times 3$ matrix whose columns are $(0,0)^\top$, $(1,0)^\top$, and $(0,2)^\top$. Equivalently, its first row is $(0,1,0)$ and its second row is $(0,0,2)$. For a general polynomial \begin{align*} p(x)=a+bx+cx^2, \end{align*} the coordinate column in $\mathcal{B}$ is $[p]_{\mathcal{B}}=(a,b,c)^\top$, and differentiating term by term gives \begin{align*} D(p)=D(a\cdot 1+bx+cx^2)=0+b+2cx. \end{align*} Thus \begin{align*} [D(p)]_{\mathcal{C}}=(b,2c)^\top. \end{align*} The matrix record of differentiation therefore sends the coefficient column $(a,b,c)^\top$ to $(b,2c)^\top$, exactly matching the passage from $a+bx+cx^2$ to $b+2cx$. [/example] ### Kernel and Image A linear map can lose directions, and it can miss directions in the codomain. Kernel and image name these two phenomena so that dimension can measure them. [definition: Kernel] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $T:V \to W$ be a linear map. The kernel of $T$ is $\ker(T)=\{v \in V:T(v)=0\}$. [/definition] The kernel is the part of the domain that becomes invisible under the map. Loss of information is only half the story: a map may also fail to reach some vectors in its codomain, so we need a term for the set of outputs that actually occur. [definition: Image of a Linear Map] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $T:V \to W$ be a linear map. The image of $T$ is $\operatorname{im}(T)=\{T(v):v \in V\}$. [/definition] Kernel and image measure different failures of a linear map: directions can collapse to zero, and only some directions may appear as outputs. The dimension problem is to account for all independent directions in the domain at once. In finite dimension, a direction cannot disappear except by entering the kernel, and the remaining independent directions are exactly what can contribute to the image. This forces a single dimension balance between loss and output. [quotetheorem:385] Rank-nullity is dimension conservation for linear maps. It is the reason solving a homogeneous system and describing the possible outputs are two parts of the same calculation. [example: Rank-Nullity for a Projection] Let $T:\mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(x_1,x_2,x_3)=(x_1,x_2,0)$. To find the kernel, solve $T(x_1,x_2,x_3)=(0,0,0)$. This means \begin{align*} (x_1,x_2,0)=(0,0,0). \end{align*} Equality of coordinates gives $x_1=0$ and $x_2=0$, while $x_3$ is unrestricted. Hence \begin{align*} \ker(T)=\{(0,0,x_3):x_3\in\mathbb{R}\}. \end{align*} Every vector in the kernel has the form \begin{align*} (0,0,x_3)=x_3(0,0,1), \end{align*} and if $a(0,0,1)=(0,0,0)$, then $(0,0,a)=(0,0,0)$, so $a=0$. Thus $(0,0,1)$ is a basis of $\ker(T)$, and $\dim\ker(T)=1$. To find the image, observe that for every $(x_1,x_2,x_3)\in\mathbb{R}^3$, \begin{align*} T(x_1,x_2,x_3)=(x_1,x_2,0). \end{align*} Therefore every output lies in $\{(u,v,0):u,v\in\mathbb{R}\}$. Conversely, if $(u,v,0)$ is any vector in that set, then \begin{align*} T(u,v,0)=(u,v,0), \end{align*} so it is an output of $T$. Hence \begin{align*} \operatorname{im}(T)=\{(u,v,0):u,v\in\mathbb{R}\}. \end{align*} Every vector in the image satisfies \begin{align*} (u,v,0)=u(1,0,0)+v(0,1,0), \end{align*} and if $a(1,0,0)+b(0,1,0)=(0,0,0)$, then $(a,b,0)=(0,0,0)$, so $a=0$ and $b=0$. Thus $(1,0,0),(0,1,0)$ is a basis of $\operatorname{im}(T)$, and $\dim\operatorname{im}(T)=2$. Since $\dim\mathbb{R}^3=3$, the *[Rank-Nullity Theorem](/theorems/916)* gives \begin{align*} \dim\mathbb{R}^3=\dim\ker(T)+\dim\operatorname{im}(T). \end{align*} Substituting the computed dimensions gives \begin{align*} 3=1+2. \end{align*} The projection loses exactly the one vertical direction and keeps the two horizontal directions. [/example] ### Isomorphism Tests For maps between finite-dimensional spaces of the same dimension, injectivity and surjectivity become equivalent. This is a finite-dimensional phenomenon, not a general fact about all vector spaces. [definition: Isomorphism of Vector Spaces] Let $k$ be a field, and let $V$ and $W$ be vector spaces over $k$. An isomorphism from $V$ to $W$ is a bijective linear map $T:V \to W$. [/definition] An isomorphism is the correct notion of sameness for vector spaces, but checking bijectivity directly can duplicate work. The practical test is whether one half of bijectivity is enough. If two finite-dimensional spaces have the same dimension, rank-nullity turns this into a dimension-counting obstruction: a map that loses no directions has no dimension left to miss outputs, and a map that reaches every output has no dimension left for a nonzero kernel. [quotetheorem:386] The finite-dimensional hypothesis cannot be removed. Infinite-dimensional spaces can contain a proper copy of themselves. [example: Injective Need Not Mean Surjective in Infinite Dimension] Let $k[x]$ be the vector space of polynomials over $k$, and define $S:k[x]\to k[x]$ by $S(p)(x)=xp(x)$. We first show that $S$ is injective. Suppose $S(p)=0$. If $p$ were nonzero, write \begin{align*} p(x)=a_0+a_1x+\cdots+a_dx^d \end{align*} with $a_d\ne 0$. Multiplying by $x$ gives \begin{align*} S(p)(x)=xp(x)=a_0x+a_1x^2+\cdots+a_dx^{d+1}. \end{align*} The coefficient of $x^{d+1}$ in $S(p)$ is $a_d$, which is nonzero, so $S(p)$ cannot be the zero polynomial. This contradicts $S(p)=0$, hence $p=0$. Therefore $\ker(S)=\{0\}$, so $S$ is injective. Now consider the polynomial $1\in k[x]$. For any $p(x)=a_0+a_1x+\cdots+a_dx^d$, we have \begin{align*} xp(x)=a_0x+a_1x^2+\cdots+a_dx^{d+1}. \end{align*} The constant coefficient of $xp(x)$ is $0$, while the constant coefficient of $1$ is $1$. Hence no polynomial $p$ satisfies $S(p)=1$, so $1\notin\operatorname{im}(S)$. Thus $S$ is injective but not surjective, showing that the equal-dimension isomorphism principle is genuinely finite-dimensional. [/example] ## Duality and Linear Functionals Finite-dimensional vector spaces also have a controlled dual theory. Linear functionals are scalar measurements, and a basis gives exactly enough coordinate measurements to describe all of them. A linear functional extracts a scalar from a vector while respecting the vector-space operations. Coordinate projection is the guiding example: after a basis is chosen, a functional can read off a coordinate or a linear combination of coordinates. [definition: Linear Functional] Let $k$ be a field and let $V$ be a vector space over $k$. A linear functional on $V$ is a linear map $f:V \to k$. [/definition] Instead of studying one measurement at a time, we need a single object that contains all linear scalar measurements on $V$. Because sums and scalar multiples of linear measurements are again linear measurements, these functionals naturally form a vector space of their own. [definition: Dual Space] Let $k$ be a field and let $V$ be a vector space over $k$. The [dual space](/page/Dual%20Space) of $V$ is $V^*=\{f:V \to k:f \text{ is linear}\}$, with pointwise addition and scalar multiplication. [/definition] A basis of $V$ lets every vector be described by coordinates, so the most basic measurements are the ones that extract a single coordinate. To use these coordinate readings systematically in the dual space, we need to specify the functionals by how they act on the original basis vectors. [definition: Dual Basis] Let $k$ be a field, let $V$ be a finite-dimensional vector space over $k$, and let $\mathcal{B}=(v_1, \ldots, v_n)$ be an ordered basis of $V$. The [dual basis](/theorems/414) of $\mathcal{B}$ is the ordered list $\mathcal{B}^*=(\varepsilon_1, \ldots, \varepsilon_n)$ in $V^*$ such that $\varepsilon_i(v_j)=1$ when $i=j$ and $\varepsilon_i(v_j)=0$ when $i \ne j$. [/definition] The functional $\varepsilon_i$ reads the $i$th coordinate in the basis $\mathcal{B}$. The remaining issue is whether coordinate readings are merely convenient examples or whether they exhaust all possible linear measurements. A functional is determined by its values on a basis, so any hidden measurement would have to give information not already captured by those basis values. [quotetheorem:9924] Finite-dimensional duality says that no hidden linear measurements exist beyond coordinate measurements and their linear combinations. This is one of the sharp differences from infinite-dimensional functional analysis. [example: Dual Basis in $\mathbb{R}^2$] Let $\mathcal{B}=(v_1,v_2)$ be the ordered basis of $\mathbb{R}^2$ with $v_1=(1,1)$ and $v_2=(1,-1)$. For $x=(x_1,x_2)$, write \begin{align*} x=a_1v_1+a_2v_2. \end{align*} Substituting the two basis vectors gives \begin{align*} (x_1,x_2)=a_1(1,1)+a_2(1,-1). \end{align*} Using coordinatewise scalar multiplication and addition, the right side is \begin{align*} a_1(1,1)+a_2(1,-1)=(a_1,a_1)+(a_2,-a_2)=(a_1+a_2,a_1-a_2). \end{align*} Thus \begin{align*} (x_1,x_2)=(a_1+a_2,a_1-a_2). \end{align*} Equality of coordinates gives \begin{align*} x_1=a_1+a_2 \text{ and } x_2=a_1-a_2. \end{align*} Adding these two equations gives \begin{align*} x_1+x_2=(a_1+a_2)+(a_1-a_2)=2a_1. \end{align*} Since the scalar field is $\mathbb{R}$, divide by $2$ to get \begin{align*} a_1=\frac{x_1+x_2}{2}. \end{align*} Subtracting the second equation from the first gives \begin{align*} x_1-x_2=(a_1+a_2)-(a_1-a_2)=2a_2. \end{align*} Dividing by $2$ gives \begin{align*} a_2=\frac{x_1-x_2}{2}. \end{align*} The dual basis functionals read off these two coordinates, so \begin{align*} \varepsilon_1(x_1,x_2)=\frac{x_1+x_2}{2} \end{align*} and \begin{align*} \varepsilon_2(x_1,x_2)=\frac{x_1-x_2}{2}. \end{align*} For example, $\varepsilon_1(v_1)=1$, $\varepsilon_1(v_2)=0$, $\varepsilon_2(v_1)=0$, and $\varepsilon_2(v_2)=1$, so these functionals are exactly the coordinate readers for the basis $\mathcal{B}$. [/example] ## Classification and Direct Sums Finite-dimensional vector spaces over a fixed field are classified by one number: their dimension. This is the structural payoff of the theory. ### Classification by Dimension If two finite-dimensional vector spaces over $k$ have the same dimension, choose a basis in each and send the first basis to the second. The basis theorem for linear maps then produces an isomorphism. [quotetheorem:7839] The theorem does not say that every $n$-dimensional vector space is literally $k^n$. It says that each one becomes indistinguishable from $k^n$ after a basis is chosen. [example: Matrix Spaces] Let $M_{m \times n}(k)$ be the vector space of $m \times n$ matrices over $k$. For each pair $(i,j)$ with $1\le i\le m$ and $1\le j\le n$, let $E_{ij}$ be the matrix whose $(i,j)$ entry is $1$ and whose other entries are $0$. We show that the $mn$ matrices $E_{ij}$ form a basis of $M_{m \times n}(k)$. Let $A=(A_{ij})$ be any $m\times n$ matrix. For fixed indices $r,s$, the $(r,s)$ entry of $A_{ij}E_{ij}$ is $A_{ij}$ when $(r,s)=(i,j)$ and is $0$ otherwise. Therefore the $(r,s)$ entry of the finite sum \begin{align*} \sum_{i=1}^m\sum_{j=1}^n A_{ij}E_{ij} \end{align*} is exactly $A_{rs}$, because all summands have $(r,s)$ entry $0$ except the summand $A_{rs}E_{rs}$. Hence \begin{align*} A=\sum_{i=1}^m\sum_{j=1}^n A_{ij}E_{ij}. \end{align*} So the matrices $E_{ij}$ span $M_{m \times n}(k)$. To check linear independence, suppose \begin{align*} \sum_{i=1}^m\sum_{j=1}^n c_{ij}E_{ij}=0 \end{align*} for scalars $c_{ij}\in k$. Looking at the $(r,s)$ entry of both sides gives \begin{align*} c_{rs}=0 \end{align*} because the $(r,s)$ entry of $c_{rs}E_{rs}$ is $c_{rs}$ and every other summand has $(r,s)$ entry $0$. Since this holds for every pair $(r,s)$, all coefficients $c_{ij}$ are $0$. Thus the matrices $E_{ij}$ are linearly independent. The matrices $E_{ij}$ therefore form a basis of $M_{m \times n}(k)$. There is one such matrix for each of the $mn$ positions in an $m\times n$ matrix, so $\dim_k M_{m \times n}(k)=mn$. [/example] ### Direct Sums Classification tells us that dimension determines a finite-dimensional vector space up to isomorphism. Direct sums explain how such spaces can be assembled from independent components. [definition: Direct Sum of Vector Spaces] Let $k$ be a field, and let $V$ and $W$ be vector spaces over $k$. The [direct sum](/page/Direct%20Sum) $V \oplus W$ is the vector space whose elements are ordered pairs $(v,w)$ with $v \in V$ and $w \in W$, with addition and scalar multiplication defined componentwise. [/definition] The external direct sum keeps the two components separate by construction. There is also an internal version inside a single ambient vector space $X$: subspaces $U$ and $W$ form an internal direct sum, written $X=U\oplus W$, when every vector of $X$ can be written uniquely as $u+w$ with $u\in U$ and $w\in W$. The word "uniquely" is the point; it says the two subspaces do not overlap in a way that creates competing decompositions. With that bridge in place, the dimension question becomes the same one as before. If an ambient space is built from two independent subspaces, then basis directions from each component should combine without creating new relations. The formal result below is the internal direct-sum version of that counting principle. [quotetheorem:3273] Direct sums appear whenever a vector has independent components: blocks of a matrix, real and imaginary parts, or decompositions into a kernel and a complementary subspace. [example: A Line Plus a Plane] Let $L=\{(t,0,0):t \in \mathbb{R}\}$ and $P=\{(0,y,z):y,z \in \mathbb{R}\}$. For any $(x,y,z)\in\mathbb{R}^3$, we have $(x,0,0)\in L$ and $(0,y,z)\in P$, and coordinatewise addition gives \begin{align*} (x,0,0)+(0,y,z)=(x+0,0+y,0+z)=(x,y,z). \end{align*} Thus every vector in $\mathbb{R}^3$ is the sum of one vector from $L$ and one vector from $P$. The decomposition is unique. Suppose \begin{align*} (x,y,z)=(t,0,0)+(0,u,v) \end{align*} with $(t,0,0)\in L$ and $(0,u,v)\in P$. The right side is \begin{align*} (t,0,0)+(0,u,v)=(t,u,v). \end{align*} Therefore $(x,y,z)=(t,u,v)$, so equality of coordinates gives $t=x$, $u=y$, and $v=z$. Hence the only possible decomposition is \begin{align*} (x,y,z)=(x,0,0)+(0,y,z). \end{align*} Define $\Phi:L\oplus P\to\mathbb{R}^3$ by $\Phi(\ell,p)=\ell+p$. For $\ell_1,\ell_2\in L$, $p_1,p_2\in P$, and $a\in\mathbb{R}$, \begin{align*} \Phi((\ell_1,p_1)+(\ell_2,p_2))=\Phi(\ell_1+\ell_2,p_1+p_2)=(\ell_1+\ell_2)+(p_1+p_2). \end{align*} By associativity and commutativity of addition in $\mathbb{R}^3$, this equals \begin{align*} (\ell_1+p_1)+(\ell_2+p_2)=\Phi(\ell_1,p_1)+\Phi(\ell_2,p_2). \end{align*} Also, \begin{align*} \Phi(a(\ell,p))=\Phi(a\ell,ap)=a\ell+ap=a(\ell+p)=a\Phi(\ell,p). \end{align*} So $\Phi$ is linear. The existence of the decomposition above shows that $\Phi$ is surjective, and uniqueness shows that $\Phi$ is injective. Hence $\Phi$ is an isomorphism, so $\mathbb{R}^3$ is isomorphic to $L\oplus P$. The line $L$ has basis $(1,0,0)$, so $\dim L=1$. The plane $P$ has basis $(0,1,0),(0,0,1)$, so $\dim P=2$. Since $\mathbb{R}^3$ has the standard basis $(1,0,0),(0,1,0),(0,0,1)$, its dimension is $3$, and the decomposition records the dimension count \begin{align*} 3=1+2. \end{align*} The point is that the $x$-axis component and the $yz$-plane component are independent pieces of the vector. [/example] ## Beyond and Connected Topics Finite-dimensional vector spaces are the starting point for most of linear algebra. Once bases and dimension are under control, the next layer is the study of linear operators $T:V \to V$, where changing basis becomes a tool for simplifying a matrix. Eigenvalues, diagonalization, [Jordan normal form](/theorems/864), and invariant subspaces all ask how much structure can be revealed by choosing a better basis. The theory also leads to bilinear forms and [inner product](/page/Inner%20Product) spaces. Over $\mathbb{R}$ or $\mathbb{C}$, an inner product adds length, angle, orthogonality, adjoints, and [orthogonal projection](/theorems/437). These ideas connect finite-dimensional algebra with Euclidean geometry and later Hilbert-space methods. In algebra, finite-dimensional vector spaces support field extensions, representations, modules over a field, and Lie algebras. A finite-dimensional [Lie algebra](/page/Lie%20Algebra) is first a finite-dimensional vector space, then a vector space equipped with a bracket operation. The Androma notes [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations) and [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification) build on this linear foundation. The finite-dimensional theory also marks a boundary. Infinite-dimensional vector spaces may lack finite bases, injective maps may fail to be surjective, and proper subspaces can behave like the whole space. Functional analysis studies what can be recovered by adding topology, norm, completeness, and continuity. For a course-level continuation, the natural internal references are [Cambridge IA Vectors and Matrices](/page/Cambridge%20IA%20Vectors%20and%20Matrices), where matrices and Euclidean vector spaces are developed concretely, and [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra), where bases, quotient spaces, duality, and canonical forms are treated systematically. ## References Androma, [Cambridge IA Vectors and Matrices](/page/Cambridge%20IA%20Vectors%20and%20Matrices). Androma, [Cambridge IB Linear Algebra](/page/Cambridge%20IB%20Linear%20Algebra). Androma, [Lie Algebras I: Foundations](/page/Lie%20Algebras%20I%3A%20Foundations). Androma, [Lie Algebras II: Structure and Classification](/page/Lie%20Algebras%20II%3A%20Structure%20and%20Classification). Sheldon Axler, *Linear Algebra Done Right* (2015). Kenneth Hoffman and Ray Kunze, *Linear Algebra* (1971). Paul Halmos, *Finite-Dimensional Vector Spaces* (1942).