A bilinear form may look like a scalar-valued function of two variables, but its useful structure is linear. If one input is fixed, the other input becomes a linear functional. The rank of the bilinear form measures how many independent functionals are produced in this way. It separates the directions detected by the pairing from the directions that pair to zero against everything. This invariant connects [Bilinear Form](/page/Bilinear%20Form), [Vector Space](/page/Vector%20Space), [Linear Map](/page/Linear%20Map), and [Dual Space](/page/Dual%20Space). In coordinates it is ordinary matrix rank, but its meaning is coordinate-free: it measures the effective dimension of the pairing. It is the basic numerical invariant behind radicals, nondegeneracy, symmetric forms, alternating forms, and quadratic forms. ## Definition The page is about a numerical invariant, so the first definition isolates that invariant directly. Fixing the left input of a bilinear form produces a linear functional on the right-hand space; the rank counts the dimension of the space of functionals obtained in this way. [definition: Rank of a Bilinear Form] Let $V$ and $W$ be finite-dimensional vector spaces over a field $k$, and let $B: V \times W \to k$ be a bilinear form. The rank of $B$ is \begin{align*} \operatorname{rank}(B) := \dim \{\lambda \in W^* : \text{there exists } v \in V \text{ such that } \lambda(w)=B(v,w) \text{ for all } w \in W\}. \end{align*} [/definition] The standard dot product shows the intended meaning of the definition before any machinery is introduced. Every coordinate direction produces an independent test functional, so the rank reaches the full ambient dimension. Since the associated maps and radicals have not yet been named, the example describes zero-pairing directions directly in terms of vectors that pair to zero against every vector on the opposite side. [example: Standard Dot Product Has Full Rank] Let $k$ be a field and let $V=k^n$. Define $B:k^n \times k^n \to k$ by \begin{align*} B(x,y)=\sum_{i=1}^n x_i y_i. \end{align*} For the standard basis vectors $e_1,\ldots,e_n$, if $i=j$ then $B(e_i,e_j)=1$, and if $i \neq j$ then $B(e_i,e_j)=0$. Thus the matrix of $B$ in the standard basis is $I_n$. For each $i$, the functional on $k^n$ produced by fixing $e_i$ is \begin{align*} \lambda_i(y)=B(e_i,y)=y_i. \end{align*} Therefore $\lambda_i$ is the $i$th coordinate functional on $k^n$. These $n$ coordinate functionals form the standard basis of the dual space $(k^n)^*$, so the functionals obtained by fixing the first input span all of $(k^n)^*$. Hence \begin{align*} \operatorname{rank}(B)=\dim (k^n)^*=n. \end{align*} Now suppose a vector $x \in k^n$ pairs to zero against every vector $y \in k^n$. Taking $y=e_i$ gives \begin{align*} 0=B(x,e_i)=x_i. \end{align*} This holds for every $i$, so $x=0$. The same argument on the right gives: if $y \in k^n$ pairs to zero against every vector $x \in k^n$, then taking $x=e_i$ gives \begin{align*} 0=B(e_i,y)=y_i. \end{align*} Thus $y=0$. The standard dot product detects every nonzero vector on both sides; later definitions will call this two-sided detection nondegeneracy. [/example] ## Background and Associated Maps The rank definition assumes that the two-variable pairing can be probed by fixing one input and reading off a linear functional in the other. Bilinearity is the precise condition that guarantees this probe is legitimate. Without it, the values of a two-variable function need not assemble into linear maps into dual spaces, so the matrix-style notion of rank would not survive a change of coordinates. [definition: Bilinear Form] Let $k$ be a field, and let $V$ and $W$ be vector spaces over $k$. A bilinear form on $V \times W$ is a function $B: V \times W \to k$ such that for every $v, v_1, v_2 \in V$, every $w, w_1, w_2 \in W$, and every $a,b \in k$, \begin{align*} B(av_1+bv_2,w)=aB(v_1,w)+bB(v_2,w) \end{align*} and \begin{align*} B(v,aw_1+bw_2)=aB(v,w_1)+bB(v,w_2). \end{align*} [/definition] The values of $B$ are scalars, so they cannot be counted directly as vectors in $V$ or $W$. The usable object is obtained by holding the left input fixed: each $v \in V$ then determines a linear functional on $W$, and the collection of all such functionals records exactly which tests on $W$ are produced by vectors of $V$. To make rank a property of the pairing rather than a loose collection of scalar evaluations, this assignment has to be treated as an actual linear map. Naming that map gives a precise object whose image measures how many independent tests on $W$ come from the left input. [definition: Left Linear Map of a Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The left linear map associated to $B$ is the linear map $L_B: V \to W^*$ defined by \begin{align*} L_B(v)(w)=B(v,w) \end{align*} for all $v \in V$ and $w \in W$. [/definition] A complete rank theory must not privilege the first input without checking the second. Rectangular pairings may detect $V$ and $W$ in different ways, so we need the companion map from $W$ into $V^*$. This map records which functionals on $V$ arise from vectors in $W$. [definition: Right Linear Map of a Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The right linear map associated to $B$ is the linear map $R_B: W \to V^*$ defined by \begin{align*} R_B(w)(v)=B(v,w) \end{align*} for all $v \in V$ and $w \in W$. [/definition] Rank is meaningful because some directions may be invisible to all tests supplied by the opposite space. We need a name for the invisible vectors on the left side. This subspace will later become the kernel term in the rank-nullity formula for $B$. [definition: Left Radical of a Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The left radical of $B$ is the subspace \begin{align*} \operatorname{rad}_L(B) := \{v \in V : B(v,w)=0 \text{ for all } w \in W\}. \end{align*} [/definition] The right input can have its own invisible vectors. This separate definition is necessary because a form can be nondegenerate on one side and degenerate on the other when the spaces have different dimensions. [definition: Right Radical of a Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The right radical of $B$ is the subspace \begin{align*} \operatorname{rad}_R(B) := \{w \in W : B(v,w)=0 \text{ for all } v \in V\}. \end{align*} [/definition] The initial definition of rank can now be rewritten using the associated left map: \begin{align*} \operatorname{rank}(B)=\dim \operatorname{Range}(L_B). \end{align*} This formula uses the left associated map, so it needs a compatibility result with the right associated map. The next theorem explains why the rank belongs to the bilinear form itself rather than to a chosen side. [quotetheorem:7655] The equality is necessary because the two associated maps are not literally the same object: $L_B$ has domain $V$ and codomain $W^*$, while $R_B$ has domain $W$ and codomain $V^*$. The theorem says that these different probes detect the same number of independent scalar tests. For example, a rectangular pairing can have a nonzero right radical while its left radical is zero, but the number of independent rows and the number of independent columns are still equal. Thus rank measures the amount of information carried by the pairing itself, not the side from which one chooses to inspect it. This limitation is also important. Equality of ranks does not say that the two radicals have the same dimension, and it does not by itself imply nondegeneracy. To use rank for nondegeneracy, we must compare this common rank with the dimensions of the original spaces. Rank is most useful when compared with the dimensions of the original spaces. To make that comparison precise, we name the condition that no nonzero vector on the left is invisible to the pairing. [definition: Left Nondegenerate Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The form $B$ is left nondegenerate if \begin{align*} \operatorname{rad}_L(B)=\{0\}. \end{align*} [/definition] The corresponding right-sided condition records perfect detection of vectors in $W$. It must be named separately because left and right nondegeneracy need not agree for pairings between unequal-dimensional spaces. [definition: Right Nondegenerate Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The form $B$ is right nondegenerate if \begin{align*} \operatorname{rad}_R(B)=\{0\}. \end{align*} [/definition] Many applications require a pairing that loses no information from either input. The two one-sided notions combine into the usual notion of a nondegenerate bilinear form. [definition: Nondegenerate Bilinear Form] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. The form $B$ is nondegenerate if it is both left nondegenerate and right nondegenerate. [/definition] ## Equivalent Characterisations Computations usually begin by choosing bases. Once bases are chosen, the values of a bilinear form on basis vectors form a matrix, and every value of the form is recovered from that matrix and the coordinate vectors. [definition: Matrix of a Bilinear Form] Let $k$ be a field, let $V$ and $W$ be finite-dimensional vector spaces over $k$, let $B: V \times W \to k$ be a bilinear form, let $(v_1,\ldots,v_m)$ be a basis of $V$, and let $(w_1,\ldots,w_n)$ be a basis of $W$. The matrix of $B$ in these bases is the matrix $A \in k^{m \times n}$ with entries \begin{align*} A_{ij} := B(v_i,w_j). \end{align*} [/definition] Once a matrix has been attached to $B$, there is a possible ambiguity: the rank defined from the associated map might depend on the abstract pairing, while the rank computed from entries might depend on the chosen bases. The needed comparison is that these two measurements give the same number, so coordinate calculations can be trusted. The next issue is therefore not how to form the matrix, but whether the act of choosing coordinates preserves the invariant that rank is meant to measure. Establishing this bridge lets later rank formulas move freely between the coordinate-free pairing and ordinary row-column computations, without creating two competing notions of rank. [quotetheorem:7656] The matrix characterisation is necessary because the definition of rank was phrased without coordinates, while most computations begin with entries. The theorem says that choosing bases does not create a second notion of rank: the coordinate matrix records exactly the same independent functionals that the associated map records. This lets one compute with row reduction or minors and then interpret the result back on the original bilinear form. For instance, in the standard dot product example the matrix $I_n$ has rank $n$, matching the fact that the coordinate functionals obtained from the first input span all of $(k^n)^*$. In a rectangular example, the same theorem allows the rank to be read from the number of independent rows or columns even though the two associated maps have different domains and codomains. This is why the matrix theorem is the computational bridge: it turns an abstract invariant into a basis-dependent calculation without changing the value being measured. Its limitation is that it depends on finite-dimensional coordinate choices and says nothing by itself about which vectors are invisible. A matrix of rank $r$ tells us the size of the detected part, but it does not identify the lost directions until the corresponding nullspaces are examined. That is why the next formula translates the same rank into radical dimensions. Matrix rank is not the only way to measure the size of the pairing. Directions in the radicals are precisely the inputs that all scalar tests fail to see, so the missing dimensions should subtract from the ambient dimensions. A coordinate-free formula is needed to make that loss visible without choosing bases. [quotetheorem:7657] This criterion turns nondegeneracy into a rank test. On the left, nondegeneracy means that the associated map $L_B:V \to W^*$ has zero kernel, so its rank must reach the full dimension of $V$. On the right, the analogous condition requires rank $\dim W$. Therefore a bilinear form on unequal-dimensional spaces cannot be nondegenerate on both sides: one common rank cannot simultaneously equal two different dimensions. In practice, this is the fastest way to test nondegeneracy after computing a matrix rank, while also warning that full row rank and full column rank are different requirements for rectangular matrices. The criterion is necessary, not just convenient: a zero radical is an injectivity condition on the associated map, and injectivity in finite dimensions is measured exactly by reaching the dimension of the domain. Its limitation is that it is one-sided unless the two spaces have the same dimension. For a rectangular form, rank $\dim V$ proves that the left input is fully detected, but it says nothing by itself about whether every vector of $W$ is detected. In a square situation the two tests coincide, so full rank is equivalent to nondegeneracy on both sides; in a rectangular situation, one must still check which side has the possible invisible directions. This is why later examples separate active coordinates from radical coordinates instead of merely reporting one number. The same rank can mean different detection behavior depending on the dimensions of $V$ and $W$, and restrictions can turn a previously detected direction into an invisible one by removing the vectors that paired with it. The rank criterion will therefore be used as a diagnostic: first compute the rank, then compare it with the relevant source dimension to decide exactly which nondegeneracy statement is justified. There is another way to view rank: as the number of independent scalar channels needed to build the pairing. This perspective is useful when a bilinear form is assembled from simpler linear measurements. [definition: Factorisation of a Bilinear Form Through a Finite-Dimensional Space] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, and let $B: V \times W \to k$ be a bilinear form. A factorisation of $B$ through a finite-dimensional vector space $U$ over $k$ consists of linear maps $S: V \to U$ and $T: W \to U^*$ such that \begin{align*} B(v,w)=T(w)(S(v)) \end{align*} for all $v \in V$ and $w \in W$. [/definition] A factorisation through $U$ may contain redundant coordinates: extra dimensions of $U$ can be present even if they do not contribute to the values of $B$. To make factorisations measure rank rather than bookkeeping choices, one must identify the smallest possible intermediate dimension that still reproduces the whole pairing. [quotetheorem:7658] This characterization explains why rank is the intrinsic size of the pairing, not merely the size of a displayed matrix. Any factorisation through a larger space may contain unused or repeated scalar channels, while a factorisation through a smaller space cannot carry enough independent tests to reproduce $B$. Thus the rank is exactly the least number of intermediate coordinates needed to build the form from linear measurements on $V$ and $W$. The result is especially useful when a bilinear form is presented as a sum of simple products, because each product contributes one possible scalar channel through an intermediate space. It also explains what compression can and cannot do: one may change coordinates or remove redundant channels, but one cannot lower the intermediate dimension below the rank without losing some values of the pairing. The examples below make this concrete by showing active coordinates, invisible directions, and field-dependent rank changes. ## Coordinate Computations and Radical Behavior ### Degenerate Coordinates and Characteristic A diagonal form with a zero coefficient shows that rank counts active coordinates rather than ambient dimension. It also shows that the field can matter when a coefficient vanishes in that field. [example: Degenerate Diagonal Form] Let $B:k^3 \times k^3 \to k$ be defined by \begin{align*} B(x,y)=x_1y_1+2x_2y_2 \end{align*} for $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$. In the standard basis, $B(e_1,e_1)=1$, $B(e_2,e_2)=2$, $B(e_3,e_3)=0$, and $B(e_i,e_j)=0$ whenever $i \neq j$, so the only possibly nonzero diagonal entries are $1$ and $2$. Let $\varepsilon_i$ be the $i$th coordinate functional on $k^3$. For any $x \in k^3$, \begin{align*} L_B(x)(y)=B(x,y)=x_1y_1+2x_2y_2=(x_1\varepsilon_1+2x_2\varepsilon_2)(y) \end{align*} for every $y \in k^3$. Hence \begin{align*} \operatorname{Range}(L_B)=\{x_1\varepsilon_1+2x_2\varepsilon_2:x_1,x_2 \in k\}. \end{align*} If $2 \neq 0$ in $k$, then as $x_1$ and $x_2$ vary, the coefficients $x_1$ and $2x_2$ range over all of $k$. Therefore \begin{align*} \operatorname{Range}(L_B)=\operatorname{span}\{\varepsilon_1,\varepsilon_2\} \end{align*} and $\operatorname{rank}(B)=2$. For the left radical, $x \in \operatorname{rad}_L(B)$ means $B(x,y)=0$ for every $y \in k^3$. Taking $y=(1,0,0)$ gives \begin{align*} 0=B(x,(1,0,0))=x_1. \end{align*} Taking $y=(0,1,0)$ gives \begin{align*} 0=B(x,(0,1,0))=2x_2. \end{align*} Since $2 \neq 0$, this implies $x_2=0$, while $x_3$ is unrestricted because $B((0,0,x_3),y)=0$ for every $y$. Hence \begin{align*} \operatorname{rad}_L(B)=\operatorname{span}\{(0,0,1)\}. \end{align*} The same coordinate tests on the first input give \begin{align*} \operatorname{rad}_R(B)=\operatorname{span}\{(0,0,1)\}. \end{align*} If $2=0$ in $k$, then \begin{align*} B(x,y)=x_1y_1 \end{align*} for all $x,y \in k^3$. In this case \begin{align*} \operatorname{Range}(L_B)=\operatorname{span}\{\varepsilon_1\} \end{align*} so $\operatorname{rank}(B)=1$. Also, $x \in \operatorname{rad}_L(B)$ exactly when $x_1=0$, because taking $y=(1,0,0)$ gives $0=x_1$, and if $x_1=0$ then $B(x,y)=0$ for every $y$. Thus \begin{align*} \operatorname{rad}_L(B)=\operatorname{span}\{(0,1,0),(0,0,1)\}. \end{align*} The same argument on the right gives \begin{align*} \operatorname{rad}_R(B)=\operatorname{span}\{(0,1,0),(0,0,1)\}. \end{align*} The coefficient $2$ contributes an active coordinate only when it is nonzero in the field, so the rank depends on the characteristic of $k$. [/example] ### Rectangular Detection Pairings between spaces of different dimensions show why left and right radicals were defined separately. The smaller side can be fully detected while the larger side still has unavoidable invisible directions. [example: Rectangular Pairing] Let $B: k^2 \times k^3 \to k$ be defined by \begin{align*} B((x_1,x_2),(y_1,y_2,y_3))=x_1y_1+x_2y_2. \end{align*} Let $\varepsilon_1,\varepsilon_2,\varepsilon_3$ be the coordinate functionals on $k^3$. For $x=(x_1,x_2) \in k^2$, \begin{align*} L_B(x)(y)=B(x,y)=x_1y_1+x_2y_2=(x_1\varepsilon_1+x_2\varepsilon_2)(y). \end{align*} Thus \begin{align*} \operatorname{Range}(L_B)=\operatorname{span}\{\varepsilon_1,\varepsilon_2\}. \end{align*} The functionals $\varepsilon_1$ and $\varepsilon_2$ are independent, since if $a\varepsilon_1+b\varepsilon_2=0$, then evaluating at $(1,0,0)$ gives $a=0$, and evaluating at $(0,1,0)$ gives $b=0$. Therefore \begin{align*} \operatorname{rank}(B)=\dim \operatorname{Range}(L_B)=2. \end{align*} Now let $x=(x_1,x_2) \in \operatorname{rad}_L(B)$. Then $B(x,y)=0$ for every $y \in k^3$. Taking $y=(1,0,0)$ gives \begin{align*} 0=B((x_1,x_2),(1,0,0))=x_1. \end{align*} Taking $y=(0,1,0)$ gives \begin{align*} 0=B((x_1,x_2),(0,1,0))=x_2. \end{align*} Hence $x=(0,0)$, so \begin{align*} \operatorname{rad}_L(B)=\{0\}. \end{align*} For the right radical, let $y=(y_1,y_2,y_3) \in \operatorname{rad}_R(B)$. Then $B(x,y)=0$ for every $x \in k^2$. Taking $x=(1,0)$ gives \begin{align*} 0=B((1,0),(y_1,y_2,y_3))=y_1. \end{align*} Taking $x=(0,1)$ gives \begin{align*} 0=B((0,1),(y_1,y_2,y_3))=y_2. \end{align*} The coordinate $y_3$ is unrestricted, and if $y=(0,0,y_3)$ then \begin{align*} B((x_1,x_2),(0,0,y_3))=x_1 \cdot 0+x_2 \cdot 0=0 \end{align*} for every $x \in k^2$. Therefore \begin{align*} \operatorname{rad}_R(B)=\operatorname{span}\{(0,0,1)\}. \end{align*} Thus $B$ detects every nonzero vector on the $k^2$ side, but it misses the third coordinate direction on the $k^3$ side; it is left nondegenerate and not right nondegenerate. [/example] ### Alternating Rank in Odd Dimension Not every bilinear form is arbitrary. In many geometric settings the pairing of a vector with itself is forced to vanish, so the diagonal values carry no information at all. This constraint is strong enough to impose restrictions on rank, and it must be named before those restrictions can be discussed. [definition: Alternating Bilinear Form] Let $V$ be a vector space over a field $k$. A bilinear form $B: V \times V \to k$ is alternating if \begin{align*} B(v,v)=0 \end{align*} for every $v \in V$. [/definition] The next example displays a forced radical for an alternating form in dimension $3$. The lost direction is not caused by a poor choice of basis; it reflects the parity restriction on alternating rank. [example: Alternating Form in Odd Dimension] Let $k$ be a field with $2 \neq 0$ in $k$, and define $B: k^3 \times k^3 \to k$ by \begin{align*} B(x,y)=x_1y_2-x_2y_1. \end{align*} For $x=y$, we get \begin{align*} B(x,x)=x_1x_2-x_2x_1=0, \end{align*} so $B$ is alternating. In the standard basis $e_1,e_2,e_3$, the nonzero basis values are \begin{align*} B(e_1,e_2)=1 \end{align*} and \begin{align*} B(e_2,e_1)=-1. \end{align*} Also, \begin{align*} B(e_1,e_1)=B(e_2,e_2)=B(e_3,e_3)=B(e_1,e_3)=B(e_3,e_1)=B(e_2,e_3)=B(e_3,e_2)=0. \end{align*} Thus the matrix of $B$ in the standard basis has first row $(0,1,0)$, second row $(-1,0,0)$, and third row $(0,0,0)$. Let $\varepsilon_1,\varepsilon_2,\varepsilon_3$ be the coordinate functionals on $k^3$. For $x=(x_1,x_2,x_3)$, \begin{align*} L_B(x)(y)=B(x,y)=x_1y_2-x_2y_1=(-x_2\varepsilon_1+x_1\varepsilon_2)(y) \end{align*} for every $y \in k^3$. Hence every functional in $\operatorname{Range}(L_B)$ lies in $\operatorname{span}\{\varepsilon_1,\varepsilon_2\}$. Conversely, for any $a,b \in k$, choosing $x=(b,-a,0)$ gives \begin{align*} L_B(b,-a,0)=a\varepsilon_1+b\varepsilon_2. \end{align*} Therefore \begin{align*} \operatorname{Range}(L_B)=\operatorname{span}\{\varepsilon_1,\varepsilon_2\}. \end{align*} The functionals $\varepsilon_1$ and $\varepsilon_2$ are independent because evaluating $a\varepsilon_1+b\varepsilon_2=0$ at $e_1$ gives $a=0$, and evaluating at $e_2$ gives $b=0$. Thus \begin{align*} \operatorname{rank}(B)=2. \end{align*} Now let $x=(x_1,x_2,x_3) \in \operatorname{rad}_L(B)$. Taking $y=e_2$ gives \begin{align*} 0=B(x,e_2)=x_1. \end{align*} Taking $y=e_1$ gives \begin{align*} 0=B(x,e_1)=-x_2. \end{align*} Since $-1 \neq 0$ in any field, $x_2=0$. The coordinate $x_3$ is unrestricted, and if $x=(0,0,x_3)$ then \begin{align*} B((0,0,x_3),(y_1,y_2,y_3))=0 \cdot y_2-0 \cdot y_1=0 \end{align*} for every $y \in k^3$. Hence \begin{align*} \operatorname{rad}_L(B)=\operatorname{span}\{(0,0,1)\}. \end{align*} The same computation on the right gives the right radical. If $y=(y_1,y_2,y_3) \in \operatorname{rad}_R(B)$, then taking $x=e_1$ gives \begin{align*} 0=B(e_1,y)=y_2, \end{align*} and taking $x=e_2$ gives \begin{align*} 0=B(e_2,y)=-y_1. \end{align*} Thus $y_1=y_2=0$, while $y_3$ is unrestricted, so \begin{align*} \operatorname{rad}_R(B)=\operatorname{span}\{(0,0,1)\}. \end{align*} This alternating form has one forced invisible direction in $k^3$, so it has rank $2$ rather than $3$ and is not nondegenerate. [/example] ## Invariance and Rank Criteria ### Change of Basis The same bilinear form can have many different matrices, one for each choice of bases. If rank changed under this choice, it would describe the coordinates rather than the pairing itself. The essential invariance question is whether replacing the bases changes only the displayed matrix and not the amount of independent pairing data. In the coordinate formula below, $\mathrm{GL}_n(\mathbb{F})$ denotes the group of invertible $n \times n$ matrices over the field $\mathbb{F}$, and $P^\top$ denotes the transpose of $P$. The matrices $P$ and $Q$ record invertible changes of basis on the two input spaces. [quotetheorem:391] This change-of-basis rule is the reason rank is an invariant of the bilinear form, not an artifact of coordinates. Multiplying the matrix on the left and right by invertible change-of-basis matrices can reorganize the rows and columns, but it cannot create or destroy independent pairing information. Thus computations may be done in whichever bases make the matrix simplest, while the resulting rank still belongs to $B$ itself. ### Nondegeneracy by Rank With rank now established as coordinate-independent, it can be used to test intrinsic properties. Nondegeneracy says that no nonzero input is invisible, but checking this directly means solving for radicals. Rank gives a single numerical substitute: it reaches the full dimension of the tested side exactly when that side has no invisible direction. For rectangular pairings, the two sides can impose different maximum-rank requirements, so the thresholds must be stated separately. The next criterion turns the qualitative condition of having no radical into a dimension check. This is especially useful because it separates the left and right tests, making clear which ambient dimension supplies the relevant full-rank threshold. [quotetheorem:7659] The criterion is useful because it turns the qualitative phrase "no invisible vectors" into a rank computation. Its necessity comes from rank-nullity: if the left radical is zero, then the associated map $L_B:V \to W^*$ is injective and has rank $\dim V$, and conversely rank $\dim V$ leaves no room for a nonzero kernel. The right-hand condition is the same argument applied to $R_B:W \to V^*$. The limitation is that the two tests have different thresholds unless $\dim V=\dim W$. A rectangular form may be left nondegenerate without being right nondegenerate, or conversely, because one common rank is being compared with two different ambient dimensions. This is why later restrictions must track which side loses vectors or tests: full rank on one side is not automatically full rank on the other. ### Restriction to Subspaces Applications often restrict a pairing to a smaller collection of vectors. To state the corresponding rank comparison, we first name the restricted bilinear form. [definition: Restriction of a Bilinear Form to Subspaces] Let $k$ be a field, let $V$ and $W$ be vector spaces over $k$, let $B: V \times W \to k$ be a bilinear form, and let $V_0 \subset V$ and $W_0 \subset W$ be subspaces. The restriction of $B$ to $V_0 \times W_0$ is the bilinear form $B|_{V_0 \times W_0}: V_0 \times W_0 \to k$ defined by \begin{align*} B|_{V_0 \times W_0}(v,w)=B(v,w) \end{align*} for all $v \in V_0$ and $w \in W_0$. [/definition] Restriction can remove available vectors and tests, so it should not increase the amount of independent pairing data. The following monotonicity statement formalises that expectation. [quotetheorem:7660] The inequality is a one-way control statement. Restricting $V$ or $W$ can only discard inputs or tests, so the restricted associated map has no more independent functionals than the original one. Equality can occur when the chosen subspaces still contain all directions needed to realize the original image, but rank drops as soon as the restriction removes a direction that contributed an independent row, column, or scalar channel. This is the basic tool for comparing local or constrained versions of a pairing with the ambient form. For example, passing to a subspace may preserve nondegeneracy on that subspace, or it may introduce a new radical even when the original form was nondegenerate. The theorem therefore explains why restriction arguments must be accompanied by a rank check rather than by an assumption that nondegeneracy is inherited unchanged. ### Square Forms and Determinants When $V$ and $W$ have the same finite dimension and bases have been chosen, the representing matrix is square. In that special case, the question of whether any nonzero vector is invisible becomes the familiar question of whether the matrix is invertible. The determinant is the coordinate test that detects this invertibility. [quotetheorem:7661] The determinant criterion is compact, but it applies only in the square finite-dimensional case. Radical formulas and associated maps are more flexible because they continue to describe rectangular pairings and restrictions. ## Relationship to Other Concepts The rank of a bilinear form is the rank of a linear map obtained by currying the form. The radicals are the kernels of the associated maps: \begin{align*} \operatorname{rad}_L(B)=\ker(L_B). \end{align*} Also, \begin{align*} \operatorname{rad}_R(B)=\ker(R_B). \end{align*} Thus rank, kernel, image, and duality are the same structural package viewed from different angles. For a symmetric bilinear form on one vector space, rank controls the dimension of the nondegenerate quotient by the radical. This is central for symmetric bilinear forms and quadratic forms, where the radical records the part of the space on which the quadratic data loses information. For an alternating bilinear form, the rank is even over any field under the convention $B(v,v)=0$. Nondegenerate alternating forms are therefore possible only in even dimension, which is the linear-algebraic starting point for symplectic vector spaces and the theory of alternating bilinear forms. In algebraic geometry, after bases are chosen, the condition $\operatorname{rank}(B) \le r$ is expressed by the vanishing of all minors of size $r+1$. Rank therefore stratifies families of bilinear forms by polynomial equations. In differential geometry, the rank of a smoothly varying bilinear form distinguishes genuine metrics from degenerate tensor fields. ## Beyond and Connected Topics The associated-map viewpoint places the rank of a bilinear form inside the general theory of [Linear Map](/page/Linear%20Map) and [Dual Space](/page/Dual%20Space). Once $B$ is replaced by $L_B:V \to W^*$ or $R_B:W \to V^*$, rank formulas, injectivity tests, and quotient constructions become ordinary statements about linear transformations. The same invariant also guides later study of [Bilinear Form](/page/Bilinear%20Form) and [Vector Space](/page/Vector%20Space) by showing which coordinates or subspaces are detected by a pairing. This is why the rank appears in classifications of symmetric, alternating, and quadratic structures even when those later subjects add extra hypotheses beyond bilinearity. ## References Sheldon Axler, *Linear Algebra Done Right* (2015). Serge Lang, *Algebra* (2002). Friedberg, Insel, and Spence, *Linear Algebra* (2018). [Bilinear Form](/page/Bilinear%20Form). [Dual Space](/page/Dual%20Space).