Abstract

We give a formula for determining when a set of k vectors in n-space is linearly independent, and if so, what is the volume of the parallelepiped with these vectors as its sides. This function, the gramian, allows one to partially apply the determinant when the number of vectors you have is less than the dimension of the ambient space. These results were classically known, but are not part of the standard linear algebra curriculum. Prerequisites: familiarity with matrices and determinants.

Introduction

Given k vectors in n-dimensional space, when are they linearly independent, and what is the k-volume of the parallelepiped that they define? When k = n, the answer is familiar: arrange the vectors in a matrix and take the determinant. The vectors are then linearly dependent iff the determinant is zero, and otherwise the volume of the parallelepiped, Vol(v₁,...,v_k), is the absolute value of the determinant. For k < n, the widely known answers are piecemeal: for k = 2, n = 3, one can use the cross product; to determine linear dependence, one can use gaussian elimination, though this is an algorithm, rather than a formula. Ideally, one would have some analog for the determinant which applies when the number of vectors is not equal to the dimension of the ambient space. This answer is provided in the gramian, which we introduce below; we loosely follow [1].

If { v₁,...,v_k} is linearly dependent, this definition doesn't make sense; more formally,

When { v₁,...,v_k} are linearly independent, P(v₁,...,v_k) has a well-defined k-volume (volume as a subset of k-dimensional space); when they are linearly dependent, this volume is zero. Note that if you have, for instance, 3 vectors in 2-space, then P(v₁,...,v₃) has non-zero volume as a subset of 2-space, but it is "flat" and thus has zero volume as a subset of 3-space.

Recall that an inner product is a choice of distance (and also volume) in a vector space; the main example is the dot product, defined as

which is a scalar, and where v^* = conj(v)^T is the complex conjugate of the transpose. We take complex conjugates so that complex vectors have real length; if K = R, this reduces to v·w = v^Tw. We define the norm (length) of v as

Matrices that preserve distances and angles, i.e., Ov·Ow = v·w, are of particular interest, and are called orthogonal;² these correspond to rigid motions. For instance, orthogonal transforms of R² are rotations and reflections.

Definition 4 [gramian] The gramian of v₁,...,v_k is G(v₁,...,v_k) = detM^*M, where

and M^* = [conj(M)]^T is the conjugate transpose of M.

Note the similarity to the norm (3); in fact, G(v) = ||v||, so the gramian of a single vector is the length, i.e., the 1-dimensional volume, of that vector.

If we calculate the entries of M^*M, we obtain the following characterization of the gramian:

Definition 6 [gramian-alternate] G(v₁,...,v_k) = detA, where a_ij = v_i·v_j.

This presentation shows that the matrix M^*M captures all the geometric information about the vectors { v₁,...,v_k}: the length of the vectors and the angles between them. Further, this is the only information that it contains; we've lost the particular orientations of the vectors and their embedding in n-space. We claim that this is enough information to easily determine the volume.

Consider the case where k = n; then det M^*M = [conj(det M)](det M) = |det M|². In particular, G(v₁,...,v_k) >= 0. Since Vol(v₁,...,v_k) = |det M|, we obtain:

For any collection of k vectors, the above holds by simply restricting to a k-dimensional subspace containing them; that is, Vol(v₁,...,v_k) = detA. By using (6), we obtain the following geometric characterization of the gramian:

Proposition 8 Vol(v₁,...,v_k) = G(v₁,...,v_k)^1/2.

Note that in case k =< n, we still have G(v₁,...,v_k) >= 0, so this is well-defined.

Proof.   The proof was sketched above; we fill in the details here.

Given v₁,...,v_k, where k =< n, assume that they are linearly independent³ and pick an orthonormal basis w₁,...,w_k for their span, W. Extend to an orthonormal basis w₁,...,w_n for Kⁿ; the matrix O sending w_i --> e_i is an orthogonal change of coordinates, so it does not change inner products: O(v)·O(w) = v·w. Let v'_i = O(v_i); then

since v_i is in the span of w₁,...,w_k. The set { v'₁,...,v'_k} visibly lies in the k-dimensional subspace of vectors with last n-k coordinates zero, so the k-volume of P(v'₁,...,v'_k) is |det M|, where m_ij is defined as above. Effectively, we are restricting to the subspace W. By the discussion before this proposition, Vol(v'₁,...,v'_k) = {det M^*M}^1/2 (we're just dropping the trailing zeros). Since O is orthogonal,

as desired.

When k > n, the vectors are linearly dependent, so the k-volume of P(v₁,...,v_k) is zero. In this case the gramian is zero, by the argument in the proof of (11), below. [X]

In particular, the k-volume of P(v₁,...,v_k) is zero iff { v₁,...,v_k} don't form the sides of a k-parallelepiped; that is, if P(v₁,...,v_k) has dimension less than k. This yields:

Corollary 11 [Gram's criterion for linear dependence of vectors] The set of vectors { v₁,...,v_k } is linearly dependent iff G(v₁,...,v_k) = 0.

Proof.   We just dealt with the case k =< n; it remains to show that if k > n, the gramian is zero. Consider A = M^*M; each row of the k×k matrix A is a linear combination of rows of M, of which there are n. Thus, the row-rank of A is at most n < k, so its determinant is zero. [X]

Note that the gramian is always nonnegative,⁴ and equals zero iff the vectors are linearly dependent. As a consequence, we obtain the familiar:

Corollary 12 [Bunyakovskii-Cauchy-Schwarz inequality] (v·w)² =< ||v|| ||w||, with equality iff v,w are linearly dependent (one is a multiple of the other).

Proof.   G(v,w) => 0, with equality iff v,w are linearly dependent. Now

the result follows.

the composition of which is an endomorphism of V (linear operator on V), and thus has a determinant. The gramian of T (with respect to the inner products on V, W) is then the determinant of this map.

Recall that the inner product of v, w with respect to an inner product other than the dot product is given by [conj(v)]^TPw, where P is some invertible matrix. Similarly, M^*PM is the gramian with respect to this inner product, which follows from expanding (6).

The key part of the restriction on K is that the characteristic be zero; the above doesn't work in positive characteristic. For instance, suppose charK = p > 0, and consider

Then

but { v₁,v₂ } is clearly linearly independent. The problem is that nonzero vectors can have zero norms; worse, there are non-trivial subspaces (such as the span of { v₁,v₂ }) in which the inner product of any two vectors is zero. This prevents us from naively following our geometric intuition from euclidean space, and is one of the reasons that the study of quadratic forms (generalized inner products) over positive characteristic is interesting.

Footnotes

¹ Date: July 16, 1999.
1991 Mathematics Subject Classification. 15A15
The author wishes to thank Antonia Bluher for suggesting this topic.

² If the vector space is complex, they are instead called unitary.

³ If they are not linearly independent, instead take W to be some k-dimensional subspace containing them.

⁴ In particular, it is real; this eliminates the need to take the absolute value to obtain the volume. This is more significant in the complex case, where it shows that there is a well-defined notion of real-valued volume.
 

[1]

F. R. Gantmacher, Matrix theory, vol. 1, Chelsea Publishing Company, New York, 1959, Translation of TEORIYA MATRITS by G AHTMAXEP.