Linear Algebra via Exterior Products

Subspace

A subspace of a vector space \(V\) is a subset

\(S \subset V\) such that \(S\) itself is a vector space. (i.e. it must be closed under addition, and scaling and thus also include \(0\))

Projector

A linear operator \(\hat{P} : V → V\) is called a projector if \(\hat{P} \hat{P} = \hat{P}\)

Two vector spaces are isomorphic if there exists a one-to-one linear map between them. The linear map is called the isomorphism.

• Any vector space \(V\) of dimension \(n\) is isomorphic to the space \(K^n\) of \(n\) -tuples.
• Vector spaces \(K^m\) and \(K^n\) are isomorphic only if they have equal dimension, \(m = n\) . The reason they are not isomorphic for \(m \neq n\) is that they have different numbers of vectors in a basis, while one-to-one linear maps must preserve linear independence and map a basis to a basis.

If \(V\) and \(W\) are two given vector spaces over a field \(K\), we define a new vector space \(V \oplus W\) as the space of pairs \((v,w)\) , where \(v \in V\) \(w \in W\).

Dual vectors = co-vectors = linear forms

Definition: A co-vector is a linear map \(V \to \mathbb K\) . The set of all co-vectors is the dual space to the vector space \(V\) . The zero covector is the linear function that maps all vectors into zero. This "space of all linear functions" is the space we denote by \(V^\ast\) . In our earlier notation, this space is the same as Hom\((V,\mathbb K)\) .

The hyperplane (i.e. subspace of codimension 1) annihilated by a covector \(f^\ast \in V^\ast\) is the set of all vector \(x \in V\) such that \(f^\ast(x) = 0\) .

The hyperplane annihilated by a nonzero covector \(\astrk f\) is a subspace of \(V\) of dimension \(N-1\) (where \(N \equiv \text{dim} V\))

The motivation is that we would like to define a product of vectors, \(u \otimes v\), which behaves as we expect a product to behave, e.g. \[(a + \lambda b) \otimes c = a \otimes c + \lambda b \otimes c, \forall \lambda \in \mathbb K, \forall a, b, c \in V\] and the same with respect to the second vector. This property is called bilinearity. It turns out to be impossible to define a nontrivial ( a "trivial" product would be \(a \otimes b = 0\) for all \(a,b\)) product of vectors in a general vector space, such that the result is again a vector in the same space. The solution is to define a product of vectors so that the resulting object is not a vector from \(V\) but an element of another space.

Definition: Suppose \(V\) and \(W\) are two vector spaces over a field \(\mathbb K\) ; then one defines a new vector space, which is called the tensor product of \(V\) and \(W\) and denoted by \(V \otimes W\). This is the space of expressions of the form

\begin{equation} \label{eqn:1.16} v_1 \otimes w_1 + ... + v_n \otimes w_n \tag{1.16} \end{equation}

where \(v_i \in V, w_i \in W\). The plus sign behaves as usual (commutative and associative). The symbol \(\otimes\) is a special separator symbol Further, we postulate that the following combinations are equal,

\begin{equation} \label{eqn:1.17} \lambda ( v \otimes w) = (\lambda v) \otimes w = v \otimes (\lambda w) \tag{1.17} \end{equation} \begin{equation} \label{eqn:1.18} (v_1 + v_2) \otimes w = v_1 \otimes w + v_2 \otimes w \tag{1.18} \end{equation} \begin{equation} \label{eqn:1.19} v \otimes ( w_1 =w_2) = v \otimes w_1 + v \otimes w_2 \tag{1.19} \end{equation}

for any vectors \(v,w, v_{1,2}, w_{1,2}\) and for any constant \(\lambda\). Once could say that the symbol \(\otimes\) behaves as a noncommutative product sign. This expression \(v \otimes w\) which is by definition an element of \(V \otimes W\), is called the tensor product of vectors \(v\) and \(w\).

The representation of a tensor \(A \in V \otimes W\) in the form \eqref{eqn:1.16} is not unique, i.e there may be many possible choice of the vectors \(v_j\) and \(w_j\) that give the same tensor \(A\) . For example,

\begin{equation} v_1 \otimes w_1+ v_2 \otimes w_2 = (v_1 - v_2) \otimes w_1 + v_2 \otimes (w_1 + w_2) \end{equation}

The tensor product of several spaces is defined similarly, e.g. \(U \otimes V \otimes W\) is the space of expressions of the form

\[u_1 \otimes v_1 \otimes w_1 + ... + u_n \otimes v_n \otimes w_n\]

Alternatively (and equivalently) one can define the space \(U \otimes V \otimes W\) as the tensor product of the spaces \(U \otimes V\) and \(W\)

(Also \((U \otimes V) \otimes W \cong U \otimes (V \otimes W)\) )

Definition: If we only work with one space \(V\) and if all other spaces are constructed out of \(V\) and \(V^\ast\) using the tensor product, then we only need the spaces of the form

\[ V\otimes ... \otimes V \otimes V^\ast \otimes .. \otimes V^\ast\]

Elements of such space are called tensor of rank \((m,n)\). For example, vectors \(v \in V\) have rank \((1,0)\) , covector \(f^\ast \in V^\ast\) have rank \((0,1)\). Scalars from \(\mathbb K\) have ran \((0,0)\)

Linear operators can be thought of as elements of the space \(V \otimes V^\ast\). This gives a convenient way to represent a linear operator by a coordinate-free formula. We will see that the space Hom\((V,W)\) of linear maps \(V \to W\) is canonically isomorphic to \(W \otimes V^\ast\).

• \(\text{End}V \cong V \otimes V^\ast\)
• In other words, linear operators are canonically (without choosing a basis) and uniquely mapped into tensors \(\in V \otimes V^\ast\)
• From now on, I will not use the map explicitly. Rather, I will simply not distinguish between the spaces \(\text{End}V\) and \(V \otimes V^\ast\). I will write things like \(v \otimes w^\ast \in \text{End}V\) or \(\hat{A} = x \otimes y^\ast\). The space implied in each case will be clear from the context.

Comes from considering the proeprties of areas and voumens in the framework of elementatry Euclidean geometry.

2D Oriented Area: The oriented area \(A(a,b)\) of a parallelogram spanned by the vectors \(a\) and \(b\) in the two-dimensional Euclidean space is an antisymmetric and bilinear function of the vectors:

\begin{align*} A(a,b) = -A(b,a) \\ A(\lambda a,b) = \lambda A(a,b) \\ A(a, b+c) = A(a,b) + A(a,c) \end{align*}

The oriented area is defined as

\(A(a,b) = \pm |a| . |b|. \sin\alpha\)

where the sign i positive when angle \(\alpha\) is measured from the vector \(a\) to \(b\) in the counterclockwise direction, and negative otherwise.

The antisymmetric property comes from the definition of \(A(a,b)\) while the bilinearity property can be proved from geometrical constructions.

Computation of oriented area is straighforward due to its algebraic properties. Say \(a\) and \(b\) are given through their components in standard basis \(\{e_1, e_2\}\). We assume, of course, that the vectors \(e_1\) adn \(e_2\) are orthogonal to each other and have unit length. Also, assume the right angle is measured from \(e_1\) to \(e_2\) in counterclockwise direction, so that \(A(e_1,e_2) = +1\). Then, (by using \(A(e_1,e_1) = 0\) and \(A(e_2,e_2) = 0\))

\begin{equation*} A(a,b) = A(\alpha_1 e_1 + \alpha_2 e_2, \beta_1 e_1 + \beta_2 e_2) = \alpha_1 \beta_2 - \alpha_2 \beta_1 \end{equation*}

At this point, could we also define something else, say a symmetric product instead of the exterior product? We could try to define a product say \(a \odot b\) such that \(a \odot b = 2 b \odot a\). But this won't work because, we would have

\begin{equation*} b \odot a = 2 a \odot b = 4 b \odot a \end{equation*}

so all the `\(\odot\)' products would have to vanish.

We can define a symmetric tensor product, \(\otimes_S\) with the property

\begin{equation*} a \otimes_S b = b \otimes_S a \end{equation*}

but it is impossible to define anything else in a similar fashion. (Grassmann has proved it)

The antisymmetric tensor product is the eignespace (within \(V \otimes V\)) of the exchange operator \(\hat{T}\) with the eigenvalue \(-1\). This operator has only eigenvalues \(\pm 1\) , so the only other possibility is to consider the eignespace with eignevalue \(+1\). This eignespace is spanned by symmetric tensors of the form \(u \otimes v+ v \otimes u\), and can be considered as the space of symmetric tensor products. We could write

\begin{equation*} a \otimes_s b = a \otimes b + b \otimes a \end{equation*}

and develop the properties of this product. However it turns out that the symmetric tensor product is much less useful for the purposes of linear algebra that the antisymmetric subspace.

Similar to tensors, n-vectors are not uniquely respresentable by linear combinations of exterior products. There can be alternative representation for the same $n$-vector.

Statement: Any $(N-1)$-vector in an $N$-dimensional space can be written as a single-term exterior product of the form \(a_1 \wedge ... \wedge a_{N-1}\)

This can be proved by induction.

A simplest example is a map

\begin{equation*} L_a : \omega \to a \wedge \omega \end{equation*}

mapping \(\wedge^k V \to \wedge^{k+1}V\); here the vector \(a\) is fixed. It is important to chect that \(L_a\) is a linear map between these spaces.

\begin{align} L_a (\omega + \lambda \omega') &= a \wedge (\omega + \lambda \omega') \\ &= a \wedge \omega + \lambda a \wedge \omega' \\ &= L_a(\omega) + \lambda L_a(\omega') \end{align}

Let us now fix a covector \(a^\ast\) . A covector maps from vector space to field of scalar. Now we will use covector to construct a map \(V \wedge V \to V\). Lets denote this map by \(l_{a^\ast}\).

It would be incorrect to define the map \(l_{a^\ast}\) by the formula \(l_{a^\ast}(v\wedge w) = a^\ast (v)w\) because such a definition does not respect the antisymmetry of the wedge product and thus violates the linearity condition (How? I get that it violates antisymmetric, but it doesn't violate linearity. No! it does, because the linearity must also hold when the representation of the exterior product changes i.e \(\omega = v \wedge w\) may also be expressed as \(\omega = (v - w) \wedge w\) but since \(l_\astar\) violates antisymmetry, and the different representations of \(\omega\) are obtained by using the anitsymmetry and linearity properties, it also violates linearity. (See proof of page 88))

\begin{align} l_{a^\ast} (w \wedge v) &\overset{!}{=} l_{a^\ast} ((-1) v \wedge w) \\ &= -l_{a^\ast}(v \wedge w) \neq a^\ast(v) w \end{align}

So we need to act with \(a^\ast\) on each of the vectors in a wedge product and make sure that the correct minus sign comes outside.

\begin{equation*} l_a^\ast (v \wedge w) \equiv a^\ast(v) w - a^\ast(w) v \end{equation*}

is an acceptable formula.

Let us now extent \(l_{a^\ast}\) to a map

\begin{equation*} l_{a^\ast} : \wedge^k V \to \wedge^{k-1} V \end{equation*}

defined as follows:

\begin{align} l_\astar v &\equiv \astar (v)\\ l_\astar (v \wedge \omega) &\equiv \astar(v) \omega - v \wedge l_\astar (\omega) \end{align}

This definition is inductive. The action of \(l_{a^\ast}\) on a sum of terns is defined by requiring linearity,

\begin{align} l_\astar (A + \lambda B) \equiv l_\astar(A) + \lambda l_\astar(B) \end{align}

We can convert this inductive definition to a more explicit formula.

\begin{equation*} l_\astar(v_1 \wedge ... \wedge v_k) \equiv \astar (v_1) v_2 \wedge ... \wedge v_k - \astar(v_2) v_1 \wedge v_3 \wedge ... \wedge v_k + ... + (-1)^{k-1} \astar(v_k) v_1 \wedge ... \wedge v_{k-1} \end{equation*}

This map is called the interior product or the insertion mapcar. The insertion map \(l_\astar \psi\) `inserts' the covector \(\astar\) into the tensor \(\psi \in \wedge^k V\) by acting with \(\astar\) on each of the vectors in the exterior product that makes up \(\psi\).

One powerful property of the exterior product is its close realtion to linear independence of sets of vectors. E.g. if \(u = \lambda v\) then \(u \wedge v = 0\).

Suppose \(\{v_1, ... , v_N\}\) is a given basis; we would like to compute its dual basis. For instance the covector \(v_1^\ast\) of the dual basis is a linear function such that

\begin{equation*} x = \sum_{i=1}^N x_i v_i; v_i^\ast(x) = x_i \end{equation*}

We start from the observation that tensor \(\omega \equiv v_1 \wedge ... \wedge v_N\) is nonzero. The exterior product \(x \wedge v_2 \wedge ... \wedge v_N\) is qual to zero if \(x_1 = 0\). This suggest that the exterior product of \(x\) with the $(N-1)$-vector is quite similar to the covector \(v_1^\ast\).

Lets introduce a special complement operation denoted by a star: \(*(v_1) \equiv v_2 \wedge ... \wedge v_N\) Then \(v_1^\ast(x) \omega = x \wedge *(v_1)\). This equation can be used to computing \(v_1^\ast\).

For \(v_2\), if we would like to have \(x_2 \omega = x \wedge *(v_2)\) we need to add an extra minus sign and define

\begin{equation*} *(v_2) \equiv -v_1 \wedge v_3 \wedge ... \wedge v_N \end{equation*}

It is clear that we can define the tensors \(*(v_i)\) for \(i=1,...,N\) in this way. The complement map \(*: V \to \wedge^{N-1} V\) satisfies \(v_j \wedge *(v_j) = \omega\) for each basis vector \(v_j\).

Remark: \(*(v_i)\) suggest that the complement operator is a some operation applied to \(v_i\) but it is not so. The complement of \(v_i\) depends on the entire basis and not merely on the single vector! Also, the property \(v_1 \wedge (v_1) = \omega\) is not sufficient to define the tensor \(*(v_1)\).

Remark: When the vector space is equipped with a scalar product, on usually chooses an orthonormal basis to define the complement map; then the complement map is call Hodge Star. It turns outh that the Hodge star is indpendent of the choice of the basis as long as the basis is orthonormal with respect to the given scalar product, and as long as the orientation of the basis is unchanged (i.e. as long as the tensor \(\omega\) doesn't change sign). In other words, Hodge star operation is invariant under orthogonal and orientation-preserving transformations of the basis;

is a procedure for computing the exterior product of $n$-vectors.

We have defined the rank of a map as the dimension of the image of the map, and we have seen that the rank is equal to the minimum number of tensor product terms needed to represent the map as a tensor. An analogous concept can be introduced for sets of vectors.

Definition: if \(S = \{v_1, ... ,v_n\}\) is a set of vectors (where \(n\) is not necessarily smaller than the dimension \(N\) of space), the rank of the set \(S\) is the dimension of the subspace spanned by the vectors in \(S\).

\begin{equation*} rank(S) = \dim Span S \end{equation*}

We will now show how to use the exterior product for computing the rank of a given (finite) set \(S\).

First, compute the tensor \(v_1 \wedge ... \wedge v_n\). If this tensor is non zero then the set S is linearly independent and the rank of \(S\) is equal to \(n\).

If rank is less than \(n\). First assusem that all \(v_j \neq 0\) (i.e. any zero vectors can be omitted without changing the rank of \(S\)). Then we compute \(v_1 \wedge v_2\) ; if the result is zero, we may omit \(v_2\) and try \(v_1 \wedge v_3\). If it is not 0, we try \(v_1 \wedge v_2 \wedge v_3\) and so on. Eventually we arrive at a subset \(\{v_{i_1}, ..., v_{i_k}\} \subset S\) such that \(v_{i_1} \wedge ... \wedge v_{i_k} \neq 0\) but \(v_{i_1} \wedge ... \wedge v_{i_k} \wedge v_j = 0\) for any other \(v_j\). Then the rank of \(S\) is equal to \(k\).

Let us choose a basis \(\{e_j\}\) in \(V\); then the dual basis \(\{e^\ast_j\}\) in \(V\) and the basis \(\{e_{k_1} \wedge ... \wedge e_{k_m} \}\) in \(\wedge^m V\) are fixed.

Then exterior product \(A \equiv u \wedge v\) is written in index notation as \(A^{ij} = u^iv^j - u^jv^i\). Not that the matrix \(A^{ij} = - A^{ji}\) is antisymmetric.

Another example: The 3-vector \(u\wedge v \wedge w\) can be expanded in the basis as v\begin{equation} u∧ v ∧ w = ∑_i,j,k=1^N B^ijke_i ∧ e_j ∧ e_k \end{equation}

And the components \(B^{ijk}\) would be given by

\begin{equation} B^{ijk} = u^i v^j w^k - u^i v^k w^j + ... \end{equation}

where every premutation of the set \((i,j,k)\) of indices enters with the sign corresponding to the parity of that premutation.

Remark: The `three-dimensional array' \(B^{ijk}\) is antisymmetici to any pair of indices. Such arrays are called totally antisymmetric.

The formula for the components \(B^{ijk}\) of \(u\wedge v \wedge w\) is not particularly convenient, so lets rewrite it in suitable form for generalization.

Let us first consider the exterior product of three vectors as a map \(\hat{E} : V \otimes V \otimes V \to \wedge^3 V\) This map is linear and can be represented, in the index notation as

\begin{equation*} (u \wedge v \wedge w)^{ijk} = E^{ijk}_{lmn} u^l v^m w^n \end{equation*}

where the array \(E^{ijk}_{lmn}\) is the component representation of the map \(E\).

By analogy, the map \(\hat{E} : V \otimes ... \otimes V \to \wedge^n V\) can be represent in the index notation by the array of components \(E^{i_1...i_n}_{j_1...j_n}\). This array is totally antisymmetric with respect to all the indices \(\{i_s\}\) and separately with respect to all \(\{j_s\}\).

Using this array, the exterior product of two general antisymmetric tensors, say \(\phi \in \wedge^m V\) and \(\psi \in \wedge^n V\) such that \(m+n \leq N\) can be represented in the index notation by

\begin{equation} (\phi \wedge \psi)^{i_1...i_{m+n}} = \frac 1 {m!n!} \sum_{(j_s,k_s)} E^{i_1...i_{m+n}}_{j_1..j_m k_1...k_n} \phi^{j_1...j_m}\psi^{k_1...k_n} \end{equation}

The combinatorial factor \(m!n!\) is needed to compensate from the \(m!\) equal terms arising from the summation over \((j_1,...,j_m)\) due to the fact that \(\phi^{j_1...j_m}\) is totally antisymmetric and similarly form the \(n!\) equal terms.

The general formula for \(E^{i_1...i_n}_{j_1...j_n}\) is that it is \((-1)^{|\sigma|}\) when \((i_1,...,i_n)\) is a permutation \(\sigma\) of \((j_1,...,j_n)\) and is \(0\) otherwise.

The Levi-Civita symbol is defined as a totally antisymmetric array with \(N\) indices, whose values are \(0\) or \(\pm 1\) according to the formula

\begin{equation} \epsilon^{i_1...i_N} = (-1)^{|\sigma|} \text{if $(i_1,...,i_n)$ is a permutation $\sigma$ of $(1,...,N)$; 0 otherwise} \end{equation}

Depending on convenicence, we may write \(\epsilon\) with upper or lower indices since \(\epsilon\) is just an array of numbers.

Notice that

\begin{equation*} \epsilon^{i_1...i_N} = E^{i_1...i_N}_{1...N} \end{equation*}

Lets consider the expression

\begin{equation} \label{eqn:2.11} \tilde{E}^{i_1...i_n}_{j1...j_n} \equiv \sum_{k1,...,k_{N-n}} \epsilon^{i_1...i_nk_1...k_{N-n}} \epsilon_{j_1...j_nk_1...k_{N-n}} \tag{eqn:2.11} \end{equation}

This expression has \(2n\) free indices and is totally antisymmetric in these free indices (since \(\epsilon\) is totally antisymmetic in all indices).

Statement: The exterior product operator \(E^{i_1...i_n}_{j_1...j_n}\) is expressed through Levi-Civita symbol as

\begin{equation} E^{i_1...i_n}_{j_1...j_n} = \frac 1 {(N-n)!} \tilde{E}^{i_1...i_n}_{j_1...j_n} \end{equation}

Proof: Let us compare the values of \(E^{i_1...i_n}_{j_1...j_n}\) and \(\tilde{E}^{i_1...i_n}_{j_1...j_n}\) where the indices have some fixed values. There are two cases either the set \((i_1,...,i_n)\) is a permutation of the set \((j_1,...,j_n)\); in that case we may denote this peruation by \(\sigma\); or it is not a permutation.

Considering the case when a permuation \(\sigma\) brings \((j_1,...,j_n)\) into \((i_1,...,i_n)\), we find that the symbols \(\epsilon\) in \eqref{eqn:2.11} will be nonzero only if the indices \((k_1,...,k_{N-n})\) are a permuatation of the complemnt of the set \((i_1,...,i_n)\). There are \((N-n)!\) such permutations, each contributing the same value to the sum in \eqref{eqn:2.11}. Hence we may write the sum as

\begin{align} \tilde{E}^{i_1...i_n}_{j1...j_n} &\equiv (N-n)! \epsilon^{i_1...i_nk_1...k_{N-n}} \epsilon_{j_1...j_nk_1...k_{N-n}} \text{no implicit sums!} \\ &= (N-n)! (-1)^{|\sigma|} \end{align}

where the indices \({k_s}\) are chosen such that the values of \(\epsilon\) are nonzero. Thus the required formula is valid for first case. In case the permuatation \(\sigma\) doesn't exist, we note that

\begin{equation} \tilde{E}^{i_1...i_n}_{j_1...j_n} = 0 \end{equation}

And therefore \(E\) and \(\tilde{E}\) both will be equal to zero.

Note that the formula for the top exterior power \((n=N)\) is simple and involves no summation

\begin{equation*} E^{i_1...i_N}_{j_1...j_N} = \epsilon^{i_1...i_N}\epsilon_{j_1...j_N} \end{equation*}

Remark: As a rule, a summation of the Levi-Civita symbol \(\epsilon\) with any anitsymmetric tensor (e.g. another \(\epsilon\)) gives rise to a combinatorial factor \(n!\) when the summation goes over \(n\) indices.

The formalism of exterior algebra is used e.g. in physical theories of quantum fermionic fields and supersymmetry.

Definition: An algebra is a vector space with a distributive multiplication.

In other words, \(A\) is an algebra if its a vector space over a field \(\mathbb K\) and if for any \(a,b \in A\) their product \(ab \in A\) is defined, such taht the product is distributive over addition and \(\lambda (ab) = (\lambda a)b = a (\lambda b)\) for \(\lambda \in \mathbb K\).

Examples of Algerbas:

1. All \(N \times N\) matrices with coefficients from \(mathbb K\) are $N²$-dimensional algebra.
2. The filed \(\mathbb K\) is a one-dimensional algerba over itself. This algebra is commutative.

Statement: If \(\omega \in \wedge^m V\) then we can define the map \(L_{\omega}: \wedge^k V \to \wedge^{k+m} V\) by the formula

\begin{equation*} L_{\omega} (v_1 \wedge ... \wedge v_k) = \omega \wedge v_1 \wedge ... \wedge v_k \end{equation*}

For elements of \(\wedge^0V \equiv \mathbb K\) we define \(L_\lambda \omega \equiv \lambda \omega\) and also, \(L_\omega \lambda \equiv \lambda \omega\) for \(\lambda \in \mathbb K\). Then the map \(L_\omega\) is linear for any \(\omega \in \wedge^m V\), \(0 \leq m \leq N\)

Definition: The exterior algebra (also called the Grassmann algebra) based on a vector space \(V\) is the space \(\wedge V\) defind as the direct sum,

\begin{equation*} \wedge V \equiv \mathbb K \oplus V \oplus \wedge^2 V \oplus ... \oplus \wedge^N V \end{equation*}

with the multiplication defined by the map \(L\), which is exteded to the whole of \(\wedge V\) by linearity.

Definition D0: The determinant of a square \(N \times N\) matrix \(A_{ij}\) is the number

\begin{equation*} \label{eqn:3.1} det(A_{ij}) \equiv \sum_{\sigma} (-1)^{|\sigma|} A_{\sigma(1)1} ... A_{\sigma(N)N} \tag{eqn:3.1} \end{equation*}

where the summation goes over all permutations \(\sigma : (1, ..., N) \to (k_1, ..., k_N)\), and the parity function \(|\sigma|\) is equal to \(0\) if the permutation \(\sigma\) is even and to \(1\) if it is odd.

If a \(2 \times 2\) matrix, there are only two permutations of the set \((1,2)\), namely \((1,2)\) and \((2,1)\) so, the explicit formula for determinant is:

\begin{equation*} a_{11}a_{22} - a_{21}a_{12} \end{equation*}

Note that an \(N \times N\) matrix has \(N!\) terms in this type of formula.

The space \(\wedge^N V\) is one dimensional. Therefore, all nonzero tensors from \(\wedge^N V\) are proportional to each other.

Example: In two dimensional space \(V\) lets choose a basis \({e_1, e_2}\) and cosider two arbitrary vectors \(v_1\) and \(v_2\). \(v_1 = a_{11} e_1 + a_{12} e_2\), Then

\begin{equation*} v_1 \wedge v_2 = (a_{11} a_{22} - a_{12} a_{21}) e_1 \wedge e_2 \end{equation*}

We may observe that firstly, the 2-vector \(v_1 \wedge v_2\) is proportional to \(e_1 \wedge e_2\) and secondly, the proportionality coefficient is equal to the determinant of matrix \(A_{ij}\)

Lemma: In an $N$-dimensional space

1. The volume of parallelepiped spanned by \(\{\lambda v_1, v_2, ..., v_N\}\) is \(\lambda\) times greater than that of \(\{v_1,v_2,...,v_N\}\) .
2. Two parallelepipeds spanned by the sets of vectors \(\{v_1,v_2,...,v_N\}\) and \(\{v_1+\lambda v_2, v_2, ..., v_N\}\) have equal volume.

Statement: Consider an $N$-dimensional space \(V\), Then:

1. Two parallelepipeds spanned by the sets of vectors \(\{u_1, ..., u_n\}\) and \(\{v_1,v_2,...,v_N\}\) have equal volumes if and only if the corresponding tensors from \(\wedge^N V\) are equal upto a sign,

\begin{equation*} u_1 \wedge ... \wedge u_N = \pm v_1 \wedge ... \wedge v_N \end{equation*}

1. If \(u_1 \wedge ... \wedge u_N = \lambda v_1 \wedge ... \wedge v_N\) , where \(\lambda \in \mathbb K\) is a number, \(\lambda \neq 0\), then the volumes of the two parallelepipeds differ by a factor of \(|\lambda|\).

Proof of Statement 1: We can transform from the first basis \(\{u_1, ..., u_n\}\) into the second basis \(\{v_1, ... , v_N\}\) (upto ordering of vectors) by a sequence of transformations of two types: either we multiply one of the vectors \(u_j\) by a number \(\lambda\) or we add \(\lambda u_j\) to another vector \(u_k\). Using above lemmas and properties of exterior product we can show that these transformation change the volume in the same way as the exterior product \(u_1 \wedge ... \wedge u_N\) changes.

A tensor such as \(v_1 \wedge ... \wedge v_n\) can be used to determine the numerical value of the volume only if we can compare it to another given tensor, \(u_1 \wedge ... \wedge u_N\), which by assumption corresponds to a prallelepiped of unit volume. A choice of a reference tensor \(u_1 \wedge ... \wedge u_N\) can be made, for instance, if we are given a basis in \(V\). Thus there is no natural map from \(\wedge^N V\) to \(\mathbb K\). In other words the sapce \(\wedge^N V\) is not canonically isomorphic to the space \(\mathbb K\).

Remark: When a scalar product is defined in \(V\), there is a preffered choice of basis, namely an orthonormal basis \(\{e_j\}\) such that \( = \delta_{ij}\)

Let \(\hat{A} \in \text{End}V\) be a linear operator. Consider this action on tesnors from the sapce \(\wedge^N V\) defiend in the following way, \(v_1 \wedge ... \wedge v_N \to \hat{A} v_1 \wedge ... \wedge \hat{A} v_N\). I denote this operation by \(\wedge^N \hat{A}^N\).

We just defined \(\wedge^N \hat{A}^N\) on single term products; the action of \(\wedge^N \hat{A}^N\) on linear combinations of such products is obtained by requiring linearity.

Homework: Verify \(\wedge^N \hat{A}^N\) is a linear map.

Since, \(\wedge^N V\) is one-dimensional. So, \(\wedge^N \hat{A}^N\) being a linear operator in a one-dimensional space, must act simply as multiplication by a number. Thus we can write

\begin{equation*} \wedge^N \hat{A}^N = \alpha \hat{1}_{\wedge^N V} \end{equation*}

where \(\alpha \in \mathbb K\) is a number which is somehow associated with the oprator \(\hat{A}\). This number is actually equal to the determinant of the operator \(\hat A\) as given by Definition D0.

Definition D2: The determinant \(det{\hat A}\) of a linear transformation \(\hat A\) is the nuber by with the oriented volume of any parallelepiped grows after the transformation.

Statement 3: If \(\{e_j\}\) is any basis in \(V\), \(\{e_j^\ast\}\) is the dual basis, and a linear operator \(\hat A\) is represented by a tensor,

\begin{equation*} \hat A = \sum_{j,k=1}^N A_{jk} e_j \otimes e_k^\ast \end{equation*}

then the determinant of \(\hat A\) is given by the formula \eqref{eqn:3.1}

Proof: The action of \(\wedge^N \hat{A}^N\) on the basis element \(e_1 \wedge ... \wedge e_N \in \wedge^NV\) is

\begin{align} &\wedge^N \hat{A}^N (e_1 \wedge ... \wedge e_N) \\ &= \hat A e_1 \wedge ... \wedge \hat A e_N \\ &= \big(\sum_{j_1 = 1}^N A_{j_1 1} e_{j_1}\big) \wedge ... \wedge \big( \sum_{j_N = 1}^N A_{j_N N}e_{j_N} \big) \\ &= \sum_{j_1} ... \sum_{j_N}(A_{j_1 1} ... A_{j_N N}) e_{j_1} \wedge ... \wedge e_{j_N} \end{align}

In the last sum, the only nonzero terms are those in which the indices \(j_1, ... , j_N\) do not repeat; in other words, \((j_1,...,j_N)\) is a permutation of the set \((1,...,N)\). From the antisymmetry of the exterior product and the definition of the parity of the permutation, we can express above expression as in \eqref{eqn:3.1}