Two of the most important invariants of an operator on a finite-dimensional vector space, its trace and its determinant, were defined through matrices (Definitions 5.3.11 and 5.4.20), and their independence of the basis needed a computation. Tensor products and exterior powers produce both of them without choosing a basis. Throughout this section is finite-dimensional.
9.11.1 Trace as Contraction
Under the isomorphism of Proposition 9.8.10, the trace becomes a natural pairing.
Definition 9.11.1 (Contraction). The contraction is the linear functional it exists by the universal property, since is bilinear.
Proposition 9.11.2 (Trace as contraction). Let , and let be the tensor corresponding to under Proposition 9.8.10. Then In particular, the trace of the rank-one map is .
Proof. Choose a basis of , with dual basis , and let be the matrix of , so . The tensor corresponds to the map sending to and the other basis vectors to , whose matrix has a single , in row and column . Hence For the rank-one map, and . ∎
Note. Since involves no basis, this gives a basis-free definition of the trace: it is the composite . The formula is what the contraction looks like in coordinates, and the invariance of the trace under change of basis becomes automatic.
9.11.2 The Determinant via the Top Exterior Power
Let . An operator on induces an operator on the one-dimensional space (Proposition 9.10.6 and Corollary 9.10.9). Every linear map from a one-dimensional space to itself is multiplication by a scalar: if spans the space and , then . For , that scalar is the determinant.
Theorem 9.11.3 (The determinant via ). Let be a vector space of dimension and . Then is multiplication by : where is the determinant of the matrix of in any basis (Definition 5.4.20). More generally, if is a basis of and for , then
Proof. Fix the basis . Since is a basis of , each is a unique multiple , and we regard as a function of the matrix whose th column is the coordinate vector of . It is multilinear in the columns, because the wedge product is multilinear and taking the coefficient of is linear; it is alternating in the columns, because a wedge product with two equal factors vanishes; and , since then . By Theorem 5.4.4(b), . This proves the second statement.
Applying it to , whose coordinate vectors are the columns of the matrix of , gives . As spans , is multiplication by , which is . ∎
Note. The Leibniz formula can also be read off directly. With , multilinearity gives A term with a repeated index vanishes. The others have for a permutation , and by Lemma 9.10.2(ii). Hence which is the Leibniz formula of Definition 5.4.3.
Note. We could have taken Theorem 9.11.3 as the definition: is the scalar by which acts on the determinant line . This definition needs no basis, no permutations and no cofactor expansions, so the independence of from the basis is manifest. It also has a geometric meaning: is where signed volumes live, and is the factor by which scales them (see the end of this section).
Corollary 9.11.4 (Properties of the determinant). Let , and .
- .
- .
- is invertible if and only if .
- .
Proof. (i) (Proposition 9.10.6), and multiplication by followed by multiplication by is multiplication by .
(ii) is the identity of .
(iii) Let be a basis of . Then is invertible if and only if is a basis of , that is, if and only if these vectors are linearly independent; by Proposition 9.10.10, this happens if and only if is nonzero.
(iv) by multilinearity. ∎
Applied to the operators of given by two matrices and , part (i) gives , which is Theorem 5.4.7, now without any computation; part (iii) is Theorem 5.4.8. Neither argument is circular: the proof of Theorem 9.11.3 used only the characterization of the determinant by its columns (Theorem 5.4.4), which does not depend on multiplicativity.
9.11.3 Exterior Powers of an Operator
The coefficients of a wedge product of fewer than vectors are determinants as well.
Lemma 9.11.5 (Coordinates of a wedge product). Let be a basis of , let , and let with , so that is an matrix. For an increasing tuple , let be the matrix formed by the rows of . Then
Proof. Let send to , and let be the standard basis of . The map sends to , which is if and otherwise (if , some is not among the , and ). So extracts the coefficient of . On the other hand, the vectors are the columns of , so by Theorem 9.11.3 in , ∎
Combined with Proposition 9.10.10, the lemma says that the columns of an matrix are linearly independent if and only if some minor of the matrix is nonzero.
Proposition 9.11.6 (Traces of exterior powers and the characteristic polynomial). Let , and let have matrix in a basis . For an increasing -tuple , let be the submatrix of with rows and columns ; its determinant is a principal minor of .
- , the sum of the principal minors of .
- The characteristic polynomial of (Definition 7.4.7) is
- If is upper triangular with diagonal entries , then , the th elementary symmetric polynomial in .
Proof. (i) The diagonal entry of the matrix of in the basis at position is the coefficient of in . The coordinate vectors of are the columns of , and rows of these columns form ; so the coefficient is by Lemma 9.11.5. Summing over gives the trace.
(ii) The determinant is given by the Leibniz formula, so it is the same polynomial whether we regard as a matrix with entries in or in the field of rational functions (Example 5.4.25). We work over , in with standard basis . Let be the th column of ; the th column of is . By Theorem 9.11.3 and multilinearity, where for and for . Fix , and let be the increasing tuple listing the complement of , with entries. In we may replace each () by , because the rest of is a combination of the with , which already occur as factors. Let list the elements of followed by . Reordering the factors according to multiplies both and by (Lemma 9.10.2(ii)). After reordering, the wedge product is by Lemma 9.11.5: the coordinates of vanish outside the rows in , and in those rows they form . Hence , and by (i). (For the inner sum is , and has trace .)
(iii) Each is upper triangular with diagonal entries , so its determinant is their product (Proposition 5.4.10); apply (i). ∎
When is algebraically closed, Theorem 7.5.2 provides a basis in which is upper triangular; then is upper triangular too, and , so the are the eigenvalues of , each repeated according to its algebraic multiplicity, and (iii) says that is the th elementary symmetric polynomial in the eigenvalues. The cases and are and .
Let . Over , think of the wedge product as the signed volume of the parallelepiped . It behaves as a signed volume should: it is multilinear and alternating, and it vanishes exactly when the parallelepiped is flat, that is, when the are dependent (Proposition 9.10.10). But it lives in the line , and turning it into a number requires an isomorphism .
Definition 9.11.7 (Volume form). Let . A volume form on is a nonzero alternating -linear form . By Proposition 9.10.5, volume forms correspond to the nonzero elements of , that is, to the isomorphisms , via .
Since is one-dimensional, any two volume forms differ by a nonzero scalar factor. A vector space has no preferred volume form until we give it more structure, such as an inner product.
Proposition 9.11.8 (Volume forms and determinants). Let be a volume form on .
- if and only if is a basis of .
- for every .
- If is a basis with , then , where the th column of is the coordinate vector of in this basis.
Proof. Write with an isomorphism. (i) follows from Proposition 9.10.10, since independent vectors in form a basis. (ii) and (iii) follow by applying to the two formulas of Theorem 9.11.3; in (iii), . ∎
So is the factor by which multiplies all signed volumes, whichever volume form is used to measure them. Volume forms can also be written as wedge products of linear functionals.
Proposition 9.11.9 (Exterior powers of the dual space). For there is an isomorphism under which corresponds to the functional If is a basis of , it sends to the dual basis of the basis of . In particular, the alternating -linear forms on are the elements of .
Proof. For fixed , the function is multilinear and alternating, because enters only the th column of the matrix and the determinant is multilinear and alternating in the columns (Theorem 5.4.4(a)); so it defines an element of . For fixed , the value is multilinear and alternating in the , because enters only the th row and the determinant is also multilinear and alternating in the rows (Proposition 5.4.6). Since is determined by its values on the spanning set of wedge products, is multilinear and alternating, and it induces a linear map . For increasing tuples and , the value of on is . This is when ; when , some is not among the , the th row is zero, and the determinant is . So the basis of is sent to the dual basis of , and the map is an isomorphism. The last statement follows from Proposition 9.10.5. ∎
Example 9.11.10 (The volume form of a real inner product space). Let be a real inner product space (Definition 9.1.11) with an orthonormal basis . The volume form satisfies : the unit cube has volume . By Proposition 9.11.8(iii), is the determinant of the matrix of coordinates of the .
Let be another orthonormal basis, with . Then , so is orthogonal, and by Proposition 5.4.6 and Theorem 5.4.7. Hence , and the volume form built from is . So the inner product determines the volume form up to sign. Reordering an orthonormal basis can already change the sign: the basis gives . Choosing the sign is choosing an orientation.
Definition 9.11.11 (Orientation). Let be a real vector space of dimension . Two ordered bases and of have the same orientation if with . An orientation of is an equivalence class of ordered bases under this relation, and the bases in the chosen class are called positively oriented.
Proposition 9.11.12 (Orientations). Let be a real vector space of dimension .
- Having the same orientation is an equivalence relation with exactly two classes. The constant of Definition 9.11.11 is , where .
- An orientation amounts to a choice of one of the two half-lines of , or of a volume form up to multiplication by a positive scalar.
- An invertible sends positively oriented bases to positively oriented bases if and only if .
Proof. (i) The formula is Theorem 9.11.3. The relation is reflexive (), symmetric ( becomes ) and transitive (the constants multiply), since inverses and products of positive numbers are positive. Fix a basis . For every basis , the constant is nonzero (Proposition 9.10.10), so it is positive or negative, and both signs occur, because replacing by changes the sign of . So there are exactly two classes.
(ii) The class of is determined by the half-line of containing . A volume form determines the class of the bases with , and and determine the same class if and only if .
(iii) . ∎
Example 9.11.13 (Orientations of the plane). In , the bases and have opposite orientations, since . Rotations have determinant and preserve orientation; reflections have determinant and reverse it, which matches the description of in Proposition 9.5.16.
Exercise 9.11.14 (Tensor products and multilinear algebra exercises).
Let be a two-dimensional vector space over with basis .
- Write down a basis of , and find bases of the subspaces and of symmetric and alternating tensors when . Verify that , and find the alternating tensor corresponding to under Proposition 9.10.12.
- Suppose . Show that is three-dimensional, whereas .
Starting from the expression for obtained from wedge products in the note following Theorem 9.11.3, prove that for every matrix .
Let .
- Show that is an isomorphism ; it involves no choices.
- Deduce that each nonzero determines an isomorphism , where when . How does change when is replaced by ?
Let with the standard inner product and standard basis , and let . Compose from (3) with the isomorphism given by the inner product (, where ). Show that the resulting isomorphism sends to the cross product .
Show solution
(1) (a) A basis of is , so . The group has one nontrivial element, acting by the swap . For we have , so is symmetric if and only if . A basis of is and . The tensor is alternating if and only if , that is, , and . When this means and , so has basis and dimension . Indeed , as Corollary 9.10.13 predicts. Under Proposition 9.10.12, corresponds to .
(b) In characteristic we have , so the condition is the condition : , which is three-dimensional by the computation in (a) (that computation did not use the characteristic). But by Proposition 9.10.8, which holds in every characteristic. So the subspace of alternating tensors is not a model of in characteristic . The subspace of symmetric tensors fails as a model of as well, although its dimension is right: the quotient map sends to , so it is not injective on .
(2) By the note following Theorem 9.11.3, The entry of is , so For fixed , put . As runs over , so does , and ; hence . Since and is a bijection of ,
(3) (a) The product from to is bilinear (Definition 9.10.17). So for each the map is linear, and is a linear map . No choices were made. To see that is injective, choose a basis . The increasing -tuples are the tuples that omit one index . Write . For , , because occurs in ; and , by moving past factors. So , and forces every . Both spaces have dimension (Proposition 9.10.8), so is an isomorphism. Finally by Proposition 9.8.10.
(b) A nonzero gives an isomorphism , , and hence . Composing with (a) gives . Since , we get . So the isomorphism depends on the choice of ; the isomorphism of (a) does not.
(4) Let and . Using and , For we have , so , which corresponds to because . Similarly , so ; and (moving past two factors), so . Therefore The construction needs , that is, , which for holds only when . This is why it produces a cross product only in dimension . (A bilinear cross product also exists on , but it comes from a different construction.)