S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 5: Multilinear Algebra / From my book

B5.4 Determinants

The determinant of a square matrix is a single scalar that detects whether the matrix is invertible, and that measures how the corresponding linear map scales volumes. We define it by an explicit formula, the Leibniz formula, and characterize it as the unique function of the columns of a matrix with three simple properties. Most of its other properties follow from this characterization. Throughout this section .

5.4.1 Alternating Multilinear Forms

Definition 5.4.1 (Multilinear and alternating forms). Let be a vector space over and . A function is multilinear if it is linear in each argument when the others are held fixed: for every position. It is alternating if whenever for some .

We use the sign homomorphism of Chapter 3 (Definition 3.4.10). We need three facts about it from there: is a group homomorphism, for every transposition , and every permutation is a product of transpositions. Recall that permutations compose right to left: applies first.

Lemma 5.4.2 (Properties of alternating forms). Let be multilinear and alternating, and .

  1. Swapping two arguments changes the sign of .
  2. For every , .
  3. If are linearly dependent, then .
  4. Adding a multiple of one argument to a different argument does not change the value of .

Proof. (a) Put in positions and , and expand by multilinearity in both positions (we display only these two): The first and last terms vanish because is alternating, so .

(b) Write as a product of transpositions; we induct on . For , . If , write with a transposition and a product of transpositions. Put ; then is with the entries in positions and swapped, and . By induction applied to the vectors , and by (a), and .

(c) Some is a linear combination of the others, (Proposition 4.6.5). By linearity in the -th argument, where in the -th term occupies position as well as position . Each term vanishes because is alternating.

(d) Replacing by () changes by , with in positions and , which is . ∎

Note. Property (a) is often taken as the definition of “alternating”. If the two versions agree: if swapping two arguments changes the sign, then , so . In characteristic , where , property (a) says nothing, and the definition we have adopted is the right one.

5.4.2 The Determinant

We regard a matrix as the list of its columns , and functions on as functions of vectors in . Accordingly, “multilinear and alternating in the columns” means multilinear and alternating as a function on .

Definition 5.4.3 (Determinant). The determinant of is This is the Leibniz formula. Each term picks one entry from every column, from pairwise different rows.

Theorem 5.4.4 (Characterization of the determinant).

  1. The determinant is multilinear and alternating in the columns, and .
  2. If is multilinear and alternating in the columns, then for all .

In particular, is the unique function that is multilinear and alternating in the columns and takes the value at .

Proof. (b) Let be multilinear and alternating in the columns, and write each column as . Expanding by linearity in each column in turn, The terms are indexed by the functions , . If is not injective, two arguments of coincide and the term vanishes. An injective map from to itself is a permutation , with , and by Lemma 5.4.2(b), . Hence

(a) Multilinearity: each product contains exactly one entry of the -th column, so it is a linear function of when the other columns are fixed; hence so is the sum.

Value at : the entries of are , so is if and otherwise. Thus .

Alternating: suppose columns of are equal, and let . For , compare the terms for and for . For we have , while and , because columns and are equal. So the two products are equal, while . Since is a homomorphism, is a bijection from the even permutations to the odd ones, so

The final statement combines (a) and (b). ∎

Example 5.4.5.

  • For : .
  • For : , so
  • For : the even permutations are , , and the odd ones are , , . For instance has , , and contributes . Altogether,

In general the Leibniz formula has terms, so it is a poor way to compute large determinants; the properties below give much better methods.

Note. Over , is the area (for ) or volume (for ) of the parallelogram or parallelepiped spanned by the columns of , and the sign records orientation. Multilinearity and the alternating property are exactly how a signed volume should behave: scaling one edge scales the volume, and a degenerate parallelepiped, with two equal edges, has volume . Chapter 9 develops this point of view with volume forms.

5.4.3 Properties of the Determinant

Proposition 5.4.6 (Determinant of the transpose). For every , .

Proof. By the Leibniz formula applied to , whose entry is , Reindex each product by , so : then . Also , and is a bijection of . Hence ∎

Since the rows of are the columns of , everything we know about columns holds for rows as well: is multilinear and alternating in the rows, swapping two rows changes the sign, adding a multiple of one row to another row does not change the determinant, and is the unique function with these properties and . By multilinearity in all columns, for .

Theorem 5.4.7 (Multiplicativity). For all ,

Proof. Fix and define . The -th column of is , and is linear; since is multilinear in the columns, is multilinear in the columns of . If two columns of are equal, so are the corresponding columns of , and . By Theorem 5.4.4(b), . ∎

Theorem 5.4.8 (Invertibility criterion). A matrix is invertible if and only if . In that case .

Proof. If is invertible, then by multiplicativity, so and . If is not invertible, its columns do not form a basis of (Proposition 5.2.14). Then they are linearly dependent, since independent vectors in would form a basis (Corollary 4.7.8). Hence by Lemma 5.4.2(c). ∎

Corollary 5.4.9 (One-sided inverses). If and , then and are invertible and ; in particular .

Proof. We have , so and is invertible. Then . ∎

Thus restricts to a group homomorphism . Its kernel is the special linear group .

Proposition 5.4.10 (Triangular matrices). If is upper triangular ( for ) or lower triangular ( for ), then is the product of the diagonal entries.

Proof. Let be upper triangular. The term of the Leibniz formula for is nonzero only if for every . Such a is the identity: forces , and if for all , then and force . So only the term survives, and it is . A lower triangular matrix is the transpose of an upper triangular one with the same diagonal. ∎

Proposition 5.4.11 (Block triangular matrices). Let , , and let be and be . Then

Proof. Let be the first matrix. Its first columns are , which depend linearly on the columns of , and two equal columns of give two equal columns of . So, for fixed and , the function is multilinear and alternating in the columns of , and Theorem 5.4.4(b) gives The -th column of is . Subtracting from it times the -th column , for , turns it into , and by Lemma 5.4.2(d) this does not change the determinant. So . This is multilinear and alternating in the columns of , so by Theorem 5.4.4(b) again it equals . The second equality follows by taking transposes. ∎

By induction, the determinant of a block triangular matrix with square diagonal blocks is .

Example 5.4.12. To compute a determinant efficiently, reduce to a triangular matrix with row operations, keeping track of their effect. For subtract times row 1 from row 2 and times row 1 from row 3, then times the new row 2 from the new row 3: None of these operations changes the determinant. The Leibniz formula for gives the same value: .

5.4.4 Cofactor Expansion and the Adjugate

For and , let be the matrix obtained from by deleting row and column .

Lemma 5.4.13 (Determinants with a standard basis column). Let , and let be the matrix obtained from by replacing its -th column with . Then .

Proof. Move the -th column of to the first position by swapping it successively with columns ; this is swaps of adjacent columns, and the other columns keep their order. Then move the -th row to the first position by swaps of adjacent rows. Each swap multiplies the determinant by . The resulting matrix has first column , and deleting its first row and column leaves , since we deleted row and column of and kept the others in order: By Proposition 5.4.11 its determinant is . Hence ∎

Theorem 5.4.14 (Cofactor expansion). Let and .

  1. For each column index , (expansion along column ).
  2. For each row index , (expansion along row ).
  3. For and for , respectively,

Proof. (a) The -th column of is . By linearity of in the -th column and Lemma 5.4.13, .

(b) Apply (a) to , expanding along its -th column, and use and Proposition 5.4.6.

(c) Let be the matrix obtained from by replacing its -th column with its -th column. Then has two equal columns, so , and because column is deleted. Expanding along column by (a) gives the first identity. The second follows in the same way from (b), replacing row by row . ∎

The scalar is the cofactor of . The signs form a checkerboard pattern starting with in the top left corner.

Definition 5.4.15 (Adjugate). The adjugate of , , is the matrix with entries that is, is the transpose of the matrix of cofactors. For we set .

Theorem 5.4.16 (The adjugate identity). For every ,

Proof. For this reads . Let . The entry of is which is if by Theorem 5.4.14(a), and if by Theorem 5.4.14(c). Similarly, the entry of is , which is if by (b) and otherwise by (c). ∎

Corollary 5.4.17 (Formula for the inverse). If , then . In particular, for ,

Proposition 5.4.18 (Cramer’s rule). Let with , and . The unique solution of is given by where is the matrix obtained from by replacing its -th column with .

Proof. The matrix is invertible, so is the unique solution. Then . By linearity of in the -th column, For the matrix in the -th term has two equal columns, so only the term survives, and . ∎

5.4.5 The Determinant of an Operator

Proposition 5.4.19 (Determinant is a similarity invariant). If , then .

Proof. If , then by multiplicativity ∎

Definition 5.4.20 (Determinant of an operator). Let be finite-dimensional. The determinant of is for any basis of . It does not depend on , because the matrices of in two bases are similar (Corollary 5.3.4).

Proposition 5.4.21 (Properties of the determinant of an operator). Let be finite-dimensional of dimension , and , . Then:

  1. and ;
  2. is invertible if and only if , and then ;
  3. .

Proof. Fix a basis. By Corollary 5.2.10, and , so (a) follows from Theorem 5.4.7. By Proposition 5.2.13, is invertible if and only if is, so (b) follows from Theorem 5.4.8. Finally and . ∎

Note. In Chapter 9 we see the determinant of an operator without choosing a basis: acts on the top exterior power as multiplication by (Theorem 9.11.3).

Example 5.4.22. Trace and determinant do not determine the similarity class of a matrix. Let over . Both have trace and determinant . But for every invertible , so the similarity class of is , and is not similar to .

Note. Finer similarity invariants come from eigenvalues and eigenvectors, studied in Chapter 7. In the example, and have the same eigenvalue , but the vectors with form all of , while those with form only the line spanned by . Over an algebraically closed field, the Jordan normal form of Chapter 7 (Theorem 7.8.4) classifies matrices up to similarity completely.

5.4.6 Permutation Matrices and the Sign

Definition 5.4.23 (Permutation matrix). For , let be the linear map with for all (Proposition 5.1.2). Its matrix in the standard basis is the permutation matrix , whose -th column is ; that is, if and otherwise.

Proposition 5.4.24 (Permutation matrices). For :

  1. and , so is an injective group homomorphism ;
  2. .

Proof. (a) For each , , so , and taking matrices gives (Corollary 5.2.10). Clearly , so and each is invertible. If , then for all , so .

(b) The columns of are , so by Lemma 5.4.2(b) applied to , . ∎

Note. Over one can see without computation that : the permutation has some finite order , so by (a), hence , and the only real roots of unity are . Part (b) says which sign occurs.

5.4.7 Determinants over Commutative Rings

The Leibniz formula makes sense for square matrices with entries in any commutative ring (Definition 4.1.1), and so do the definitions of the matrix product, the transpose and the adjugate. The proofs of Lemma 5.4.2(a), (b) and (d), Theorem 5.4.4, Proposition 5.4.6, Theorem 5.4.7, Propositions 5.4.10 and 5.4.11, Lemma 5.4.13, Theorem 5.4.14 and Theorem 5.4.16 use only the ring axioms and commutativity: no division and no linear independence. (Associativity of matrix multiplication over is a direct computation: the entry of both and is .) So all these results hold verbatim for matrices over , where “multilinear” means linear in each column with coefficients in . In place of Theorem 5.4.8, a matrix with entries in has an inverse with entries in if and only if is a unit of : if , then is an inverse by the adjugate identity, and conversely .

Example 5.4.25. The case is used in Chapter 7. For , the matrix has entries in , so is a polynomial in , and holds in . Moreover, if is a ring homomorphism and denotes the matrix obtained by applying to every entry of , then , because is a sum of products of entries. Taking for evaluation at shows that the value of the polynomial at is . Taking for complex conjugation shows for complex matrices, and hence for the conjugate transpose .

Question 5.4.26.

  1. Let be the linear map with matrix with respect to the standard basis. Find the matrix of with respect to the basis .
  2. Prove that is similar to a diagonal matrix if and only if has a basis with for some scalars . (Such vectors are called eigenvectors of ; see Chapter 7.)
  3. We know . Show that in general .
  4. Compute the determinant and, using the adjugate, the inverse of
Show solution

(a) The change-of-basis matrix from to the standard basis has the vectors of as its columns: using the formula for the inverse of a matrix. By Corollary 5.3.4, the matrix of with respect to is . First, and then As a check, and , as similarity invariance requires.

(b) () Suppose is diagonal, , and let be the -th column of . Then , and comparing -th columns gives . Since is invertible, its columns form a basis of (Proposition 5.2.14).

() Suppose is a basis with , and let be the matrix with columns . It is invertible by Proposition 5.2.14, and the -th column of is , which is also the -th column of for . So and .

(c) Take Then , but , so . So the determinant, although multiplicative and multilinear in the columns, is not additive.

(d) Expanding along the first row, The cofactors are for example, the cofactor is . This matrix is symmetric, so it equals its transpose , and One checks directly that ; for instance, the first row of times the first column of is , and times the second column is .