Lecture 23: Determinants (1)

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1 Lecture 23: Determinants (1) Travis Schedler Thurs, Dec 8, 2011 (version: Thurs, Dec 8, 9:35 PM)

2 Goals (2) Warm-up: minimal and characteristic polynomials of Jordan form matrices Snapshot: Generalizations of Jordan form over R and other fields Determinants: their unique properties, and how to compute them Characteristic polynomial of linear transformations Volume formula and the change-of-variable formula in multivariable integrals Read Chapters 8 and 10, do PS 11.

3 Warm-up: Minimal and char. polys of Jordan form matrices (3) Let A be a Jordan form matrix. Prove that: (a) The characteristic polynomial of A is (x λ) d λ, λ where d λ is the sum of the sizes of all Jordan blocks with λ on the diagonal. (b) The minimal polynomial of A is (x λ) m λ, λ where m λ is the size of the largest Jordan block with λ on the diagonal (the product is only taken over λ that occur, i.e., over eigenvalues of A). Hint: Note that p(a) = 0 if and only if p(j) = 0 for all Jordan blocks J occurring in A. Now find the minimal polynomial of a single Jordan block.

4 Solution to warm-up (4) (a) We know that the formula is valid if d λ equals the number of times that λ occurs on the diagonal. But this is just the sum of the sizes of all Jordan blocks with λ on the diagonal. (b) Since A is block diagonal with Jordan blocks on the diagonal, p(a) = 0 if and only if p(j) = 0 for every Jordan block J occurring in A. For each Jordan block J = λi + N n, the only eigenvalue is λ, so the minimal polynomial p J (x) is a power of (x λ), and this power is at most n since χ J (x) = (x λ) n. But we know that J λi = N n has the property that 0 (but Nn n = 0). So p J (x) = (x λ) n. Then p A (J) = 0 for all Jordan blocks if and only if (x λ) n divides p A whenever a Jordan block λi + N n occurs. Hence p A (J) = 0 for all Jordan blocks if and only if the given polynomial is a factor of p A, i.e., the given polynomial is the minimal polynomial. N n 1 n

5 Snapshot: Gen. of Jordan form for F = R [opt] (5) Over R, not all transformations have Jordan form because they need not have eigenvalues.

6 Snapshot: Gen. of Jordan form for F = R [opt] (5) Over R, not all transformations have Jordan form because they need not have eigenvalues. That is, they don t all have upper-triangular matrices.

7 Snapshot: Gen. of Jordan form for F = R [opt] (5) Over R, not all transformations have Jordan form because they need not have eigenvalues. That is, they don t all have upper-triangular matrices. But they have block upper-triangular matrices. So they have a block upper-tri. Jordan form : block-diagonal with Jordan blocks of the form X I 0 X I......, X I 0 X ( where X ) is either 1 1 (an eigenvalue), or the matrix a b for some a, b R, b 0. b a

8 Snapshot: Gen. of Jordan form for F = R [opt] (5) Over R, not all transformations have Jordan form because they need not have eigenvalues. That is, they don t all have upper-triangular matrices. But they have block upper-triangular matrices. So they have a block upper-tri. Jordan form : block-diagonal with Jordan blocks of the form X I 0 X I......, X I 0 X ( where X ) is either 1 1 (an eigenvalue), or the matrix a b for some a, b R, b 0. b a This matrix is unique up to permuting the Jordan blocks, and replacing b with b for all the X s inside any given block.

9 Snapshot: Gen. of Jordan form for arbitrary F [opt] (6) Suppose that F contains Q (i.e., all nonzero integers are invertible). Then we can still put every transformation in block-diagonal form with blocks X I 0 X I......, X I 0 X now with X a matrix with irreducible char. poly.. This is unique up to rearranging blocks and replacing X with another matrix with the same char. poly.

10 Snapshot: Gen. of Jordan form for arbitrary F [opt] (6) Suppose that F contains Q (i.e., all nonzero integers are invertible). Then we can still put every transformation in block-diagonal form with blocks X I 0 X I......, X I 0 X now with X a matrix with irreducible char. poly.. This is unique up to rearranging blocks and replacing X with another matrix with the same char. poly. Remark: This works also for some cases where F does not contain Q: the char. poly. χ T of T must be Galois (all irreducible factors must have distinct roots: e.g., x over R has distinct roots ±i).

11 Determinants (7) Consider the properties of a function φ : Mat(n, n, F) F: (a) φ(ab) = φ(a)φ(b) and φ(e) = det(e) when E is an (b) (c) elementary matrix (single row op.) (i) If A is obtained from A by adding a mult. of one row to another, then φ(a ) = φ(a). (ii) If A is obtained from A by rescaling a row by λ, then φ(a ) = λφ(a). (iii) If A is obtained from A by permuting two rows, then φ(a ) = φ(a). (i ) Multiadditivity: If A, B, and C are identical except in one row, where C is the sum of that row in A and B, then φ(c) = φ(a) + φ(b). (ii ) Multihomogeneity: part (ii) above. (iii ) Alternation: If A has two identical rows, then φ(a) = 0. Theorem The function φ = det is: the unique nonzero function satisfying (a); the unique function up to scaling satisfying (b); the unique function up to scaling satisfying (c).

12 Equivalence of (a) and (b) up to scaling (8) Note: φ = 0 satisfies all (a), (b), and (c). Lemma For φ nonzero, (a) is equivalent to: (b) together with φ(i ) = 1. Proof. (b) is equivalent to: if E is an elementary matrix (corr. to row op), then φ(eb) = det(e)φ(b) for all B. So (a) implies (b). Also (a) implies φ(i ) = φ(i 2 ) = φ(i )φ(i ), so φ(i ) = 1 or 0; but if φ(i ) = 0 then φ(a) = φ(ai ) = 0 for all A. So (a) implies φ(i ) = 1. Conversely, given (b), write any A = E 1 E n C where C is reduced row-echelon and E i elementary. If A is invertible, C = I. Then φ(i ) = 1 implies φ(a) = det(e 1 ) det(e n ). In particular φ(e) = det(e) for E elementary. Also, if A and B are invertible, we get φ(ab) = φ(a)φ(b). On the other hand, if A is not invertible, C has a zero row, and then φ(c) = 0 by homogeneity, and hence φ(a) = 0 as well by row ops. So if A or B is noninvertible, φ(ab) = 0 = φ(a)φ(b).

13 (c) implies (b) (9) Lemma Condition (c) implies (b). Proof. First we prove (i). This follows immediately from (i ) and (iii ). Note that (ii)=(ii ). Finally we prove (iii). Take A = (u 1 u n ) t. Consider the matrix C obtained by replacing the i-th and j-th rows both by u i + u j. Then (iii ) implies φ(c) = 0. But by (i ) and (iii ), φ(c) also equals φ(a) + φ(a ) where A is obtained from A by swapping the i-th and j-th rows. So φ(a ) = φ(a), i.e., (iii). So it remains to show: det satisfies (c); a function satisfying (b) is unique up to scaling. We showed det satisfies (c) on the HW!

14 Uniqueness (10) Lemma There is at most one function φ, up to scaling, satisfying (b).

15 Uniqueness (10) Lemma There is at most one function φ, up to scaling, satisfying (b). Proof. If φ satisfies (b), then φ(a) is determined from its row-echelon form matrix using row ops. But we saw that φ(c) = 0 for C row-echelon and noninvertible. So then φ(a) is determined uniquely by φ(i ), and changing φ(i ) just rescales φ. Examples! Use Gaussian elimination to compute determinant! (And hence char. poly as well!)

16 Characteristic polynomials (11) Recall that for a matrix A, χ A (x) = det(xi A). (For upper-tri matrices, this is obviously the usual formula.) Theorem For T L(V ), χ M(T ) (x) does not depend on basis.

17 Characteristic polynomials (11) Recall that for a matrix A, χ A (x) = det(xi A). (For upper-tri matrices, this is obviously the usual formula.) Theorem For T L(V ), χ M(T ) (x) does not depend on basis. Proof. Let A and A be two different matrices of T. Then A = BAB 1 for some B. So χ A (x) = det(xi BAB 1 ) = det(b(xi A)B 1 ) = det(b) det(xi A) det(b 1 ) = det(xi A) = χ A (x).

18 Characteristic polynomials (11) Recall that for a matrix A, χ A (x) = det(xi A). (For upper-tri matrices, this is obviously the usual formula.) Theorem For T L(V ), χ M(T ) (x) does not depend on basis. Proof. Let A and A be two different matrices of T. Then A = BAB 1 for some B. So χ A (x) = det(xi BAB 1 ) = det(b(xi A)B 1 ) = det(b) det(xi A) det(b 1 ) = det(xi A) = χ A (x). Same argument shows: det(m(t )) doesn t depend on the basis. Alternatively, note, for A Mat(n, n, F), χ A (x) = x n tr(a)x n ( 1) n det(a), and independence of χ M(T ) (x) implies that of tr and det.

19 Characteristic polynomial for arbitary F (12) Thus we get: Definition The char. poly. of T is det(xi M(T )), independent of basis!

20 Characteristic polynomial for arbitary F (12) Thus we get: Definition The char. poly. of T is det(xi M(T )), independent of basis! I already used this definition earlier when talking about Jordan form for general F.

21 Volumes (13) Theorem (Theorem 10.38) Let F = R and V = R n. Let R V be a solid region and T L(V ). Then vol(t (R)) = det T vol(r). Proof. Write T = SP for S an isometry and P positive. Then T (R) = S(P(R)). Also det T = det(s) det(p) = det S det P = det P. So this reduces to the isometry and positive cases, i.e., it is enough to show that vol(p(r)) = det P vol(r) and that vol(s(r )) = vol(r ) for R = P(R). We do this next.

22 Volumes for positive operators and isometries (14) Lemma For S an isometry, vol(s(r)) = vol(r).

23 Volumes for positive operators and isometries (14) Lemma For S an isometry, vol(s(r)) = vol(r). Proof. We use that: a rotation in R 2 preserves area. Then if we have a rotation of the x 1, x 2 plane doing nothing to other coords, volume is preserved. Then the result follows from the next lemma. Lemma Every isometry S is a product of rotations in planes (in x 1, x 2 plane up to choice of orthonormal coords) and reflections (up to coords, x 1 x 1, other coords the same).

24 Volumes for positive operators and isometries (14) Lemma For S an isometry, vol(s(r)) = vol(r). Proof. We use that: a rotation in R 2 preserves area. Then if we have a rotation of the x 1, x 2 plane doing nothing to other coords, volume is preserved. Then the result follows from the next lemma. Lemma Every isometry S is a product of rotations in planes (in x 1, x 2 plane up to choice of orthonormal coords) and reflections (up to coords, x 1 x 1, other coords the same). Proof: by induction; if we have time on board; otherwise exercise!

25 Volumes for positive operators and isometries (14) Lemma For S an isometry, vol(s(r)) = vol(r). Proof. We use that: a rotation in R 2 preserves area. Then if we have a rotation of the x 1, x 2 plane doing nothing to other coords, volume is preserved. Then the result follows from the next lemma. Lemma Every isometry S is a product of rotations in planes (in x 1, x 2 plane up to choice of orthonormal coords) and reflections (up to coords, x 1 x 1, other coords the same). Proof: by induction; if we have time on board; otherwise exercise! Corollary: Volume does not depend on choice of orthonormal coordinates.

26 Volumes under positive operators (15) Lemma For P positive, vol(p(r)) = det P vol(r).

27 Volumes under positive operators (15) Lemma For P positive, vol(p(r)) = det P vol(r). Proof. Choose orthonormal coordinates so that. P rescales each coordinate direction by a nonneg. number. This rescales volume by the product of these numbers, det(p). So changing coords rescales volume by det(t ).

28 Change of variable for integrals [opt] (16) Let T L(R n ) be invertible, and f : R n R (or to C). Then f (T (v)) d vol = det T 1 f (T (v)) d(vol T ) = det T 1 f (v) d vol. Moreover, we can replace T with an arbitrary nonlinear but invertible F : R n R n. Then at every x 0 R n, we have F (x 0 ) L(V ), which is the multivar. derivative of F at x 0 (matrix of partials). So at x = x 0, vol(d F (x)) = vol(f (x 0 ) d x). Thus f (F (v)) d vol = det F 1 f (F (v)) d(vol F ) = ( det F 1 F 1 )f (v) d vol.

Lecture 23: Trace and determinants! (1) (Final lecture)

Lecture 23: Trace and determinants! (1) (Final lecture) Travis Schedler Thurs, Dec 9, 2010 (version: Monday, Dec 13, 3:52 PM) Goals (2) Recall χ T (x) = (x λ 1 ) (x λ n ) = x n tr(t )x n 1 + +( 1) n det(t