Linear algebra and differential equations (Math 54): Lecture 8

Size: px

Start display at page:

Download "Linear algebra and differential equations (Math 54): Lecture 8"

Clinton Newman
5 years ago
Views:

1 Linear algebra and differential equations (Math 54): Lecture 8 Vivek Shende February 11, 2016

2 Hello and welcome to class! Last time We studied the formal properties of determinants, and how to compute them by row reduction. Today We ll see some more formulas involving the determinant minor expansion and Cramer s rule and discuss the interpretation of the determinant as a signed volume.

3 There is a midterm next time The only tools allowed for the midterm are pen or pencil. I will post a practice midterm later today. On the course website, you can also find a link to many midterms and practice midterms from this class in previous semesters. The midterm will be very similar to those, and very similar to the practice midterm.

4 There is a midterm next time I very strongly suggest that you try solving the practice midterms and previous midterms without first looking at the answers. It is very easy to trick yourself about how well you understand things if you just read the solutions.

5 Review: computing determinants by row reduction To compute the determinant of a matrix, row reduce it, and keep track of any row switches or rescalings of rows. At the end, multiply together: the inverses of the row rescaling factors the diagonal entries of the final echelon matrix ( 1) #rowswaps That s the determinant of the original matrix. This method is much much faster than summing all the terms.

6 Example Let s compute the determinant of this matrix First, we row reduce, keeping track of rescalings and row switches

7 Example / So the determinant is ( 1) 2 ( 7) ( ) = 7.

8 Try it yourself! Compute the determinant of this matrix: Row reduce, keeping track of rescalings and row switches:

9 Try it yourself! The determinant is

10 Review: terms in the determinant In the 2x2 case: ] ] c [ a b d c [ a b d +ad bc

11 Review: terms in the determinant In the 3x3 case: a b c d e f g h i a b c d e f g h i a b c d e f g h i +aei +bfg +cdh a b c d e f g h i a b c d e f g h i a b c d e f g h i afh bdi ceg

12 Another perspective a b c d e f g h i a b c d e f g h i a b c d e f g h i +aei afh bdi + bfg +cdh ceg +a e h f i b d g f i +c d g e h no orange-green inversions one orange-green inversions two orange-green inversions

13 Minor expansion For a matrix A, I ll write A i j for the matrix formed by omitting row i and column j. For example, if A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 We have: A = a 11 a 22 a 23 a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 = a 11 A 1 1 a 12 A a 13 A 1 3

14 Minor expansion More generally, by the same argument, for a square n n matrix A with entry a i,j in row i and column j, for any k in 1,..., n, there is a minor expansion along the k th row A = n ( 1) j+k a kj A k j j=1 and a minor expansion along the k th column A = n ( 1) j+k a jk A j k j=1

15 The sign ( 1) row+column

16 Example Compute by minor expansion along the second row:

17 Example =

18 Example Now we minor-expand each of these 3 3 determinants. We ll use the second row for each (to catch the zero).

19 Example = ( 1) = = = = = = = = 0 ( 1)

20 Example ( 2 5) + (0?) + ( 3 1) + (1 0) = 7 That s the same as we got doing this the other way. Which was easier?

21 Try it yourself! Compute by minor expansion the determinant of the matrix

22 A formula for the inverse A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Adj(A) = A 1 1 A 2 1 A 3 1 A 1 2 A 2 2 A 3 2 A 1 3 A 2 3 A 3 3 A Adj(A) = a 11 A 1 1 a 12 A a 13 A 1 3 a 11 A a 12 A 2 2 a 13 A 2 3 a 11 A 3 1 a 12 A a 13 A 3 3 a 21 A 1 1 a 22 A a 23 A 1 3 a 21 A a 22 A 2 2 a 23 A 2 3 a 21 A 3 1 a 22 A a 23 A 3 3 a 31 A 1 1 a 32 A a 33 A 1 3 a 31 A a 32 A 2 2 a 33 A 2 3 a 31 A 3 1 a 32 A a 33 A 3 3 The diagonal terms, e.g., a 11 A 1 1 a 12 A a 13 A 1 3, are minor expansions of det(a).

23 A formula for the inverse A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Adj(A) = A 1 1 A 2 1 A 3 1 A 1 2 A 2 2 A 3 2 A 1 3 A 2 3 A 3 3 Let s look at an off-diagonal term of A Adj(A), say a 21 A 1 1 a 22 A a 23 A 1 3 Expanding this out from the definition, a a 22 a a 32 a 33 a 22 a 21 a 23 a 31 a 33 + a 23 a 22 a 23 a 32 a 33

24 A formula for the inverse The quantity a 21 a 22 a 23 a 32 a 33 a 22 a 21 a 23 a 31 a 33 + a 23 a 22 a 23 a 32 a 33 is the minor expansion of the determinant a 21 a 22 a 23 a 21 a 22 a 23 a 31 a 32 a 33 The matrix has a repeated row, so the determinant is zero! The same is true for all the off diagonal terms.

25 A formula for the inverse A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Adj(A) = A 1 1 A 2 1 A 3 1 A 1 2 A 2 2 A 3 2 A 1 3 A 2 3 A 3 3 A Adj(A) = det(a) I = Adj(A) A This holds for any square matrix A, where Adj(A) ij = ( 1) i+j A j i The entry in row i, column j of Adj(A) is the determinant of the matrix formed by removing column i and row j of A, times ( 1) i+j.

26 Try it yourself! [ a b For the 2 2 matrix c d ], determine Adj(A), and verify A Adj(A) = det(a) I = Adj(A) A [ d b Adj(A) = c a ] [ a b c d ] [ d b c a ] [ ad bc ab + ba = cd dc cb + da ] [ 1 0 = (ad bc) 0 1 ]

27 Cramer s rule Consider a matrix equation Ax = b where A is square. Then det(a) x = (Adj(A) A)x = Adj(A) b Take the i th row of the column vector on both sides: det(a) x i = j Adj(A) ij b j = j ( 1) i+j A j i b j I.e., the minor expansion along the i th column of the determinant of the matrix formed by replacing the i th column of A by b.

28 Cramer s rule Consider a matrix equation Ax = b where A is square. Then if det(a) 0, x i = det(replace column i of A by b) det(a)

29 Never use these formulas to compute As we saw, taking the determinant of a 4 4 matrix by minor expansion was more difficult than by row reduction. It only gets worse as the size of the matrix grows. Likewise, row reduction beats computing Adj for inverting matrices, and beats Cramer s rule for solving systems.

30 Why learn these formulas at all? It s conceptually satisfying to know that, not only is there a procedure for solving systems or inverting matrices, there s in fact a closed form formula. The properties of the formula reveal facts about the solutions.

31 Integer inverses and solutions Say you have an invertible matrix M with integer entries. Does its inverse also have integer entries? It does, if and only det(m) = ±1. Observe det(m) det(m 1 ) = det(mm 1 ) = 1. The determinant of an integer matrix is always an integer it s made by additions and multiplications. If M 1 has integer entries, then det(m) and det(m 1 ) are two integers which multiply to 1, hence both ±1. Similarly, the Adj of an integer matrix is an integer: it s made by additions and multiplications. So, if det M = ±1, then M 1 = Adj(M)/ det M is an integer matrix as well.

32 Integer inverses and solutions Similarly, consider the equation Ax = b. Assume A is square and has integer entries b has integer entries det(a) = ±1 We saw that A 1 has integer entries, so the (unique) solution x = A 1 b also has integer entries.

33 Volumes You probably have an intuitive notion of what volume means: the amount of stuff that can fit inside something. For our purposes, the stuff is going to be cubes of a fixed side length: Volume(S) number of cubes that fit inside S To be more precise, Volume(S) = lim ɛ dim ɛ 0 number of cubes of side length ɛ that fit inside S By this we mean: for S R n, we overlay the ɛ-mesh grid on S, and count the number of cubes which fall completely inside.

34 Linear transformations and volumes Let T : R n R n be a linear transformation. Given a set X, we want to think about how the volumes of X and T (X ) compare. The key observation is that the number of cubes in X is the same as the number of T -transformed such cubes in T (X ).

35 Linear transformations and volumes Filling the transformed cubes with yet smaller regular cubes, and observing that the failure of these to pack correctly at the boundary is washed out as ɛ 0, we conclude: Volume(X ) Volume(T (X )) = Volume(cube) Volume(T (cube)) Rearranging, Volume(T (X )) = Volume(T (cube)) Volume(X ) But what s Volume(T (cube))?

36 Linear transformations and volumes Suppose now given two linear transformations, T, S : R n R n. We apply the formula Volume(T (X )) = Volume(T (cube)) Volume(X ) to the set X = S(cube): Volume(T (S(cube))) = Volume(T (cube)) Volume(S(cube))

37 Linear transformations and volumes In other words, the function V : linear transformations R T Volume(T (unit cube)) respects multiplication in the sense that V (T S) = V (T )V (S) Note also that V (Identity) = 1, and V (non invertible matrix) = 0.

38 Linear transformations and volumes Now consider any linear transformation T. If it s not invertible, V (T ) = 0. If it is invertible, then by row reduction we can expand it as a product of elementary matrices, T = E n E 1 Since volume scaling is multiplicative, V (T ) = V (E n ) V (E 1 )

39 Linear transformations and volumes It remains to understand volume scaling of elementary matrices. Rescaling a row stretching a coordinate rescales volume by the same factor: the volume of a box is the product of its side lengths, and we rescaled one of them. Switching two rows doesn t change volume at all we re just renaming the sides of the box. Adding a multiple of one row to another takes a box to a parallelopiped with the same base and the same height, so again doesn t change volume.

40 Determinants and volumes So for an elementary linear transformation, Volume(T (unit cube)) = det(t ) Volume scaling is multiplicative, so this holds for any linear transformation. Collecting these observations, for any linear transformation T : R n R n and any set X R n, Volume(T (X )) = det(t ) Volume(X )

Linear algebra and differential equations (Math 54): Lecture 7

Linear algebra and differential equations (Math 54): Lecture 7 Vivek Shende February 9, 2016 Hello and welcome to class! Last time We introduced linear subspaces and bases. Today We study the determinant