NOTES FOR LINEAR ALGEBRA 133

Transcription

1 NOTES FOR LINEAR ALGEBRA 33 William J Anderson McGill University These are not official notes for Math 33 identical to the notes projected in class They are intended for Anderson s section 4, and are

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3 Contents Linear Equations and Matrices 5 Linear Equations 5 2 Homogeneous Systems 2 3 Matrices 3 4 Matrix Inverses 7 5 Elementary Matrices 2 6 Block Multiplication of Matrices 22 2 Determinants and Eigenvalues 23 2 Determinants Eigenvalues 3 3 Vector Geometry 35 3 Coordinate-Free and Coordinate Approaches to Geometry 35 3 Coordinate-Free Geometry Dot Product and Projections Lines and Planes 4 33 Lines Planes The Cross Product Matrix Transformations of R Composite of Transformations Inverse of a Matrix Transformation 53 4 The Vector Space R n 55 4 Subspaces Linear Independence Invertibility of Matrices Linear Dependence 6 43 Dimension 6 43 Existence of Bases Rank The Rank Theorem Null Space and Image Space 68 3

4 4 CONTENTS

5 Chapter Linear Equations and Matrices Linear Equations Introduction x + 2y = 4 is a linear equation A solution is a pair s, s 2 of numbers such that s + 2s 2 = 4 s =, s 2 = 5/2 is a solution So is s =, s 2 = 3/2 In fact, there are infinitely many solutions given by s = s, s 2 = s+4 2 for any number s This is called a parametrized solution The pair x + 2y = 4 2x + 3y = 4 () is called a system of linear equations Replacing the 2nd equation by the second minus twice the first gives x + 2y = 4 y = 22 so we conclude that x = 4, y = 22 What we did is called a row operation (It can be reversed to give the original pair of equations by replacing the second row by itself plus twice the first) We just did Gaussian elimination, and the last display is the system in echelon form (except for the ) In this case the system has a unique solution But if the system was x + 2y = 4 2x + 4y = 4, there would be no solution And if the system were x + 2y = 4 2x + 4y = 8, there would be infinitely many solutions as in the example at the very beginning 5

6 6 CHAPTER LINEAR EQUATIONS AND MATRICES Linear Systems An equation of the form a x + a 2 x a n x n = b where a,, a n, and b are given numbers is called a linear equation in the unknown variables x,, x n A set of numbers s,, s n is called a solution to this equation if a s + a 2 s a n s n = b Equivalently, we arrange the s i s into a column vector and call X a solution m of these linear equations s s 2 X =, s n a x + a 2 x a n x n = b a 2 x + a 22 x a 2n x n = b 2 a m x + a m2 x a mn x n = b m is called a system of m equations in n unknowns The column vector X is a solution if X is a solution to each of the equations We will see that there are three possibilities: () the system has no solution (then the system is called inconsistent), (2) the system has a unique (ie one and only one) solution, (3) the system has infinitely many solutions Thus the system has either at most one solution, or infinitely many The system is called consistent if there is at least one solution Otherwise inconsistent Equivalent Systems Two systems of linear equations are equivalent if every solution of one system is a solution of the other

7 LINEAR EQUATIONS 7 Coefficient Matrix, Augmented Matrix The matrix a a 2 a n a 2 a 22 a 2n A = a m a m2 a mn is called the coefficient matrix The matrix A b given by a a 2 a n b a 2 a 22 a 2n b 2 a m a m2 a mn is called the augmented matrix For the system (), the coefficient matrix is b m and the augmented matrix is Elementary Row Operations () Interchange two rows, (2) Multiply any row by a non-zero number, (3) Replace any row by itself plus a multiple of another row Theorem Every row operation results in an equivalent system (Because every row operation can be reversed to obtain the previous state) Echelon a formation of troops in which each unit is positioned successively to the left or right of the rear unit to form an oblique or steplike line (from a dictionary) Row Echelon Form () All zero rows are at the bottom, (2) The first nonzero entry from the left in each nonzero row is a, called the leading for that row, (3) Each leading is to the right of all leading s in the rows above it Here is an example of an augmented matrix A b reduced to echelon form, where an represents an arbitrary number:

8 8 CHAPTER LINEAR EQUATIONS AND MATRICES To the left of is (what happened to) A, to the right is (what happened to) b A zero row refers to the A part of the row being all zeros For a zero row, the last member (the b part) can be anything If it is the row has no information It will be non-zero iff there does not exist a solution to the system Moreover, it is called reduced row echelon form if each leading is the only nonzero entry in its column The variables corresponding to the leading s are called leading variables Nonleading varibles are associated with parameters Theorem Every augmented matrix can be carried to row echelon form, or further to reduced row echelon form, by a series of elementary row operations The row echelon form (or reduced row echelon form) is equivalent to the original system (because it is the last step in a series of equivalent systems) Gaussian Algorithm () If augmented matrix consists entirely of zeros, it is already in echelon form, so stop, (2) Find the first column from the left containing a nonzero entry k Move the row containing k to the top of the matrix, (3) Multiply the top row by /k to create the first leading, (4) Make each entry in the column below the leading zero by subtracting multiples of the top row from lower rows, (5) Repeat steps -4 on the matrix consisting of the remaining rows Note: It may be more convenient to leave step 3 to the end, after echelon form or reduced echelon form has been achieved This can help avoid dealing with fractions in the row reduction process In this case, the third elementary row operation "add a multiple of a row to any other row" might be better stated as "replace any given row by a non-zero multiple of that row plus a non-zero multiple of any other row" Rank Any matrix A (not just the augmented matrix) can be reduced to echelon form by elementary row operations The number of leading s in row echelon form is called the rank of the matrix Theorem Given a system of m equations in n unknowns having at least one solution (ie a consistent system) If the augmented matrix has rank r, the set of solutions has n r parameters Proof Carry the augmented matrix to a reduced row-echelon matrix R Then R has r leading s, so there are exactly r leading variables Hence there are n r nonleading variables, and each of these is assigned a parameter Note: having at least one solution is important here In reduced echelon form, a row like 5 means no solution Solve the system x + 2y w = 3 2x + 4z + 2w = 6 x + 2y z = 6 2x y + z + w = 3

9 LINEAR EQUATIONS 9 Solution The coefficient matrix is The augmented matrix is R2-2R is R3-R is R4-2R is Divide R2 by -4, and row 3 by -, and get Add 5R2 to R4, get Add 4R3 to R4, get Divide R4 by -6, get The matrix is now in echelon form Starting from the bottom, we get w =,z w = 3, so z = 2, y z w = 3, so y = 2, and x + 2y w = 3, so x =

10 CHAPTER LINEAR EQUATIONS AND MATRICES Or we can reduce the previous echelon form to reduced echelon form Replacing R3 by R3+R4 gives Replace row 2 by row 2+row 4, get Replace row by row +row 4, get Replace row 2 by row 2+row 3, get Finally, row minus 2 row 2 gives This is reduced row echelon form We can write the solution as 2 X = Solve x + y z + 2w = 2 2x y + z + w = 3x 2y + z 2w = 27 Solution The augmented matrix is R2-2R and R3-3R give

11 LINEAR EQUATIONS Divide row 2 by -3, get Replace row 3 by row 3+5 row 2, get Multiply row 3 by -, get This is echelon form The solution is therefore w 3 4w 9 X = 3w 2 w (2) where w is a parameter Let us get reduced row echelon form Replace row 2 by row 2 plus row 3, and row by row plus row 3, get Finally, replace row by row minus row 2, get for which the solution is given in () Note that this solution can be written as 3 9 X = 2 + w 4 3 Note that rank is r = 3 and number of unknowns is n = 4, so there are n r = parameters For which values of k does the system x + x 2 + x 3 = 2x + x 2 + x 3 = x + (k 2 4)x 3 = k have (i) a unique solution (ii) no solution (iii) infinitely many solutions? In case (iii), write the solutions in parametric form

12 2 CHAPTER LINEAR EQUATIONS AND MATRICES Solution The augmented matrix is 2 k 2 4 k R2 2R R2 and R3 R R3 gives R3 R2 R3 gives 3 k 2 5 k k 2 4 k 2 (i) if k 2 and k 2 (ie k 2 4), then unique solution (ii) if k = 2, then no solution (iii) if k = 2, then infinitely many solutions Augmented matrix is 2 3 giving x = 2, x 2 = 3 x 3 With t = x 3, this gives x x 2 x = 3 t = 3 + t t 2 Homogeneous Systems A system is homogeneous if the constant terms b are all zero The zero solution is always a solution and is called the trivial solution But there may be additional solutions x + 2y = 2x + 3y = Using the same row operation as before, we find x + 2y = y = so the unique solution is x = y = (the zero solution)

13 3 MATRICES 3 x + 2y w = 2x + 4z + 2w = x + 2y z = 2x y + z + w = Using the same row operations as before, we find so the unique solution is x = y = z = w = (the zero solution) Theorem If a homogeneous system of linear equations has more variables than equations, then it has nontrivial solutions Proof There is at least one solution (the trivial one) Let m be the number of equations Then n > m r, so n r There is at least one parameter, so there must be infinitely many solutions 3 Matrices Definition A matrix is a rectangular array of numbers, for example 3 4 A = 2 5, X =, Y = (5, 9, 2, 5, ) A is a 3 2 matrix, X is a 4 matrix, and Y is a 5 matrix More commonly, X is called a 4-dimensional column vector, and Y a 5-dimensional row vector More generally, a a 2 a 3 a n a 2 a 22 a 23 a 2n A = (3) a m a m2 a m3 a mn is called an m n matrix (m rows and n columns) a ij is called the i, jth component of A, where i is the row number and j the column number Operations on Matrices Let A and B be matrices, and c a scalar (ie a number) If A is as in (3), then ca ca 2 ca 3 ca n ca 2 ca 22 ca 23 ca 2n ca = ca m ca m2 ca m3 ca mn If b b 2 b 3 b n b 2 b 22 b 23 b 2n B = b m b m2 b m3 b mn

14 4 CHAPTER LINEAR EQUATIONS AND MATRICES is also an m n matrix (ie same dimensions as A), then a + b a 2 + b 2 a 3 + b 3 a n + b n a 2 + b 2 a 22 + b 22 a 23 + b 23 a 2n + b 2n A + B = a m + b m a m2 + b m2 a m3 + b m3 a mn + b mn Examples Then Let A = 2 5, B = A = 25, A + B = 2 2, A B = Definition Let A be as in (3) The transpose of A, denoted by A T is the n m matrix a a 2 a m a 2 a 22 a m2 A T = a 3 a 23 a m3 a n a 2n a mn ie the ijth component of A T is a ji For example, for the matrix in the previous example A T = ( 2 ) Matrix Multiplication Let r = (r, r 2,, r n ) be an n-dimensional row vector (ie a n matrix) and c = (c, c 2,, c n ) T an n-dimensional column vector (ie an n column matrix) The dot product of r and c is c c 2 r c = (r, r 2,, r n ) = r c + r 2 c r n c n c n Let A be an m n matrix and B an n p matrix Then the product AB is the m p matrix whose ijth component is the dot product of the ith row of A and the jth column of B Then Let 3 A = 2 5, B = AB = 2 5 =

15 3 MATRICES 5 Note that BA is not defined The identity matrix is a square matrix all of whose components are zero except for the diagonal components which are For example, the 3 3 diagonal matrix is I = I 3 = Examples Let A = Let Then DI = ID = D 2 and I = 3 4 Then AI = IA = A 3 2 D = 2 5, I = Remarks () If A, B, and C are matrices with AB = C, if B = [B B 2 B p ], and C = [C C 2 C p ] where the B j s and C j s are the columns of B and C respectively, then AB j = C j for all j =,, p That is, A times the jth column of B is the jth column of C (2) Matrix multiplication is not commutative (ie we do not always have AB = BA) For the matrices 2 A = and B = 3 4 we have AB = 3 4, BA = 3 (3) We can have A 2 = even though A For the A above, we have A 2 = = (4) We can write linear systems of equations as matrix equations AX = B, where A is the m n matrix of coefficients, X is the n column matrix of unknowns, and B is the m column matrix of constants The system x + 2y w = 3 2x + 4z + 2w = 6 x + 2y z = 6 2x y + z + w = 3 that we previously solved can be written as 2 x y 2 z = w 3

16 6 CHAPTER LINEAR EQUATIONS AND MATRICES Theorem (Properties of Matrix Multiplication) Let A, B, and C be matrices of compatible dimensions, let I be the identity matrix, and let c be a number () IA = A and BI = B, (2) A(BC) = (AB)C (associativity), (3) A(B + C) = AB + AC, A(B C) = AB AC, (4) (A + B)C = AC + BC, (A B)C = AC BC, (5) c(ab) = (ca)b = A(cB), (6) (AB) T = B T A T In (), A and B do not have to be square Properties (3) and (4) are called distributivity of multiplication over addition The proof of property (6) is (i, j) th component of (AB) T = (j, i) th component of AB = jth row of A ith column of B = ith row of B T jth column of A T = (i, j) th component of B T A T Consider the equation AX = B It may have many solutions If X satisfies AX = B, X is called a particular solution The equation AX = is called the associated homogeneous system Theorem Let X be a particular solution of the system AX = B (that is, assume AX = B) () If X is any solution of AX =, then X + X is a solution of AX = B (2) Every solution of AX = B can be written as X = X + X, where AX = Proof () Let X = X + X Then AX = AX + AX = + B = B (2) Let X be a solution of AX = B and let X = X X Then AX = AX AX =, so X = X + X as required 2 2 Let A = and B = 3 Use Gaussian elimination to express every solution to the system AX = B as the sum of a particular solution and the general solution to the homogeneous equation AX = Solution The augmented matrix is The reduced echelon form is

17 4 MATRIX INVERSES 7 Note that rank is r = 2 and number of unknowns is n = 4, so there are n r = 2 parameters The leading variables are x and x 3, and the nonleading variables are x 2 = s and x 4 = t The solution is x 3 + 2s 5t x 2 X = x 3 = s + 3t = + s + t 3 t x 4 The particular solution X and general solution X to the homogeneous equation are X =, X = s + t X = and X 2 = 3 are called basic solutions to the homogeneous system AX =, because every solution to AX = is of the form X = sx + tx 2 4 Matrix Inverses Definition Let A be a square matrix A matrix C such that CA = I and AC = I is called an inverse of A (note that C must be square with the same dimensions as A) An inverse may not exist If one does, A is called invertible If not, then A is not invertible If an inverse exists, it is unique For if C and D are inverses of A, then C = CI = C(AD) = (CA)D = ID = D If A is invertible, we denote its inverse by A (just like the inverse of a number c can be denoted by c instead of /c) Theorem (Properties of Inverses) Assume all matrices below are square () I is invertible and I = I, (2) If A is invertible, then so is A, and (A ) = A, (3) If A and B are invertible, then so is AB and (AB) = B A, (4) If A, A 2,, A k are invertible, then so is A A 2 A k, and (A A 2 A k ) = A k A k A 2 A, (5) If A is invertible, then so is A k, and (A k ) = (A ) k, (6) If A is invertible, then so is A T, and (A T ) = (A ) T, (7) If A is invertible and c is a number, then ca is invertible and (ca) = c A Proof () We have I I = I, (2) From fact that AA = I and A A = I, (3) We have (B A )(AB) = I, and (AB)(B A ) = I, (4) Same proof as for (3), (5) Take all the A i s in (3) to be A (6) (A ) T A T = (AA ) T = I T = I and A T (A ) T = (A A) T = I T = I

18 8 CHAPTER LINEAR EQUATIONS AND MATRICES The Matrix Inversion Algorithm How to Find the Inverse of a Matrix A Let X, X 2, X 3 be solutions of the equations AX =, AX 2 =, AX 3 = (4) Then the matrix C whose columns are X, X 2, and X 3 satisfies AC = I (because A times the jth column of C is the jth column of AC) In solving the three systems in (4) for X, X 2, and X 3, the reduced echelon form will be the same, so we can do simultaneous row operations on all three systems That is, we do elementary row operations on the matrix A I to reduce it to I B, and then B will the the inverse A of A More generally, let X, X 2,, X n be solutions of the equations AX =, AX 2 =,, AX n = (42) Then the matrix C whose columns are X, X 2,, X n satisfies AC = I The reduced echelon form will be the same for each of the equations, and everything proceeds as in the case n = 3 Find the inverse of A = Solution The augmented matrix is A I = R2-2R gives Replacing R by R+2R2 gives Multiply R2 by to get Hence A = You can check that AA = I and A A = I Exercise Find the inverse of a b A =, c d where a, b, c, d are numbers such that ad bc

19 4 MATRIX INVERSES 9 Solution Go through the same steps as in the previous example to get A d b =, ad bc c a or simply verify that AA = A A = I Find the inverse of A = 2 2 Solution The augmented matrix A I is A I = 2 2 from which there results A = We can use this to solve AX = B where B = 6 Multiplying AX = B by A from the left gives 3 A (AX) = A B, which is X = A B = = which is the same as previously obtained by row reduction Theorem (Conditions for Invertibility) Let A be a square matrix The following are equivalent () A is invertible, (2) The system AX = has only the trivial solution X = (3) A can be carried to I by elementary row operations, (4) The system AX = B has a solution X for every column B, (5) There exists a matrix C such that AC = I Proof () (2): multiply both sides of AX = from the left by A, to get A AX = A =, so X = (2) (3): let R be the result of reducing A to reduced echelon form Then either R has a row of zeros, or R = I If R has a row of zeros, then AX = has infinitely many solutions So R = I

20 2 CHAPTER LINEAR EQUATIONS AND MATRICES (3) (4): obvious (4) (5): define C to be the matrix obtained after (42) (5) (): Suppose CX = Then X = IX = (AC)X = A(CX) = A = Then (2) is true for C instead of A Then according to (5), there is a matrix C such that CC = I Since AC = I, then C = (AC)C = A(CC ) = A, so CA = I Remark The equivalence (5) () implies that in the definition of invertibility of A, only one of the two conditions AC = I and CA = I is needed If the matrix expression AB T (CA ) T is defined, where A is n n invertible, B is p s, and C is q t, then only one of the following is a necessary restriction on n, p, s, q, and t Indicate which one A) n = p = s = q = t B) n = s = q and p = t C) n = s = t and p = q D) p = s = n and q = t Solution Answer is C) B T is s p, so n = s AB T is n p CA is p n So t = n and q = p 5 Elementary Matrices Definition A square matrix that is obtained by doing a single elementary row operation to an identity matrix is called an elementary matrix Lemma Let A be an m n matrix Let A be the result of performing an elementary row operation on A Let E be the result of performing the same operation on I m Then A = EA Take A = 2 2 () Suppose the operation is to exchange rows and 3 Then 2 A = 2, E =, 2 and EA = 2 2

21 5 ELEMENTARY MATRICES 2 Lemma Every elementary matrix E is invertible, and E is the elementary matrix of the reverse of the operation that produced E This is because every elementary row operation is reversible Suppose a series of k elementary row operations is performed on A, and let E i be the elementary matrix corresponding to the ith operation Then the result of the k operations is UA, where U = E k E k E 2 E Note that U is invertible, since it is the product of invertible matrices Theorem Suppose a series of elementary row operations carries A to B Then () B = UA for some invertible matrix U, (2) U can be constructed by performing the same elementary row operations on I: we have [A I] [B U] Proof (2) In [A I] [B U], the same row operations that take A to UA take I to U Remarks Taking B = I, we have U = A Therefore the matrix inversion dealt with previously is a special case of the previous theorem Theorem A matrix A is invertible iff it is the product of elementary matrices Proof If A is invertible, there is a series of row operations carrying A to I By the above theorem, I = UA (so U = A ) for U of the form U = E k E k E 2 E Then A = (A ) = U = E E 2 E k Conversely, if A is the product of elementary matrices, it is invertible because the product of invertible matrices is always invertible (Ex 5(a), p 54) Express the invertible matrix A = 3 as a product of elementary matrices Do the same for A Solution First reduce A to reduced echelon form: 3 R R 2 R 2=R A = 2 3R 3 The corresponding elementary matrices are (since I = E =, E 2 = ) R 2= R 2 = I, E 3 3 = That is, E 3 E 2 E A = I, telling us that A = E 3 E 2 E (so we did A first) Then A = ( A ) = (E3 E 2 E ) = E E 2 E 3 = ( ) 3

22 22 CHAPTER LINEAR EQUATIONS AND MATRICES 6 Block Multiplication of Matrices If matrices A and B are partitioned compatibly into blocks, the product AB can be computed by matrix multiplication using the blocks as entries Then Suppose We can do this another way Let A =, A = so that Then so A = , B = B = A B + A 2 B 2 = 6 AB = (, A = 3, A 4 = 2 3 ), 2, B = A A 2 A =, B = A 3 A 4 2, 4 6 ( B B 2 6, A B + A 4 B 2 = 2 5, 6 A B + A 2 B 2 AB = = 7 63 A 3 B + A 4 B We will use the next fact several times later on in the course Theorem Let A = [C C 2 C n ] be an m n matrix with columns C, C 2,, C n If X = (x, x 2,, x n ) T is a column vector, then then x x n ) x 2 AX = [C C 2 C n ] = x C + x 2 C x n C n (*) Proof The ith component of AX is a i x + + a in x n Since the ith component of C j is a ij, this is also the ith component of x C + x 2 C x n C n

23 Chapter 2 Determinants and Eigenvalues 2 Determinants a b Definition Let A = Then we define the determinant of A to be det(a) = ad bc c d Definition Let a a 2 a n a 2 a 22 a 2n A = a n a n2 a nn be a square n n matrix Let A ij be the (n ) (n ) matrix obtained by stroking out the ith row and jth column of A, and define C ij (A) = ( ) i+j det(a ij ) () C ij (A) is the (i, j)th cofactor of A We also define the number D i = a i C i + a i2 C i2 + + a in C in (2) for each row i, and for each column j D j = a jc j + a 2j C 2j + + a nj C nj (3) Find all the cofactors for the matrix 5 7 A = 9 3 6, 4 8 and then find D, D 2, and D 3 Solution We have 3 6 C (A) = ( ) 2 det = 48, C 8 2 (A) = ( ) 3 det 9 6 = 24, C 4 3 (A) = ( ) 4 det so D = a C + a 2 C 2 + a 3 C 3 = (5 48) + ( 24) + (7 84) = C 2 (A) = ( ) 3 det = 56, C 8 22 (A) = ( ) 4 det 4 23 = 28, C 23 (A) = ( ) 5 det 9 3 = 84, = 44, 4 8

24 24 CHAPTER 2 DETERMINANTS AND EIGENVALUES so D 2 = (9 56) + ( 3 28) + (6 44) = C 3 (A) = ( ) 4 det = 5, C (A) = ( ) 5 det 5 7 = 33, C (A) = ( ) 6 det 5 = 6, 9 3 so D 3 = (4 5) + (8 33) + ( 6) = 324 Observe that D i is the same for each row i This works for columns as well as rows For our example, D = a C + a 2 C 2 + a 3 C 3 = (5 48) + (9 56) + (4 5) = 324, D 2 = a 2 C 2 + a 22 C 22 + a 32 C 32 = ( 24) + ( 3 28) + (8 33) = 324, D 3 = a 3 C 3 + a 23 C 23 + a 33 C 33 = (7 84) + (6 44) + ( 6) = 324 This common value 324 is called the determinant of A, and we write det(a) = 324 If we knew beforehand that D = D 2 = D 3 = D = D 2 = D 3, we would only have had to calculate one of them In fact, it can be proved these are all equal, for any 3 3 matrix (and in fact for any square n n matrix) If we had used D to find det(a), we would say we found det(a) by expanding around the first row of A If we used D 2, we would say we found det(a) by expanding around the second column of A The expansion is called Laplace s expansion Now we know how to calculate determinants for 2 2 and 3 3 matrices Suppose we want to calculate the determinant of a 4 4 matrix A The A ij s in () are now 3 3 matrices, so we can calculate all the C ij s and then use either (2) or (3) with n = 4 to get det(a) Theorem of A If A is any square matrix, det(a) is equal to the Laplace expansion along any row or column Let I be the n n identity matrix Then det(i) = Corollary If A has a row (or column) of zeroes, then det(a) = (Just expand along that row) Theorem Let A[r ] denote an n n matrix whose ith row is r, and let c be a scalar If r and s are n row vectors and c is a scalar, then det A[r + s] = det A[r ] + det A[s] and det A[cr ] = c det A[r ] (4) Proof Expanding det A[r + s] along the ith row gives det A[r + s] = (r + s) C i + (r + s) 2 C i2 + + (r + s) n C in = [r C i + r 2 C i2 + + r n C in ]+ + [s C i + s 2 C i2 + + s n C in ] = det A[r ] + det A[s] Similarly, expanding det A[cr ] along the ith row gives det A[cr ] = (cr ) C i + (cr ) 2 C i2 + + (cr ) n C in = c[r C i + r 2 C i2 + + r n ]C in = c det A[r ] Note that r j, s j, (r + s) j, and (cr ) j denote the jth components of the row vectors r, s, r + s, and cr respectively Remark This theorem works for columns as well That is, (4) holds if A[r ] denotes an n n matrix whose jth column is r and r and s are n column vectors Suppose A = 9 3 6, B = 9 6, C = Then det C = det A + det B

25 2 DETERMINANTS 25 Theorem (Properties of Determinants) Let A be an n n matrix () If B is obtained from A by interchanging two different rows (or two different columns), then det(b) = det(a) (2) If A has two identical rows (or columns), then det(a) =, (3) If B is obtained from A by multiplying a row (or column) by a number k, then det(b) = k det(a), { det A if i = j, (4) a i C j (A) + + a in C jn (A) = if i j (5) If B is obtained from A by adding a multiple of some row (or column) of A to another row (or column), then det(b) = det(a) Proof () If A is 2 2, this is obvious Suppose it is true for all (n ) (n ) matrices, where n 3 Now assume A is an n n matrix Let A be the result of exchanging two rows Find det(a ) by expanding around a row which is not one of the two that were exchanged Each cofactor involves the determinant of an (n ) (n ) matrix with two rows that have been exchanged By induction, each cofactor for A is minus the corresponding cofactor for A Hence det(a ) = det(a) Thus the result is true for all n n matrices (2) Obvious from () (3) Obvious (4) Obvious if i = j; hence assume i j Let A be the matrix obtained from A by replacing row j by row i Since A has two identical rows, then det(a ) = But the cofactors C jl (A ) for the jth row of A do not involve the jth row, so C jl (A ) = C jl (A) for l =,, n And also a jl = a il for l =,, n If we were to calculate det(a ) by expanding about the jth row of A, we would get = det(a ) = a j C j + + a jn C jn = a ic j + + a in C jn (5) Let the other row be r, let the some row be s, and let the multiple be c Then det B = det A[r + cs] = det A[r ] + det A[cs] = det A[r ] + c det A[s] = det A[r ] = det A Note that det A[s] = because A[s] has two identical rows Corollary If E is the elementary n n matrix corresponding to an elementary row operation on A, then det(ea) = det(e) det(a) More generally, if E, E 2,, E k are elementary row matrices, then det(e k E 2 E A) = det(e k ) det(e 2 ) det(e ) det(a) Corollary If det(a) and B is obtained from A by an elementary row operation, or a sequence of them, then det(b)

26 26 CHAPTER 2 DETERMINANTS AND EIGENVALUES Theorem Let A be an n n matrix () For any number k, det(ka) = k n det(a) (2) If A is a triangular matrix (ie a ij = if i > j or a ij = if i < j), then det(a) is the product of the diagonal elements (3) det(a T ) = det(a) Proof of (3) By induction on n It is obviously true for n = 2 Now assume true for n, where n 3 Let A be n n Then expanding det(a T ) around the first row is identical to expanding det(a) around the first column, since the cofactors are the same (being determinants of (n ) (n ) matrices which are transposes of each other) Let a x b A = y u c z v w d Then expanding around the first row gives b c det(a) = a det u c = ab det = abcd w d v w d Theorem Let A be an n n matrix A is invertible iff det(a) Proof Let R be the result of reducing A to reduced echelon form There are three facts: () R has a row of zeros, or R = I, (2) A is invertible iff R = I (from a theorem in ch), (3) det(r) = iff det(a) = (from an earlier corollary in this chapter) Remarks For the next theorem, note that if A and B are square n n matrices, then AB is invertible iff both A and B are invertible For suppose AB has an inverse, so (AB)C = I for some C Then A(BC) = I, so A has an inverse Similarly B has an inverse Theorem Let A and B be square n n matrices Then det(ab) = det(a) det(b) (5) Note that det(ab) = det(ba) Proof Case : AB is not invertible Then by the previous remark, both sides of (5) are zero Case 2: AB is invertible Then A and B are invertible, so there are elementary matrices E,, E k and F,, F m such that A = E k E and B = F m F Then AB = E k E F m F, so det(ab) = det(e k ) det(e ) det(f m ) det(f ) = det(a) det(b)

27 2 DETERMINANTS 27 Corollary Let A and A, A 2,, A k be n n matrices Then () det(a A 2 A k ) = det(a ) det(a 2 ) det(a k ), (2) det(a k ) = [det(a)] k Theorem Let A be an n n matrix If A is invertible, then det(a ) = / det(a) If A and B are 3 3 matrices with det A = 2 and det B = 4, then det(5ab T A 3 BA T B ) has the value A) 25 B) -25 C) 25 D) -25 E) none of the above Solution C) Theorem Let A and B be square matrices, possibly of different sizes Then A X A det = det(a) det(b) and det = det(a) det(b) B X B Proof We have A X = B so But det row) ( I B ) A X, I A X I A X det = det det B B I I A X = det(b) (expand around the first row) and det = det(a) (expand along the last B I Definition Let A be an n n matrix We define the adjoint of A to be adj(a) = [C ij (A)] T ; that is, the transpose of the n n matrix of cofactors of A

28 28 CHAPTER 2 DETERMINANTS AND EIGENVALUES In the first example of this chapter, we found all the cofactors for the matrix 5 7 A = They were C (A) = 48, C 2 (A) = 24, C 3 (A) = 84, C 2 (A) = 56, C 22 (A) = 28, C 23 (A) = 44, C 3 (A) = 5, C 32 (A) = 33, C 33 (A) = 6, The matrix of cofactors is therefore so , adj(a) = Theorem Let A be a square n n matrix () Then A adj(a) = det(a)i = adj(a) A, (2) If det(a), then A = det(a) adj(a) Proof Let u ij be the ijth component of A adj(a) Then by part (4) of a previous theorem (properties of determinants), ) u ij = (a i a i2 a in The proof of (2) is a direct result of () C j C j2 C jn = a ic j + + a in C jn = { det A if i = j, if i j Remark formula From (), we have A adj(a) = det(a)i, so by taking determinants on both sides, obtain the det adj(a) = det(a) n Suppose that A and B are 3 3 matrices and that det(a) = 3, det(b) = 3 Let C = A 3 BA T adj(b 2 )adj(a) Then det(c) is A) 27, B) 27, C), D) E) None of the above

29 2 DETERMINANTS 29 Solution Noting that det[adj(b 2 )] = [det(b 2 )] 2 = det(b) 4 and det[adj(a)] = det(a) 2, we have det(c) = ( ) 3 [det(a)] 6 [det(b)] 5 = 3 So answer is E) For the matrix A of the preceding example, we found det(a) = 324 Therefore A = Theorem (Cramer s Rule) Consider the system AX = B, where A is invertible n n, and write x x n x 2 X = For each i =, 2,, n, let A i (B) be the matrix obtained by replacing the ith column of A by B Then Proof So X = A B = x i = det[a i(b)], i =, 2,, n det(a) det(a) adj(a)b = det(a) C C 2 C n C 2 C 22 C n2 C n C 2n C nn x i = det(a) [b C i + b 2 C 2i + + b n C ni ] = det[a i(b)], det(a) where det[a(b i )] is expanded along the ith column b b 2 b n Solve the system x + 2y w = 3 2x + 4z + 2w = 6 x + 2y z = 6 2x y + z + w = 3 We have A = 2, B = A (B) = 6 2, A (B) = 6, A (B) = ,

30 3 CHAPTER 2 DETERMINANTS AND EIGENVALUES and We get A 4 (B) = det(a) = 24, det(a (B)) =, det(a 2 (B)) = 48, det(a 3 (B)) = 48, det(a 4 (B)) = 24, and so x = / 24 =, y = 48/ 24 = 2, z = 2, w =, the same solution previously obtained by Gaussian elimination 22 Eigenvalues Definition Let A be an n n matrix A number λ is called an eigenvalue of A if there is a non-zero column vector X such that AX = λx X is called an eigenvector corresponding to the eigenvalue λ (Note: We always have A = λ since both sides are That is why we demand X ) How to Find the Eigenvalues of A The equation AX = λx is equivalent to (λi A)X = So λ is an eigenvalue of A iff the homogeneous equation (λi A)X = has a non-zero solution, which is true iff λi A is not invertible, which is true iff det(λi A) = Definition c A (x) = det(xi A) is called the characteristic polynomial of A So λ is an eigenvalue of A iff λ is a root of the polynomial c A (x) (that is, satisfies c A (λ) = ) Useful Notation: Let C be an m n matrix, with columns C,, C n We can write C = [C,, C n ] If A and B = [B, B p ] are m n and n p matrices, we can use this notation to write matrix multiplication as A[B,, B p ] = [AB,, AB p ] Find the eigenvalues of A = Solution The characteristic polynomial of A is x 3 c A (x) = det(xi A) = det = (x )(x 2) 6 = x 2 x 2 2 3x 4 Since x 2 3x 4 = (x +)(x 4), the eigenvalues are λ = and λ 2 = 4 Let us compute an eigenvector for λ We have to solve 3 x x = 2 2 x 2 x 2 We get x = 3 2 x 2, so the solution vector is X = s ( 3 2 ),

31 22 EIGENVALUES 3 with s a parameter Let us compute an eigenvector for λ 2 We have to solve 3 x x = We get x = x 2, so the solution vector is x 2 X 2 = t, with t a parameter s and t can be anything, so take s = 2 and t =, and define 3 X =, X 2 2 = Let P be the matrix and D the diagonal matrix We will also need P = Hence AP = PD It follows that 2 4 P = [X, X 2] = D = x 2 3, 2 4 Then AP = [AX 2 3, AX 2] = [ X, 4X 2], and PD = P AP = D ( 3 2 ) = 4 We have diagonalized A We now want to see if all this works when A is n n rather than just 2 2 We have already proved the next theorem Theorem Let A be an n n matrix () The eigenvalues λ of A are the roots of the characteristic polynomial c A (x) of A (2) The eigenvectors corresponding to the eigenvalue λ are the nonzero solutions to the homogeneous system (λi A)X = of linear equations with λi A as coefficient matrix The eigenvectors can be found by Gaussian elimination Facts Let A be an n n matrix c A (x) is a polynomial of degree n It has n roots, not necessarily distinct Not all the roots have to be real They can be complex So this means A has n eigenvalues, not necessarily distinct Not all the eigenvalues have to be real They can be complex Examples () Let A = Then c A (x) = det x = (x ) x 2 = x 2 2x + So there are two roots (and therefore eigenvalues) λ = and λ = We simply say is an eigenvalue of multiplicity 2

32 32 CHAPTER 2 DETERMINANTS AND EIGENVALUES (2) Let A = Then x c A (x) = det = x x 2 + This polynomial has no real roots, and so A has no real eigenvalues Definition A square matrix D is called a diagonal matrix if the only nonzero components are on the diagonal D looks like λ λ 2 D = Diag(λ,, λ n ) = λ n For example, I is a diagonal matrix The sum and product of two diagonal n n matrices is again diagonal A number times a diagonal matrix is again diagonal Definition A square n n matrix A is called diagonalizable if there is an invertible n n matrix P such that P AP is a diagonal matrix Remark Not every square matrix can be diagonalized Why is it useful to diagonalize a matrix? Suppose we are given a square matrix A and we need to compute powers of A like A k, where k could be very large If A can be diagonalized by some matrix P, then P AP = D where D is diagonal This can also be written as A = PDP Then A 2 = PDP PDP = PDIDP = PD 2 P, and more generally A k = PD k P But powers like D k of a diagonal matrix are easy to compute Theorem Let A and P be n n matrices, and let D = Diag(d,, d n ) Write P = [C,, C n ] where C,, C n are the columns of P Then AP = PD if and only if AC i = d i C i for all i =,, n Proof Let D = [D,, D n ] We have AP = A[C,, C n ] = [AC,, AC n ], PD = [PD,, PD n ], so AP = PD iff AC j = PD j for all j But PD j = d j C j for all j, which finishes the proof Theorem If A is an n n matrix with n distinct eigenvalues, then A is diagonalizable Note: For an eigenvalue λ of A, its multiplicity as a root of the characteristic polynomial is called its algebraic multiplicity, and the number of basic eigenvectors corresponding to λ (which is the number of parameters in the solution of (λi A)X = ) is called its geometric multiplicity Theorem A square n n matrix A is diagonalizable iff the algebraic multiplicity of every eigenvalue λ of A equals its geometric multiplicity In this case, the basic solutions X,, X n of the system become the columns of the diagonalizing matrix P; that is, P = [X, X 2,, X n ], and is invertible

33 22 EIGENVALUES 33 This is example 6 on p 95 of the textbook Diagonalize A = Solution We have x xi A = x x After some work, we find c A (x) = det(xi A) = (x 2)(x + ) 2 The eigenvalues are 2 and, the latter of multiplicity 2 First consider the eigenvalue 2 We have to solve (2I A))X =, which is 2 x 2 x 2 = 2 x 3 By Gaussian elimination, the solution is X = t, so we will take X = Next, consider the eigenvalue We have to solve ( I A))X =, which is x x 2 =, or simply x + x 2 + x 3 = The solution is X = s + t, so we will take x 3 X 2 =, X 3 = These are two basic eigenvectors corresponding to the eigenvalue, which is of multiplicity 2 Hence by the theorem, A is diagonalizable with 2 P = [X, X 2, X 3 ] =, and D = Find A if A =

34 34 CHAPTER 2 DETERMINANTS AND EIGENVALUES Solution Since P = 3 2, we have 2 A = PD P = =

35 Chapter 3 Vector Geometry Definition A scalar quantity is a number, like 4 degrees centigrade A vector quantity has both magnitude and direction For example, a velocity of 4 mph in a northwesterly direction The 4 mph part is called the speed 3 Coordinate-Free and Coordinate Approaches to Geometry 3 Coordinate-Free Geometry () The notation AB means a vector (because of the arrow on top) going from A to B whose direction is A to B and whose magnitude (denoted by AB ) is the length of the line between A and B (2) Two vector quantities are the same if they both have the same magnitude (length) and direction They are parallel if they have the same or opposite direction Since a vector depends only on its direction and magnitude, and not the particular positions of its tip or tail, we will frequently denote vectors by single letters like v Operations on Vectors () Multiplication by a Scalar Suppose c is a scalar, like c = 23 or c = 58 If c >, c AB is a new vector in the same direction whose length is c times the old length If c <, c AB { is a new vector c if c, in the opposite direction whose length is c times the old length, where c = c if c < In particular, AB= BA CD, we move the vectors so that the tails coin- (2) Addition of Vectors To add two vectors AB and cide There results a parallelogram and AB + method Equivalently, we can move the tail of one (say CD is the diagonal This is called the parallelogram CD) to the head of the other AB 35

36 36 CHAPTER 3 VECTOR GEOMETRY Then AB + CD= AD This is the tip-to-tail method The two methods are obviously equivalent Show that the diagonals of a parallelogram bisect each other Solution Let M be the midpoint of AC M can be found using a compass Then AM= MC, so BM= BA + AM= CD + MC= MC + CD= MD Hence (i) M is on BD, and (ii) BM and MD have the same length, so M is the midpoint of BD If we have a great many vectors, this graphical method of combining them is pretty labour intensive Since it doesn t matter where the vectors are located (only length and direction), we can think of them as starting in the same place (which we will call the origin), and then use this origin to create a coordinate system in the space in which the vectors occur Coordinate Geometry To create a coordinate system, we first choose an origin O In two dimensions, we choose two perpendicular lines through O called the x-, and y-axes respectively In three dimensions, we choose three lines through O perpendicular to each other called the x-,y-, and z-axes Every point is completely specified by giving its coordinates with respect to these axes Thus, in two dimensions, a point P with coordinates (x, y ) would be called P(x, y ) In three dimensions, P(x, y, z )

37 3 COORDINATE-FREE AND COORDINATE APPROACHES TO GEOMETRY 37 Let P = P(x, y, z) be the point in R 3 with coordinates x, y, z The position vector corresponding to P is OP Obviously there is a correspondence between points and position vectors A vector is in its standard position if its tail is at the origin We can describe a vector v by placing it in its standard position and then recording the coordinates x, y, z of its tip Let us write this coordinate x x representation of v as the 3 matrix y In two dimensions, we would represent v as y z x x Theorem Let v and w be vectors with coordinate representations y and y respectively, and z z let c be a scalar Let v + w (tip to tail) and c v result from the coordinate-free operations Then v + w x + x cx has coordinate representation y + y and c v has representation cy z + z cz Thus the coordinate-free operations of scalar multiplication and addition of vectors give the same results as the corresponding matrix operations in the coordinate system So from now on, we will think of vectors as 3 (or 2 ) matrices x Moreover, = is the zero vector, and note that v = y z x Theorem Let v = y and w = z x y z be vectors Then () v = w as vectors (ie same magnitude and direction) iff x = x, y = y, and z = z,

38 38 CHAPTER 3 VECTOR GEOMETRY (2) v = x 2 + y 2 + z 2 (3) v = iff v =, (4) av = a v for any scalar a Proof () Place both v and w with tails at the origin Then both tips coincide, so the coordinates of the tips must coincide (2) This follows from the RHS of the previous figure and Pythagorus s theorem Theorem Let P(x, y, z ) and Q(x 2, y 2, z 2 ) be two points in R 3 Then x 2 x () PQ= y 2 y, z 2 z (2) The distance between P and Q is PQ = (x x 2 ) 2 + (y y 2 ) 2 + (z z 2 ) 2 Proof From the picture, OQ= OP + PQ, so PQ= OQ OP= x 2 y 2 z 2 x y z x 2 x = y 2 y z 2 z Theorem Let v and w be non-zero vectors The following statements are equivalent: () v and w are parallel (ie have the same or opposite direction), (2) each of v and w is a scalar multiple of the other, (3) one of v and w is a scalar multiple of the other 32 Dot Product and Projections Definition Let v and w be vectors in R 3 Then v w def = (v, v 2, v 3 ) w w 2 w 3 = v w + v 2 w 2 + v 3 w 3, is called the dot product (also called the scalar product)of v and w If v and w are vectors in R 2, then v w is similarly defined as v w = v w + v 2 w 2

39 32 DOT PRODUCT AND PROJECTIONS 39 Theorem Let u, v, w be vectors and c a number Then () v w is a number, (2) v w = w v, (3) v = v =, (4) v v = v 2, (5) (a v) w = a( v w) = v (a w), (6) v ( u + w) = v u + v w, (7) v ( u w) = v u v w Definition Let vectors v and w be positioned with a common tail Then they determine uniquely an angle φ between and π (radians) Lemma [Law of Cosines] If a, b, and h are the sides of a triangle and φ is the angle opposite h, then h 2 = a 2 + b 2 2ab cos φ Proof We have p = a sin φ and q = b a cos φ, so h 2 = p 2 + q 2 = a 2 sin 2 φ + b 2 2ab cos φ + a 2 cos 2 φ = a 2 + b 2 2ab cos φ Theorem Let φ be the angle between nonzero vectors v and w Then v w = v w cos φ Proof The law of cosines gives v w 2 = v 2 + w 2 2 v w cos φ, whereas we also have v w 2 = ( v w) ( v w) = v v v w w v + w w = v 2 2 v w + w 2 Note: If v and w are nonzero, then cos φ = v w v w

40 4 CHAPTER 3 VECTOR GEOMETRY Definition Vectors v and w are orthogonal (and we write v w) if the angle φ between them is π/2 Theorem Let v and w be nonzero vectors Then v and w are orthogonal v w = Theorem (Projection Theorem) Let v and d be vectors with d Then v can be written uniquely in the form v = v + v 2, where v is parallel to d and v 2 is orthogonal to d We have v d v = d d 2 Proof Take v = td, where t is a number to be determined Note that v is parallel to d Define v 2 = v v = v td Then we want = v 2 d = ( v t d) d = v d t d 2, so that t = v d d 2 Definition Let v and d be vectors with d The vector v d proj d ( v) = v = d d 2 is called the projection of v on d 3 Let v = 4, d = 3 Write v as a sum v + v 2 where v is parallel to d and v 2 d 9 Solution Then Since v d = = 5 and d 2 = =, we have v d v = d d = 3 d v 2 = v v = =, which has length v 2 = = 67 2

41 33 LINES AND PLANES 4 33 Lines and Planes 33 Lines Given a point P (x, y, z ), there is exactly one line L through P parallel to a given nonzero vector d, which will be called a direction vector for the line Equation of a Line: Let p = OP = x y z which is so iff p p = t d, or equivalently Then a point P(x, y, z) is on L iff p p is parallel to d, p = p + t d, (3) for some number t This is the equation of a line in vector form The two-dimensional case is the a same, except the z coordinate is missing If we write d = b, (3) can be written c or equivalently x y = z x y z a + t b, c x = x + ta y = y + tb z = z + tc These are the scalar equations of a line The point P(x, y, z) lies on the line iff there is a number t for which these equations hold Let L be the line through P (x, y, z ) with direction vector d = P(x, y, z ), find the point Q(x, y, z) in L which is closest to P a b Given the point c Solution Q will be the point in L such that the vector QP is orthogonal to d () Q belongs to L, so x = x + ta y = y + tb z = z + tc

42 42 CHAPTER 3 VECTOR GEOMETRY (2) QP is orthogonal to d, so x x a y y b = z z c Numerical Let L be the line through P (6,, 3) with direction vector d = 3 Find the point Q(x, y, z) in L which is closest to the point P(3, 5, 6) Then find the shortest distance from P to the line L Solution () x = 6 t y = + 3t z = 3 (2) QP is orthogonal to d, so 3 x 5 y 3 =, 6 z so (x 3) + 3(5 y) = (2) gives (x 3)+3(5 y) = Substituting from () gives 5 t =, so t = 3 2 Then from (), we have 9 x = 9 2, y = 2, and z = 3 Hence QP= OP OQ= = 2, so the shortest distance is QP 67 = 2 Another Method: Let v = P P= OP OP = 5 = 4 Write v as v + v 2, where v d and v 2 d 67 This was the example after the projection theorem It gave v 2 = 2 For the point Q itself, we have OQ= OP v = = Planes Equation of a Plane in R 3 Given a vector n and a point P (x, y, z ), there is exactly one plane containing P and which is perpendicular to n n is called a normal vector to the plane

43 33 LINES AND PLANES 43 Equation of a Plane: Let P(x, y, z) be any point in the plane Then ( OP OP ) n gives the vector equation for a plane, which is a Writing n = b, this becomes c ( p p ) n = x x a y y b =, z z c or equivalently a(x x ) + b(y y ) + c(z z ) = This is the scalar equation of a plane It frequently appears as ax + by + cz = k Given a point P(x, y, z ) and a plane ax + by + cz = k, find the point P (x, y, z ) in the plane closest to P Solution It will be the point P in the plane such that P P is parallel to n and passes through P a We have n = b c () P is in the plane gives ax + by + cz = k, (2) P P is parallel to n gives x x a y y = t b z z c Numerical Suppose the plane is 2x y + 3z = 5 and the point is P(, 5, 5) Then n = (2,, 3) T and we have () 2x y + 3z = 5, (2) x 2 5 y = t 5 z 3 (2) gives x = 2t, y = 5 + t, z = 5 3t Substituting into () and solving for t gives t = 2 Then x =, y = 2, and z = 7 2 Then P P= 2, so the distance from P to P is then P P = ( + ( 2 )2 + ( 3 2 )2 =

44 44 CHAPTER 3 VECTOR GEOMETRY Another method: is to use the projection theorem Let Q be any point in the plane Then PP is the projection of PQ on n; that is, PP = PQ n n n 2 In the numerical example, we can take Q(, 5, ), that is position vector OQ= (, 5, ) T, so PQ= (, 5, ) T (, 5, 5) T = (,, 5) T Then PQ n = 7 and n 2 = 4, so PP = 7 n, 4 so PP = 2 n = 2 4 Finally, P= P + P P, so just as before P = P P P= 5 2 = /2, /2 Consider the following two lines 3 L : 2 + t and L 2 : 4 + s 2 2 () Find the distance between these two lines (2) Find the equations of two parallel planes π and π 2 such that L π and L 2 π 2 Solution () The line giving the shortest distance will be perpendicular to both L and L 2 Let this line have direction vector n Either by inspection, or by solving n d = and n d 2 = ( d and d 2 are the direction vectors of L and L 2 ) we find n = (,, ) T Let A and B be any points on L and L 2 respectively, say 3 A = 2, B = 4 Then The projection of AB on n is AB= = 2 AB n n 2 with length 2 2 This is the shortest distance n = 4 2,

45 34 THE CROSS PRODUCT 45 (2) π and π 2 must have the same normal vector n = (,, ) T Thus the equations are π : x +y = k and π 2 : x + y = k 2 Since π and π 2 contain the points A and B, then k = 3 and k 2 = 7 34 The Cross Product (Also called the Vector Product) Let v = x y z and w = x 2 y 2 z 2 be vectors The vector y y 2 det z z y z 2 y 2 z ( 2 ) v w = (x z 2 x 2 z ) = x x 2 det x y 2 x 2 y z z 2 x x 2 det y y 2 (4) is called the cross product of v and w 2 Suppose v = and w = 2 Then 3 y z 2 y 2 z 7 v w = (x z 2 x 2 z ) = 5 x y 2 x 2 y 3 An Easier Formula Define the vectors i =, j =, k = These are called the standard basis vectors for R 3 Then i j k i x x 2 v w = det x y z = det j y y 2 x 2 y 2 z 2 k z z 2 Same Example Again i j k v w = det 2 3 = i det 2 3 j det k det 2 = 7i 5j + 3k =

46 46 CHAPTER 3 VECTOR GEOMETRY Proposition (Properties of the Cross Product) Let u, v, w be vectors Then () v w is a vector, (2) v = = v, (3) v v =, (4) w v = v w, (5) (a v) w = a( v w) = v (a w) for any number a, (6) v ( u + w) = ( v u) + ( v w), (7) ( u + w) v = ( u v) + ( w v) Proof These are all proved using properties of determinants In particular, for (6), we have (writing u = (x 3, y 3, z 3 ) T ), i j k i j k i j k v ( u+ w) = det x y z = det x y z +det x y z = ( v w)+( v u), x 2 + x 3 y 2 + y 3 z 2 + z 3 x 2 y 2 z 2 x 3 y 3 z 3 where the second equality comes from expanding the determinant around the third row Theorem (The Lagrange Identity) If v and w are vectors, then v w 2 = v 2 w 2 ( v w) 2 Proof From the first equality in (4), the given identity in component form is (y z 2 y 2 z ) 2 + (x z 2 x 2 z ) 2 + (x y 2 x 2 y ) 2 = (x 2 + y 2 + z)(x y2 2 + z2) 2 (x x 2 + y y 2 + z z 2 ) 2, which is easily shown to be true Theorem Let θ be the angle between the nonzero vectors v and w Then () v w = v w sin θ (=the area of the parallelogram determined by v and w) (2) v and w are parallel iff v w = Note: Area of parallelogram=2 area of triangle OAB = 2 2 bh = ab sin φ since h = a sin φ Proof () v w 2 = v 2 w 2 ( v w) 2 = v 2 w 2 ( v w cos θ) 2 = v 2 w 2 ( cos 2 θ) (2) v and w are parallel iff θ = iff v w = iff v w =