SOLUTIONS: ASSIGNMENT Use Gaussian elimination to find the determinant of the matrix. = det. = det = 1 ( 2) 3 6 = 36. v 4.


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1 SOLUTIONS: ASSIGNMENT 9 66 Use Gaussian elimination to find the determinant of the matrix det = det = det = 1 ( ) 3 6 = 36 = det = det Consider a 4 4 matrix A with rows v 1, v, v 3, v 4 If det(a) = 8, find v 4 det v v 3 v 1 This matrix is obtained from A by a row swap, so its determinant is det(a) = Using the same row vectors as in 61, 6 v 1 + v 4 det v v 3 3 v 1 + v 4 The rows of this matrix are linearly dependent, because (3 v 1 + v 4 ) = 6 v 1 + v 4 A matrix whose rows are linearly dependent must have determinant Find the determinant of the linear transformation T (f(t)) = f(3t ) from P to P A basis for P is {1, t, t } Applying T to each basis element, we have T (1) = 1, T (t) = 3t, and T (t ) = (3t ) = 9t 1t + 4 The matrix of T with respect to this basis is , which has a determinant of 7 1
2 6 Find the determinant of the linear transformation T (f(t)) = f( t) from P n to P n A basis for P n is {t k : 0 k } If k is odd, then T (t k ) = t k, and if k is even then T (t k ) = t k Therefore the determinant of T is ( 1) m where m is the number of odd numbers between 0 and n, inclusive Let N be the remainder when n is divided by 4 A quick check shows that m is even if N is 0 or 3, while m is odd if N is 1 or The n(n + 1) quantity is also even if N is 0 or 3 (because then either n or n+1 is divisible by 4) but is even if N is 1 or Thus the determinant n(n + 1) of T is ( 1) 630 Consider two distinct numbers a and b We define the function f(t) = det a b t a b t (a) Show that f(t) is a quadratic function What is the coefficient of t? Expanding down the third column, we have f(t) = D 1 D t + D 3 t, where the D i are determinants of matrices that contain no factor of t Since D 3 = b a, f(t) is a quadratic function with leading coefficient b a (b) Explain why f(a) = f(b) = 0 Conclude that f(t) = k(t a)(t b), for some constant k Find k, using your work in part (a) If t = a, the first and third columns of the matrix are the same, so it has determinant 0 Likewise, if t = b, the second and third columns of the matrix are the same This shows that f(a) = f(b) = 0 It follows that f(t) has the factors t a and t b; since f(t) is quadratic, it can have no other nonconstant factors, so f(t) = k(t a)(t b) for some constant k This constant is equal to the leading coefficient of f(t) which is b a by part (a) (c) For which values of t is the matrix invertible? A matrix is invertible if and only if its determinant is nonzero In part (b) we found that f(t) = (b a)(t a)(t b), which is nonzero for all values of t except a and b
3 631 Vandermonde determinants: Consider distinct scalars a 0, a 1,, a n We define the (n + 1) (n + 1) matrix A = Vandermonde showed that a 0 a 1 a n a 0 a 1 a n a n 0 a n 1 a n n det(a) = i>j(a i a j ), the product of all differences a i a j where i exceeds j (a) Verify this formula in the case of n = 1 [ ] 1 1 If n = 1 then A = and det(a) = a a 0 a 1 a 0 1 (b) Suppose the Vandermonde formula holds for n 1 You are asked to demonstrate it for n Consider the function a 0 a 1 a n 1 t f(t) = det a 0 a 1 a n 1 t a n 0 a n 1 a n n 1 t n Explain why f(t) is a polynomial of nth degree Find the coefficient k of t n using Vandermonde s formula for n 1 Explain why f(a 0 ) = f(a 1 ) = = f(a n 1 ) = 0 Conclude that f(t) = k(t a 0 )(t a 1 ) (t a n 1 ) for the scalar k you found above Substitute t = a n to demonstrate Vandermonde s formula Expanding down the rightmost column of the matrix, we have f(t) = ±D 0 D 1 t ± + D n t n 3
4 where the D i are determinants of n n matrices that contain no factor of t Since D n is the Vandermonde determinant of the n n matrix with coefficients a 0 through a n 1, we conclude that f(t) is an nth degree polynomial with leading coefficient k = n>i>j (a i a j ) If t = a 0, then f(t) = 0 because the leftmost and rightmost columns of the matrix are equal Applying the same reasoning to a 1 through a n 1, we can see that f(a 0 ) = f(a 1 ) = = f(a n 1 ) = 0 Since the n values a i for 0 i n are all distinct, and f(t) is an nth degree polynomial, it must be the case that f(t) = k(t a 0 )(t a 1 ) (t a n 1 ) for the value of k found above As stated, substituting t = a n into this equation demonstrates Vandermonde s formula for n 63 Find the area of the triangle defined by [ 3 7 ] and [ 8 The area of the triangle formed by two vectors is half the area of the parallelogram formed by the same two vectors Thus, the area is [ ] det = 1 (56 6 ) = Consider the area A of a triangle with vertices Express A in terms of D = det a 1 b 1 c 1 a b c [ a1 a ] ] [ b1, b ] [ c1, c The answer is A = D On one hand, note that A is the same as [ ] [ ] [ ] 0 b1 a the area of the triangle with vertices, 1 c1 a, 1 0 b a c a This area is half the absolute value of the quantity [ ] b1 a det 1 c 1 a 1 = (b b a c a 1 a 1 )(c a ) (c 1 a 1 )(b a ) 4 ]
5 On the other hand, D = (a 1 b b 1 a ) + (c 1 a a 1 c ) + (b 1 c c 1 b ) = (b 1 a 1 )c a 1 c + (a 1 c 1 )b + a 1 b = (b 1 a 1 )(c a ) (c 1 a 1 )(b a ) where 0 = a 1 a a 1 a was added to the equation in the last step 6310 Consider an n n matrix A = [ v 1 v v n ] What is the relationship between the product v 1 v v n and det(a)? When is det(a) = ṽ 1 ṽ ṽ n? Using a fact proven in the textbook (Fact 634), det(a) = ṽ 1 ṽ ṽ n = ṽ 1 ṽ ṽ n with equality if and only if v i = v i for all i; that is, if the v i form an orthogonal basis for R n 636 Consider an n n matrix A with integer entries such that det (A) = 1 Are the entries of A 1 necessarily integers? Explain Yes, A 1 must have integer entries Since det(a) = 1, A 1 = adj(a) But adj(a) has integer entries because A does 71 Let A be an invertible n n matrix and v an eigenvector of A with associated eigenvalue λ Is v an eigenvector of A 1? If so, what is the eigenvalue? We can see that v is an eigenvector of A 1 when we multiply the equation A v = λv by A 1, as follows: A 1 v = 1 λ A 1 (λ v) = 1 λ A 1 (A v) = 1 λ v, so the eigenvalue is 1 λ 713 Let A be an invertible n n matrix and v an eigenvector of A with associated eigenvalue λ Is v an eigenvector of A + I n? If so, what is the eigenvalue? Yes, because (A + I n )( v) = A v + I n v = λ v + v = (λ + ) v, so the eigenvalue is λ + 5
6 715 If a vector v is an eigenvector of both A and B, is v necessarily an eigenvector of A + B? Yes If the eigenvalues for A and B are α and β, respectively, then (A + B)( v) = A v + B v = α v + β v = (α + β) v, so the eigenvalue for A + B is α + β 716 If a vector v is an eigenvector of both A and B, is v necessarily an eigenvector of AB? Yes If the eigenvalues for A and B are α and β, respectively, then (AB)( v) = A(B v) = A(β v) = β(a v) = βα v = αβ v, so the eigenvalue for AB is αβ 7116 Arguing geometrically, find all eigenvectors and eigenvalues of the linear transformation that represents rotation by 180 degrees in R Find a basis of eigenvectors If T is the name of this transformation, then T v = v for every v in R Every vector in R is an eigenvector of T with eigenvalue 1 The standard basis for R is a basis of eigenvectors, for example 7134 Suppose v is an eigenvector of the n n matrix A, with eigenvalue 4 Explain why v is an eigenvector of A + A + 3I n What is the associated eigenvalue? By repeatedly using the equation A v = 4 v, we find that (A + A + 3I n )( v) = A v + A v + 3I n v = 16 v + 4 v + 3 v = 7 v, so the associated eigenvalue is 7 [ ] [ ] Find a matrix A such that and 1 A, with eigenvalues 5 and 10, respectively [ ] a b Write A = We obtain the equations c d are eigenvectors of 3a + b = 15 3c + d = 5 a + b = 10 c + d = 0 The solution to this system is a = 4, b = 3, c =, d = 11 6
7 7144 For m n, find the dimension of the space V of all n n matrices A for which all the vectors e 1,, e n are eigenvectors The dimension of V is m + n(n m) Suppose A is a matrix in V If the corresponding eigenvalues are λ 1,, λ m then the first m columns of A are λ 1 e 1,, λ m e m Each of these eigenvalues represents one free variable The remaining n m columns can be arbitrary vectors in R n, for a total of n(n m) free variables The number of free variables for the entire matrix is thus m + n(n m) 7
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