Math 115A: Homework 5

Math 115A: Homework 5 1 Suppose U, V, and W are finite-dimensional vector spaces over a field F, and that are linear a) Prove ker ST ) ker T ) b) Prove nullst ) nullt ) c) Prove imst ) im S T : U V, S : V W d) Prove rankst ) min{ranks), rankt )} e) Give examples where the inequalities in b) and d) are actually equality f) Give examples where the inequalities in b) and d) are strict 2 Suppose V and W are finite-dimensional vector spaces over a field F, and that are linear a) Prove ims + T ) ims) + imt ) S : V W, T : V W b) Suppose U 1 and U 2 are subspaces of W Prove that c) Prove ranks + T ) ranks) + rankt ) dimu 1 + U 2 ) dim U 1 + dim U 2 d) Give examples where the inequality in c) is equality, and when it is strict 3 Let V be a vector space over a field R, and let u, v, w V a) Suppose that β = u, v) is a basis for V Recall HW, question ) that β = u+v, u v) is also basis for V What is the change of coordinate matrix from β to β? What is the change of coordinate matrix from β to β? b) Suppose that β = u, v, w) is a basis for V Recall HW, q) that β = u+v, v+w, w+u) is also basis for V What is the change of coordinate matrix from β to β? What is the change of coordinate matrix from β to β? c) Admire how these matrices showed up in HW3, question 7 Suppose T : V W is a linear transformation, that β V = v 1, v 2 ) is a basis for V, and that β W = w 1, w 2, w 3 ) is a basis for W We know that β V = v 1 + v 2, v 1 v 2 ) is a basis for V and that β W = w 1 + w 2, w 2 + w 3, w 3 + w 1 ) is a basis for W 1 2 Suppose that [T ] β W βv = 3 What is [T ] β W β? V 5 6 1

5 a) Calculate the determinant of 2 3 5 7 11 13 17 19 1 2 3 5 6 7 8 8 8 8 8 8 8 8 8 2 2 1 6 1 1 22 26 3 38 8 8 88 88 888 888 8888 8888 b) Let a, b, c R Calculate the determinant of the following matrix 5 5 using part 1 of theorem 22 and its analog for rows 1 1 2 3 5 1 2 a b 3 1 5 1 c b Try and be as efficient as possible I only had to do non-trivial multiplications c) Consider the linear transformation T : P R) P R) defined by Calculate dett ) px) p x) + px) + p) 6 Understand the statements and proofs of theorem 261 and theorem 27 Doing question 7 first might help 7 For each of the following linear transformations T : V V, calculate c T x); list the real) eigenvalues λ 1,, λ n of T ; give their algebraic multiplicity; write down a basis for their eigenspaces; give their geometric multiplicity; say whether T is diagonalizable; if T is diagonalizable, give a basis of eigenvectors; if T is not diagonalizable, give a basis for V containing as many eigenvectors as possible a) T : R 6 R 6 defined by x 1 2 x 1 2 x 2 2 x 3 x 6 3 x 6 2

b) T : R 6 R 6 defined by c) T : R 6 R 6 defined by d) T : R 2 R 2 defined by e) T : R 8 R 8 defined by x 1 2 x 1 2 1 x 2 1 2 x 3 1 x 6 3 x 6 x 1 2 1 x 1 2 x 2 1 2 x 3 1 x 6 3 x 6 x1 ) 1 1 1 1 ) ) x1 x 1 1 1 x 1 1 1 2 1 x x 5 2 x 2 1 x 5 x 6 2 x 6 x 7 3 1 x 7 x 8 3 x 8 f) T : P 2 R) P 2 R) defined by px) p x) g) T : P 2 R) P 2 R) defined by px) p x) + px) h) T : P 2 R) P 2 R) defined by px) p x) + px) + p) 8 Recall that the set of real-valued functions on R is a vector space over R F = {f : R R} Given a R, define f a : R R by f a x) = e ax Let V = span{f a : a R} Define T : V V by f f this is well-defined) Give infinitely many eigenvalues for T, and give a corresponding eigenvector for each 3

9 Define, : R 2 R 2 R by x1 ), y1 a) Check that, is an inner product on R 2 y 2 ) = 2x 1 y 1 + x 1 y 2 + y 1 + y 2 b) Find an othonormal basis for R 2 with this inner product 1 Let V and W be inner product spaces over R Suppose T : V W is a linear transformation T is said to be an isometry iff for all v 1, v 2 V, T v 1, T v 2 = v 1, v 2 Note that on the left side of the last equation, is the inner product on W ; on the right side of the equation, it is the inner product on V a) Show that for all v 1, v 2 V, v 1, v 2 = v 1 + v 2 2 v 1 v 2 2 b) Show that T is an isometry if and only if for all v V, T v = v c) Suppose that T is an isometry Prove that T is injective 11 Consider R with its standard inner product Apply Gram-Schmidt to the -tuple: 1 1 1,, 1 12 Suppose V is a finite-dimensional vector space and that T : V V is a linear transformation 12, 2 ) a) Suppose that T is diagonalizable and that its only eigenvalues are and 1 but repetitions are allowed) i Prove that T 2 = T Hint: check it on a suitably chosen basis ii Recall from question 12 of homework, that V = im T ker T b) Suppose, in addition, that V has an inner product Suppose that T has an orthonormal basis of eigenvectors and its only eigenvalues are and 1 but repetitions are allowed) i Prove that ker T = im T ) Hint: use a suitable basis of V ii Using lecture, we know that V = im T im T ) Using i), this recovers V = im T ker T

c) The T of part a) is often called a projection The T of part b) is often called an orthogonal projection On an inner product space, a projection is an orthogonal projection if and only if it is also self-adjoint There s nothing to do here I m just telling you stuff 5

Solutions 1 Suppose U, V, and W are finite-dimensional vector spaces over a field F, and that are linear T : U V, S : V W a) Let u ker T Then ST )u) = ST u)) = S) =, so u ker ST ) b) This follows from a) c) Let w imst ) Then we can find a u U such that ST )u) = w Let v = T u) Then w = ST )u) = ST u)) = Sv) im S d) The previous part gives rankst ) ranks) b) gives dim U nullst ) dimu) nullt ) which, by rank-nullity, is the same as rankst ) rankt ) Thus, rankst ) min{ranks), rankt )} e) Take U = V = W = {} f) Take U = W = R and V = R 2 Define T : R R 2 by T x) = x, ), and S : R 2 R by Sx 1, ) = 2 Suppose V and W are finite-dimensional vector spaces over a field F, and that are linear S : V W, T : V W a) Let w ims + T ) Then we can find a v V such that S + T )v) = w Thus, w = Sv) + T v) im S + im T b) Suppose U 1 and U 2 are subspaces of W Let m 1 = dim U 1 and m 2 = dim U 2, and choose bases for U 1 and U 2, respectively: u 1,1, u 1,2,, u 1,m1 ) and u 2,1, u 2,2,, u 2,m2 ) Then span{u 1,1, u 1,2,, u 1,m1, u 2,1, u 2,2,, u 2,m2 }) is equal to span{u 1,1, u 1,2,, u 1,m1 }) + span{u 2,1, u 2,2,, u 2,m2 }) = U 1 + U 2 Thus, U 1 + U 2 has a spanning set of size m 1 + m 2 Thus, the dimension of U 1 + U 2 is less than or equal to m 1 + m 2 c) By a) and b), ranks + T ) = dimims + T )) dimims) + imt )) dimims)) + dimimt )) = ranks) + rankt ) d) Let V = W = R When S = T = R, we have equality When S = T = 1 R, we do not have equality 6

3 Let V be a vector space over a field R, and let u, v, w V ) ) 1 1 a), 1 1 1 1 1 2 1 1 1 1 1 1 1 b) 1 1, 1 2 1 1 1 1 1 1 1 1 1 2 c) Admiration happened 1 1 1 1 2 1 1 1 3 1 1 1 5 6 ) 1 1 = 1 1 1 2 5 a), since the the 7-th row is twice the 1-st row b) 1 1 1 3 1 1 1 1 1 7 1 = 1 2 1 1 1 11 1 1 1 2 3 5 1 1 2 3 5 det 2 a b 3 1 = det a b 1 2 5 1 = det a b 1 c b 1 c b 5 1 c b ) ) a b 2 5 = det det c b a b 2 1 1 2 c) dett ) = det 1 3 1 = 2 1 6 Yep 7 a) x 2) x 3) 2 2, 3, 2 e 1, e 2, e 3, e ), e 5, e 6 ), 2 Yes e 1, e 2, e 3, e, e 5, e 6 ) 1 15 1 7 1 = ab bc) 2b 5a) = 5a 2b ab + bc 7

b) x 2) x 3) 2 2, 3, 2 e 1, e 2 ), e 5 ) 2, 1 e 1, e 2, e 3, e, e 5, e 6 ) c) x 2) x 3) 2 2, 3, 2 e 1, e 3 ), e 5 ) 2, 1 e 1, e 2, e 3, e, e 5, e 6 ) d) x 1) 2 + 1 e 1, e 2 ) e) x 1) 2 + 1)x 2) x 3) 2 2, 3, 2 e 3, e 5 ), e 7 ) 2, 1 e 1, e 2, e 3, e, e 5, e 6, e 7, e 8 ) f) 3 1) 1 1, x, ) 8

g) x 1) 3 1 3 1) 1 1, x, ) h) x 1) 2 x 2) 1, 2 2, 1 x 1), 1) 1, 1 1, x 1, ) 8 Every real number a is an eigenvalue of T f a is an eigenvector with corresponding eigenvalue a because d dx eax = ae ax, ie T f a ) = af a 9 Define, : R 2 R 2 R by x1 ), y1 y 2 ) = 2x 1 y 1 + x 1 y 2 + y 1 + y 2 a) I checked privately!) that, is linear in the first variable, and that it has the requisite symmetric For the last property, note that x1 b) If x1 ), y1 y 2 ) = 2 1 + 2x 1 + 2 = 1 + x 1 + ) 2 ), then either x 1, or x 1 = and we must have x 1 + In both cases, we see that x1 ) 1, 1 ) ) 1 ), y1 y 2 ) > 9

1 Let V and W be inner product spaces over R Suppose T : V W is a linear transformation a) Let v 1, v 2 V Then Thus, v 1 + v 2 2 v 1 v 2 2 = v 1 + v 2, v 1 + v 2 v 1 v 2, v 1 v 2 = + v 1, v 1 + v 1, v 2 + v 2, v 1 + v 2, v 2 v 1, v 1 + v 1, v 2 + v 2, v 1 v 2, v 2 = v 1, v 2 b) Suppose T is an isometry and v V Then v 1, v 2 = v 1 + v 2 2 v 1 v 2 2 T v 2 = T v, T v = v, v = v 2 and so T v = v Conversely, suppose that for all v V, T v = v, and let v 1, v 2 V Then T v 1, T v 2 = T v 1 + T v 2 2 T v 1 T v 2 2 = T v 1 + v 2 ) 2 T v 1 v 2 ) 2 so T is an isometry = v 1 + v 2 2 v 1 v 2 2 c) Suppose that T is an isometry, that v V, and T v = Then v = T v = = Thus, v =, and T is injective = v 1, v 2 11 1 1 1 1 = 1 1 1 1 1 1 1 1 = 3 1 1 1 1 1 12 3 1 1 + 3 1 = 8 1 1 1 1 12 6 1 1 + 2 3 1 + 8 = 1 1 12 2 1

12 Suppose V is a finite-dimensional vector space and that T : V V is a linear transformation a) Suppose that T is diagonalizable and that its only eigenvalues are and 1 i First, we ll prove a lemma about T Lemma If v is an eigenvector of T, then T 2 v) = T v) Proof Suppose v is an eigenvector of T with eigenvalue λ Then T v = λv and T 2 v) = T T v)) = T λv) = λt v) = λλv) = λ 2 v Since λ = or λ = 1, λ 2 = λ, so T 2 v) = T v) Now we can answer the question quickly Pick a basis of T consisting of eigenvectors v 1,, v n ) By the lemma, we have T 2 v j ) = T v j ) for each j {1,, n} Thus, T 2 = T ii We can say more than V = ker T im T Let v 1,, v n ) be a basis of V consisting of eigenvectors of T where v 1,, v m E, v m+1,, v n E 1 For j {1,, m}, we have v j ker T v 1,, v m ) is a linearly independent tuple in ker T, and we conclude that nullt ) m For j {m + 1,, n}, we have v j = T v j ) im T v m+1,, v n ) is a linearly independent tuple in im T, and we conclude that rankt ) n m Rank-nullity gives nullt ) = m and rankt ) = n m, so v 1,, v m ) is a basis for ker T and v m+1,, v n ) is a basis for im T Remark By definition, E = ker T We also have E 1 im T because, for v E 1, v = T v im T Since E 1 contains a basis of im T, we have E 1 = im T b) Suppose, in addition, that V has an inner product Suppose that T has an orthonormal basis of eigenvectors and its only eigenvalues are and 1 Let v 1,, v n ) be an orthonormal basis of V consisting of eigenvectors of T where v 1,, v m E, v m+1,, v n E 1 We showed above that v 1,, v m ) is a basis for ker T and v m+1,, v n ) is a basis for im T If v ker T and w im T, we can find λ 1,, λ n F such that m n v = λ i v i, w = λ j v j, and so m v, w = λ i v i, i=1 i=1 n j=m+1 λ j v j = j=m+1 m n i=1 j=m+1 λ i λ j v i, v j = This shows ker T im T ) Finally, we know dimker T ) = dim V dimim T ) = dimim T ) ) So we have equality 11