VIEWS: 0 PAGES: 12 POSTED ON: 7/30/2012
1. (5 points each) State whether the following are true or false, and justify all of your answers. 0 1 4 a. Let A be a 3×3 matrix with eigenvectors v1 = 1 , v2 = 2 , v = 3 1 . A is 5 8 0 diagonalizable. Solution: TRUE. These vectors are linearly independent since: 0 1 4 1 2 1 1 2 1 1 2 1 ~ 0 1 4 ~ 0 1 4 the columns are linearly independent. 5 8 0 0 2 5 0 0 13 Thus we have 3 linearly independent eigenvectors for a 3×3 matrix, which implies that A is diagonalizable by The Diagonalization Theorem. b. det (ATA) ≥ 0. Solution: TRUE. det (ATA) = det (AT)(det A) since det(AB) = (det A)(det B) = (det A)(det A) since det A = det (AT) = (det A)2 ≥ 0 since det A R c. It is possible for a 3×3 matrix with real entries to have the eigenvalues 2, -3, and i. Solution: FALSE. Complex eigenvalues of a real-valued matrix come in conjugate pairs, so -i would also need to be an eigenvalue. 1. (continued) d. If J is the Jordan canonical form of a matrix A, then A is similar to J. Solution: TRUE. If J is the Jordan canonical form of a matrix A, then A = PJP-1, which implies that A is similar to J. 1 2 3 10 e. Let A = 2 3 7 . The vector v = 1 is in Nul(A). 3 6 9 2 Solution: FALSE. A vector v is in Nul(A) if Av = 0. 1 2 3 10 2 Here, Av = 2 3 7 1 = 3 0. Thus v Nul(A). 3 6 9 2 6 f. Each eigenvalue of A is also an eigenvalue of A2. Solution: FALSE. 1 0 2 1 0 If A = , then the eigenvalues of A are 1 and 2. But A = 0 4 , and its 0 2 eigenvalues are 1 and 4. Thus the eigenvalue 2 of A is not an eigenvalue of A2. In general, if is an eigenvalue of A, then 2 is an eigenvalue of A2. 1 0 5 2. Let u = 2 , v = 4 , w = 6 be vectors in R3. 3 2 7 a. (5 points) What is the distance from u to w? Solution: 4 dist(u, w) = ║u – w║ = ║ 4 ║ = 16 16 16 = 48 = 4 3 4 b. (10 points) Find the component of u orthogonal to v. Solution: Let L = Span{v}. 1 1 0 1 2 – u v v = 2 – 2 4 = 1 .6 u – projLu = vv 20 3 3 2 3 .2 c. (10 points) Find the projection of u onto the subspace of R3 spanned by v and w. [NOTE: To use the nice dot product formula, what kind of set/basis must you have??] Solution: We must have an orthogonal basis for Span{v, w} – we use Gram-Schmidt! 5 0 5 wv 6 – 10 4 = 4 . Let v1 = v; v2 = w – v = vv 20 7 2 8 Then, if W = Span{v, w}: 0 5 37 21 37 21 u v1 u v2 v 2 = .4 + 37 190 38 v1 + projWu = v1 v1 v2 v2 105 4 = 105 = 21 .2 8 275105 55 21 d. (10 points) Find the distance from u to the subspace of R3 spanned by v and w. Solution: Let W = Span{v, w}. 1 37 21 16 21 (16) 2 (4) 2 (8) 2 dist(u, W) = ║u – projWu║ = ║ 2 – 38 ║ = ║ 4 ║ = 21 21 21 3 55 21 8 21 336 4 21 = = 21 21 3. (20 points) Let u1, …, up be an orthogonal basis for a subspace W of Rn, and let T: Rn→ Rn be defined by T(x) = projW x. Show that T is a linear transformation. Solution: Since u1, …, up is an orthogonal basis, we can write: x u1 x up projW x = u1 + … + up u 1 u1 up up To show that T is a linear transformation, we must show 2 things: (i) T(x + y) = T(x) + T(y): ( x y ) u1 ( x y) u p T(x + y) = u1 + … + up u1 u 1 up up x u1 y u1 x up y up = u1 + … + up (prop’s of dot product) u1 u1 up up x u1 x up y u1 y up = u1 + … + up + u1 + … + u p (prop’s of vectors) u 1 u1 up up u 1 u1 up up = T(x) + T(y) (ii) T(cx) = cT(x): (c x ) u1 (cx) u p T(cx) = u1 + … + up u1 u1 up up c( x u1 ) c(x u p ) = u1 + … + up (prop’s of dot product) u1 u1 up up x u1 x up = c (prop’s of vectors) u1 u1 up up = cT(x) Thus, T is a linear transformation. 4. Let a subspace U of R5 be defined by: U = {(x1, x2, x3, x4, x5): x1 = 3x2, x3 = 7x4} a. (15 points) Find a basis for U. Solution: Vectors in U look like: 3 x 2 3 0 0 3 0 0 x 1 0 0 2 1 0 0 7x 4 = x2 0 + x4 7 + x5 0 . Therefore a basis is: 0 , 7 , 0 0 1 0 x4 0 1 0 x5 0 0 1 0 0 1 b. (5 points) What is the dimension of U? Solution: The dimension of U is the number of elements in the basis, which is 3. 5. Given the quadratic form on R3: Q(x) = x12 + 25x22 – 10x1x2 a. (5 points) What is the matrix of this quadratic form? Solution: 1 5 A= 5 25 b. (15 points) Make a change of variable that transforms this quadratic form into one with no cross- product terms. Tell me what the P matrix is, and write what the new quadratic form looks like in the new variable. Solution: To make a change of variable to satisfy the conditions that we want, we find a P such that x = Py, and with yT(PTAP)y a quadratic form with no cross product terms (i.e. with PTAP a diagonal matrix). So, we want to orthogonally diagonalize A! Meaning we find an orthonormal basis for R2 such that PTAP is diagonal, where P is the matrix whose columns are the orthonormal basis. The eigenvalues of A are the roots of the polynomial: (1 – )(25 – ) – 25 = 2 – 26 = ( – 26) eigenvalues are 0 and 26 Now we find bases for the eigenspaces: 1 5 1 5 5 For = 0: A – 0I = ~ basis is 1 5 25 0 0 25 5 1 1 1 For = 26: A – 26I = basis is 5 5 ~ 5 1 0 0 Since these eigenvectors correspond to different eigenvalues, they are orthogonal. So to get an orthonormal basis we just normalize. 5 26 1 u1 = 1 , u2 = 5 26 26 26 5 1 P= 1 26 5 26 2 2 and the new quadratic form is 0y1 + 26y2 . 26 26 6. (20 points) Suppose that T1, …, Tn are injective linear transformations such that T1◦…◦ Tn makes sense (i.e. the codomain of Ti+1 is the domain of Ti.) Prove that T1◦…◦ Tn is injective. [Recall: T1◦…◦ Tn(x) = T1(…(Tn(x))…) = composition of functions] Solution: We can prove this in a few ways… I’ll do 2: (i) A transformation T is one-to-one if an only if ker(T) = {0} Since T1, …, Tn are injective, ker(T1) = {0}, …, ker(Tn) = {0}. So now, what is in the kernel of T1◦…◦ Tn? Assume T1◦…◦ Tn(x) = T1(…(Tn(x))…) = 0. We show that x = 0. T1 injective T2(…(Tn(x))…) = 0. T2 injective T3(…(Tn(x))…) = 0. …………. Tn-1 injective Tn(x) = 0. Tn injective x = 0. Thus, T1◦…◦ Tn(x) = 0 x = 0, which means that ker(T1◦…◦ Tn) = {0}, and T1◦…◦ Tn is injective. (ii) A transformation T is one-to-one if and only if [T(x) = T(y) x = y] So Ti(x) = Ti(y) x = y for all i = 1, …, n. Assume T1◦…◦ Tn(x) = T1◦…◦ Tn(y). We show that x = y. T1 injective T2(…(Tn(x))…) = T2(…(Tn(y))…) T2 injective T3(…(Tn(x))…) = T3(…(Tn(y))…) …………. Tn-1 injective Tn(x) = Tn(y) Tn injective x = y Thus T1◦…◦ Tn(x) = T1◦…◦ Tn(y) x = y , which means that T1◦…◦ Tn is injective. 7. a. (5 points) What is the definition of orthogonally diagonalizable? Solution: A matrix A is orthogonally diagonalizable if there exist an orthogonal matrix P and a diagonal matrix D such that A = PDP-1 = PDPT. 3 2 4 b. (20 points) Orthogonally diagonalize the matrix A = 2 6 2 , given that its 4 2 3 eigenvalues are 7 and -2. Solution: We want to find an orthonormal basis for R3. Since we have the eigenvalues already, we find bases for the eigenspaces: 4 2 4 4 2 4 1 1 2 1 For = 7: A – 7I = 2 1 2 ~ 0 0 0 ~ 0 0 0 4 2 4 0 0 0 0 0 0 1 1 basis is 2 , 0 0 1 5 2 4 1 4 1 1 4 1 1 0 1 For = -2: A + 2I = 2 8 2 ~ 5 2 4 ~ 0 1 1 ~ 0 2 1 1 2 4 2 5 4 2 5 0 0 0 0 0 0 1 2 1 basis is 2 OR 1 1 2 The basis that we found for the eigenspace of = 7 is not orthogonal, so we use Gram- Schmidt to make it orthogonal: 1 1 1 4 5 1 2 v1 = 2 ; v2 = 0 – 2 = 5 5 0 1 0 1 2 So {v1, v2} is an orthogonal basis for the eigenspace of = 7, and {v1, v2, 1 } is an 2 2 orthogonal basis for R since 1 corresponds to a different eigenvalue than v1 and v2, 3 2 which implies that it is orthogonal to both v1 and v2. Now we just have to normalize: 1 5 4 45 2 3 u1 = 2 5 , u2 = 2 45 , u3 = 1 3 . 0 5 23 45 Thus A = PDPT for 1 5 4 45 23 7 0 0 0 7 0 . P= 2 5 2 45 1 3 and D = 0 5 2 0 0 2 45 3 8. (15 points) Show that two vectors u and v of an inner product space V are orthogonal if and only if ║u – v║2 = ║u║2 + ║v║2. Solution: First, we note that: ║u – v║2 = <u – v, u – v > (by the def of ║u║2) = <u, u> – <u, v> – <v, u> + <v, v> (by prop’s of inner product) = <u, u> – 2<u, v> + <v, v> (by prop’s of inner product) = ║u║2 – 2<u, v> + ║v║2 (by the def of ║u║2) Then ║u – v║2 = ║u║2 + ║v║2 -2<u, v> = 0 <u, v> = 0 u and v are orthogonal. 9. a. (5 points) State the Cauchy-Schwarz Inequality. Solution: For all u, v in a vector space V, <u, v> ║u║║v║. b. (5 points) Circle one: “Cauchy” is pronounced cow-shee coh-shee cow-chee a 1 ab a2 b2 2 c. (10 points) Let u = and v = 1 . Show that 2 . b 2 Solution: We use Cauchy-Schwarz! <u, v> = a + b ║u║ = a2 b2 ║v║ = 1 1 2 Then Cauchy-Schwarz a + b ( a 2 b 2 )( 2 ) (a + b)2 (a2 + b2)(2) (a b) 2 a2 + b2 2 (a b) 2 a2 b2 22 2 ab a2 b2 2 . 2 2