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201s06mt2

# 201s06mt2 - B U Department of Mathematics Math 201 Matrix...

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Unformatted text preview: B U Department of Mathematics Math 201 Matrix Theory Spring 2006 Second Midterm This archive is a property of Bo˘azi¸i University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. gc This archive is a non-proﬁt service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-proﬁt purpose may result in severe civil and criminal penalties. 1. Let T : R3 → R3 be a linear transformation satisfying; T (1, 0, 0) = (3 − k, −1, 0) T (0, 1, 0) = (−1, 2 − k, −1) T (0, 0, 1) = (0, −1, 3 − k ), where k is a real number. What should k be so that the dimension of the range of T is 2? (15 points) Solution: ⎤ 3 − k −1 0 For the matrix A = ⎣ −1 2 − k −1 ⎦ of the given transformation, R(T), the range of T is 0 −1 3 − k spanned by the column vectors of T. Since its dimension is 2, the column vectors are linearly dependent and so det(A) = 0. Using cofactor expansion along the ﬁrst row gives ⎡ detA = (3 − k ) 2 − k −1 −1 3 − k +1 −1 −1 0 3−k = (3 − k )(k 2 − 5k + 5) − (3 − k ) = (3 − k )(k 2 − 5k + 4) = (3 − k )(k − 4)(k − 1) and detA = 0 if k = 1, 3, or 4. ⎡ ⎤ 101 2. Let A = ⎣ 0 1 1 ⎦. 11b a. Find b so that A has zero as an eigenvalue;(6 points) For λ = 0 to be an eigenvalue det(A − λI ) = detA = 0 must hold. Expanding along the ﬁrst column we get 11 01 detA = + = b − 1 − 1 = b − 2 = 0 if b = 2 1b 11 b. Find the other eigenvalues and all the corresponding eigenvectors when b has the value found in part (a) above. (15 points) 1−λ 0 1 1 0 1−λ For b = 2, det(A − λI ) = 1 1 2−λ if (1 − λ)[(1 − λ)(2 − λ) − 1] − (1 − λ) = 0 if (1 − λ)(λ2 − 3λ) = λ(1 − λ)(λ − 3) = 0 So λ = 0, 1 or 3 are eigenvalues. =0 i For λ =⎤ : ⎡ ⎤ ⎡ 0 ⎤⎡ ⎤⎡ ⎡ x1 101 0 x1 A ⎣ x2 ⎦=⎣ 0 ⎦ if ⎣ 0 1 1 ⎦ ⎣ x2 ⎦=⎣ 000 0 x3 x3 x + x3 = 0 if 1 x2 + x3 = 0 if x1 = x2 = −x⎤ 3 ⎡ −1 ⇒ x = a ⎣ −1 ⎦ , a ∈ R, a = 0 (eigenvectors 1 ⎤ 0 0⎦ 0 corresponding to λ = 0). For λ = 1: ⎡ ⎤⎡ ⎤⎡ ⎤ x1 0 001 (A − I )x = ⎣ 0 0 1 ⎦ ⎣ x2 ⎦=⎣ 0 ⎦ if x3 = 0 and x1 + x2 = 0 0 1⎤ 1 1 x3 ⎡ −1 ⎣ 1 ⎦ , b ∈ R, b = 0 (eigenvectors corresponding to λ = 1). ⇒x=b 0 ⎤⎡ −2 0 1 (A − 3I )x = ⎣ 0 −2 1 ⎦ ⎣ 1 1 −1 ⎡ ⎤ ⎡ −2 0 1 −2 0 ⎣ 0 −2 1 ⎦ → ⎣ 0 0 if 0 1 −1 01 2 −2x1 + x3 = 0 if x2 − 1 x3 = 0 2 if x1 = 1 x3 = ⎤ 2 x 2⎡ ⎡ For λ = 3: ⇒ x = c⎣ 1 2 1 2 ⎤⎡ x1 x2 ⎦=⎣ x3 ⎤⎡ 1 0 ⎦⎣ −1 2 ⎤ 0 0⎦ 0 ⎤⎡ ⎤ x1 0 ⎦=⎣ 0 ⎦ x2 x3 0 ⎦ , c ∈ R, c = 0 (eigenvectors corresponding to λ = 3). 1 3. Decide whether the followings are TRUE or FALSE. If true prove; if false, give a counter example or explain.(15 points) i. For A and B be both invertible nxn matrices (A + B )−1 = A−1 + B −1 ; 10 −1 0 ,B= , A and B are invertible since detA = detB = 1 = 0 01 0 −1 00 But A + B = is not invertible since det(A + B ) = 0. 00 So (A + B )−1 is meaningless since A+B is not invertible. FALSE: For A = ii. Transformation T : R3 → R3 deﬁned by T (x, y, z ) = (x, 2) is linear. FALSE: Since T (0, 0, 0) = (0, 2), ie T (0) = 0, hence T is not linear. ii ⎡ ⎤ ⎡ ⎤ 1 1 iii. The orthogonal projection of u = ⎣ 1 ⎦ on the subspace of R3 spanned by v = ⎣ −1 ⎦ is u. −1 0 FALSE: Since uT v = 1 − 1 + 0 = 0 we have u⊥v . So the orthogonal projection of u on the subspace of R3 spanned by v (ie on v line) must be the zero vector. 4. Let A = [aij ] be an nxn matrix such that a1,n = 1, a2,n−1 = 1, . . . , an,1 = 1 and ai,j = 0 otherwise; i. Write down the matrix A. ⎤ ⎡ 0 ··· 1 ⎥ ⎢. . ⎥ ⎢. A=⎢ .⎥ .⎦ ⎣ . 1 1 ··· 0 ii. Evaluate the determinant of A in the case that n=7. detA = (−1)(−1)(−1)detI = −1 since we exchange r1 ↔ r7 , r2 ↔ r6 , r3 ↔ r5 in order to obtain the identity matrix I from A. iii. Find a formula for the determinant of A in case that n is an arbitrary positive integer. Consider the number of row exchanges to reduce A to I. For n > 2, if n is even, then n row exchanges are needed. If n is odd, then n−1 row exchanges are 2 2 needed. If the number of row exchanges is even then detA=1. If it is odd, then detA=-1. So for k ∈ Z+ if n = 4k then n = 42k = 2k even ⇒ detA = 1 2 k if n = 4k + 1, then n−1 = 42 = 2k even ⇒ detA = 1 2 if n = 4k + 2, then n = 2k + 1 odd ⇒ detA = −1 2 if n = 4k + 3, then n−1 = 2k + 1 odd ⇒ detA = −1 2 In fact, if n ≡ 0 or 1 (mod4) ⇒ detA = 1 and if n ≡ 2 or 3 (mod4) ⇒ detA = −1. ⎤ x1 ⎢x ⎥ 5. Let W be the subspace of R4 containing all vectors ⎢ 2 ⎥ satisfying x1 + x2 + x3 + x4 = 0. Find ⎣ x3 ⎦ x4 ⊥ a basis for W , the orthogonal complement of W. (16 points) ⎡ Solution: iii ⎡ −x2 − x3 − x4 ⎢ x2 W = {⎢ ⎣ x3 x ⎤ ⎡4 ⎤ ⎡ −1 −1 ⎢1⎥⎢0⎥ ⎥ ⎥⎢ = S(⎢ ⎣ 0 ⎦, ⎣ 1 ⎦, 0 0 ⎤ ⎥ ⎥ : x2 , x3 , x4 ∈ R} ⎦ ⎤ ⎡ −1 ⎢ ⎢0⎥ ⎥= R(A), columnspaceof A = ⎢ ⎢ ⎣ ⎣0⎦ 1 ⎡ ⎤ −1 −1 −1 1 0 0⎥ ⎥ 0 1 1⎦ 0 0 1 We know that R(A)⊥ = N(A⊥) . ⎡ ⎤ ⎡ ⎤ −1 1 0 0 − − − − − − − − − − − − − − − 1 −1 0 0 − − − − − − − − − − − − − −→ AT = ⎣ −1 0 1 0 ⎦ (−1)r1 → r1 , r1 + r2 → r2 , r1 + r3 → r3 ⎣ 0 −1 1 0 ⎦ −1 0 0 1 0 −1 0 1 ⎤ ⎡ − − − − − − − − − → 1 −1 0 0 −−−−−−−−− (−1)r2 → r2 , r2 + r3 → r3 ⎣ 0 1 −1 0 ⎦ 0 0 −1 1 ⎡ ⎡ ⎤ ⎤⎡ ⎤ ⎤ x1 ⎡ x1 0 1 −1 0 0 ⎢ ⎢ x2 ⎥ ⎥ ⎢0⎥ T ⎥ ⎦ ⎢ x2 ⎥ ⎢ ⎥ ⎣ So x = ⎢ ⎣ x3 ⎦ ∈ N(A ) if 0 1 −1 0 ⎣ x3 ⎦=⎣ 0 ⎦ 0 0 −1 1 x4 x4 0 if x1 − x2 = x2 − x3 = −x3 + x4 = 0 if x3 = x2 = x1 = x4 ⎤ ⎡ 1 ⎢1⎥ ⎢ Hence N(AT ) = {x4 ⎢ ⎥ : x4 ∈ R} = S(⎢ ⎣1⎦ ⎣ 1 ⎡ 6. ⎡ 01 ⎣1 0 i. Find QR-decomposition of A= 00 triangular matrix.(17 points) Solution: ⎤ 1 1⎥ ⎥)= W ⊥ . 1⎦ 1 1 √ 2 ⎤ 0 ⎦, where Q is an orthogonal matrix and R is an upper 1 √ 2 ⎤ ⎡⎤ ⎡1⎤ √ 0 1 2 The vectors a= ⎣ 1 ⎦, b= ⎣ 0 ⎦, c= ⎣ 0 ⎦ are linearly independent. 1 √ 0 0 2 ⎡ Take q1 = a and q2 = b, since a and b have unit lengths and a⊥b. iv c T T = c − (q1 c)q1 − (q2 c)q2 = c − (aT c)a − (bT c)b ⎡⎤ ⎡1⎤ √ 1 2 1 = ⎣ 0 ⎦ − √2 ⎣ 0 ⎦ 1 √ 0 ⎡ 2⎤ 0 ⎣ 0 ⎦ sinceaT c = 0, bT c = = 1 √ 2 ⇒ q3 = c c 1 √ 2 ⎡ ⎤⎡⎤ 0 0 √ = 2⎣ 0 ⎦ = ⎣ 0 ⎦ 1 √ 1 2 T T T q1 a = 1 q1 b = 0 q1 c = 0 1 T T q2 b = 1 q2 c = √2 Now 1 T q3 c = √2 ⎡ 01 ⎣1 0 Hence A= 00 ⎤⎡ 10 010 ⎦= ⎣ 1 0 0 ⎦ ⎣ 0 1 0 1 √ 00 001 2 1 √ 2 ⎤ ⎡ 0 1 √ 2 1 √ 2 ⎤ ⎦= QR ii. Find the inverse of the matrix Q found in part i) above. (4 points) Solution: Q−1 = QT ⇒ Q−1 ⎡ ⎤ 010 =⎣ 1 0 0 ⎦=Q 001 v ...
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