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Download Homework 2. Solutions 1 a) Show that (x, y) = x1y1 + x2y2 + x3y3
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Homework 2. Solutions 1 a) b) c) d) Show Show Show Show that that that that (x, y) = x1 y 1 + x2 y 2 + x3 y 3 defines a scalar product in R3 . hx, yi = x1 y 1 + x2 y 2 does not define a scalar product in R3 . (x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define a scalar product in R3 . (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 defines a scalar product in R3 . e† ) Find necessary and sufficient conditions for entries a, b, c of symmetrical matrix the formula µ 1 2 (x, y) = ( x , x ) a b b c ¶µ y1 y2 µ a b b c ¶ such that ¶ = ax1 y 1 + b(x1 y 2 + x2 y 1 ) + cx2 y 2 defines scalar product in R2 . Recall that scalar product on a vector space V is a function B(x, y) = (x, y) on a pair of vectors which takes real values and satisfies the the following conditions: 1) B(x, y) = B(y, x) (symmetricity condition) 2) B(λx + µy, z) = λB(x, z) + µB(y, z) (linearity condition (with respect to the first argument)) 3) B(x, x) ≥ 0 , B(x, x) = 0 ⇔ x = 0 (positive-definiteness condition) (The linearity condition with respect to the second argument follows from the conditions 2) and 1)) Check all these conditions for (x, y) = x1 y 1 + x2 y 2 + x3 y 3 : 1) (y, x) = y 1 x1 + y 2 x2 + y 3 x3 = x1 y 1 + x2 y 2 + x3 y 3 = (x, y). Hence it is symmetrical. 2) (λx + µy, z) = (λx1 + µy 1 )z 1 + (λx2 + µy 2 )z 2 + (λx3 + µy 3 )z 3 = = λ(x1 z 1 + x2 z 2 + x3 z 3 ) + µ(y 1 z 1 + y 2 z 2 + y 3 z 3 ) = λ(x, y) + µ(y, z). Hence it is linear. 3) (x, x) = (x1 )2 + (x2 )2 + (x3 )2 ≥ 0. It is non-negative. If x = 0 then (x, x) = 0. If (x, x) = 1 2 (x ) + (x2 )2 + (x3 )2 = 0, then x1 = x2 = x3 = 0, i.e. x = 0. This we proved positive-definiteness. All conditions are checked. Hence B(x, y) = x1 y 1 + x2 y 2 + x3 y 3 is indeed a scalar product in R3 Remark Note that x1 , x2 , x3 —are components of the vector, do not be confused with exponents! Show that B(x, y) = x1 y 1 + x2 y 2 does not define scalar product in R3 . To see that the formula (x, y) = x1 y 1 + x2 y 2 does not define scalar product check the condition 3) of positive-definiteness: (x, x) = (x1 )2 + (x2 )2 may take zero values for x 6= 0. E.g. if x = (0, 0, −1) (x, x) = 0, in spite of the fact that x 6= 0. The condition 3) of positive-definiteness is not satisfied. Hence it is not scalar product. Show that B(x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define scalar product in R3 . To see that the formula (x, y) = x1 y 1 + x2 y 2 − x3 y 3 does not define scalar product check the condition 3): (x, x) = (x1 )2 + (x2 )2 − (x3 )2 may take negative values. E.g. if x = (0, 0, −1) (x, x) = −1 < 0. The condition 3) of positive-definiteness is not satisfied. Hence it is not scalar product. Now show that (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 is a scalar product in R3 . We need to check all the conditions above for scalar product for (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 : 1) (y, x) = y 1 x1 + 3y 2 x2 + 5y 3 x3 = x1 y 1 + 3x2 y 2 + 5x3 y 3 = (x, y). Hence it is symmetrical. 2) (λx + µy, z) = (λx1 + µy 1 )z 1 + 3(λx2 + µy 2 )z 2 + 5(λx3 + µy 3 )z 3 = = λ(x1 z 1 + 3x2 z 2 + 5x3 z 3 ) + µ(y 1 z 1 + 3y 2 z 2 + 5y 3 z 3 ) = λ(x, y) + µ(y, z). Hence it is linear. 3) (x, x) = (x1 )2 + 3(x2 )2 + 5(x3 )2 ≥ 0. It is non-negative. If x = 0 then obviously (x, x) = 0. If (x, x) = (x1 )2 + 3(x2 )2 + 5(x3 )2 = 0, then x1 = x2 = x3 = 0. Hence it is positive-definite. All conditions are checked. Hence (x, y) = x1 y 1 + 3x2 y 2 + 5x3 y 3 is indeed a scalar product in R3 1 e† ) The condition of linearity and symmetricity for the bilinear form µ ¶µ 1 ¶ a b y B(x, y) = ( x1 , x2 ) = ax1 y 1 + b(x1 y 2 + x2 y 1 ) + cx2 y 2 b c y2 are evidently obeyed. The general answer on this question is: symmetric matrix is positive-definite if and only if all principal minors are positive. For matrix under consideration it means that conditions a > 0 and ac − b2 > 0 are necessary and sufficient conditions. Give a proof for this special case. Check the positive-definiteness condition. For x = (1, 0) B(x, x) = a. Hence a > 0 is necessary condition. Now consider B(x, x) = a(x1 )2 + 2bx1 x2 + c(x2 )2 = (ax1 + bx2 )2 + (ac − b2 )(x2 )2 ≥ 0 ⇔ ac − b2 ≥ 0 a We see that B(x, x) > 0µfor all ¶ x 6= 0 iff a > 0 and (ac − b2 ) > 0. T α β 2 The matrix A = obeys the conditions A A = I . Show that γ δ a) det A = ±1 b) if det A = 1 then there exists an angle ϕ : 0 ≤ ϕ < 2π such that A = Aϕ where µ ¶ cos ϕ − sin ϕ Aϕ = (rotation matrix) sin ϕ cos ϕ µ 1 0 c) if¶det A = −1 then then there exists an angle ϕ : 0 ≤ ϕ < 2π such that A = Aϕ R, where R = 0 (a reflection matrix). −1 The answers on this exercise see in lecture notes. 3 Show that for matrix Aϕ defined in the previous exercise the following relations are satisfied: T A−1 ϕ = Aϕ = A−ϕ , Aϕ+θ = Aϕ · Aθ . µ ¶ cos ϕ − sin ϕ We know (see lecture notes) that Aϕ = . Then calculate inverse matrix A−1 ϕ . One sin ϕ cos ϕ µ ¶ T T cos ϕ sin ϕ can see that Aϕ = A−1 , because Aϕ Aϕ = I. On the other hand cos ϕ = cos(−ϕ) and ϕ = − sin ϕ cos ϕ sin ϕ = − sin(−ϕ). Hence µ ¶ µ ¶ T cos ϕ sin ϕ cos(−ϕ) − sin(−ϕ) −1 Aϕ = Aϕ = = = A−ϕ . − sin ϕ cos ϕ sin(−ϕ) cos(−ϕ) Now prove that Aϕ+θ = Aϕ · Aθ : µ ¶µ ¶ µ cos ϕ − sin ϕ cos θ − sin θ (cos ϕ cos θ − sin ϕ sin θ) Aϕ · Aθ = = sin ϕ cos ϕ sin θ cos θ (cos ϕ sin θ + sin ϕ cos θ) µ cos(ϕ + θ) − sin(ϕ + θ) sin(ϕ + θ) cos(ϕ + θ) −(cos ϕ sin θ + sin ϕ cos θ) (cos ϕ cos θ − sin ϕ sin θ) ¶ = ¶ = Aϕ+θ 4 Show that under the transformation (e01 , e02 ) = (e1 , e2 ) Aϕ an orthonormal basis transforms to an orthonormal one. 2 How coordinates of vectors change if we rotate the orthonormal basis (e1 , e2 ) on the angle ϕ = anticlockwise? π 3 We have to check that scalar products (e01 , e01 ) = (e02 , e02 ) = 1 and (e01 , e02 ) = 0. Calculations show that = (cos ϕe1 + sin ϕe2 , cos ϕe1 + sin ϕe2 ) = cos2 ϕ(e1 , e1 ) + 2 cos ϕ sin ϕ(e1 , e2 ) + sin2 ϕ(e2 , e2 ) = 1, = (− sin ϕe1 +cos ϕe2 , − sin ϕe1 +cos ϕe2 ) = 1, (e01 , e02 ) = (cos ϕe1 +sin ϕe2 , − sin ϕe1 +cos ϕe2 ) = 0. (e01 , e01 ) (e02 , e02 ) Now answer the second question. If a = xex + ye2 = x0 e0x + y 0 e02 à and Aϕ = A π3 = then we have: 1 √2 3 2 − √ 3 2 1 2 ! is the matrix of bases transformation à µ 0¶ µ ¶ µ 0¶ 1 x x x = (ex , ey )A π3 = (e0x , e0y ) a = (ex , ey ) = (ex , ey ) √23 0 0 y y y 2 Hence µ ¶ µ 0¶ à 1 x x = A π3 = √23 y y0 2 and µ x0 y0 ¶ −1 = Tπ 3 − √ 3 2 1 2 !µ µ ¶ µ ¶ à 1 x x 2√ = A− π3 = y y − 3 2 x0 y0 √ 3 2 1 2 − √ 3 2 1 2 !µ x0 y0 ¶ ¶ !µ x0 y0 ¶ 5 Let {e1 , e2 , e3 } be an orthonormal basis of Euclidean space E3 . Consider the ordered set of vectors which is expressed via basis {e1 , e2 , e3 } as in the exercise 6 of the Homework 1. {e01 , e02 , e03 } Write down explicitly transition matrix from the basis {e1 , e2 , e3 } to the ordered set of the vectors What is the rank of this matrix? Is this matrix orthogonal? {e01 , e02 , e03 }. Find out is the ordered set of vectors {e01 , e02 , e03 } a basis in E3 . Is this basis an orthonormal basis of E3 ? (you have to consider all cases a),b) c) and d)). 0 0 0 Case a) The ordered set {e1 , e2 , e3 } = {e2 , e1 , e3 } is evidently orthonormal basis. Transition matrix 0 1 0 T = 1 0 0 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T . This matrix is non-degenerate, its rank is equal to 3 (det T = 0 0 1 1 6= 0). It is orthogonal because both bases are orthonormal. is not a basis because vectors are linear Case b) The ordered set {e01 , e02 , e03 } = {e1 , e1 + 3e3 , e3 } 1 1 0 dependent: e01 − e02 + 3te03 = 0. Transition matrix T = 0 0 0 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T . This matrix 0 3 1 is degenerate, its rank ≤ 2. One can see it noting that rows are linear dependent or noting that det T = 0. Vectors {e01 , e02 , e03 } are linear dependent. On the other hand vectors {e01 , e02 } are linear independent. Hence rank of the matrix T is equal to 2. This matrix is not orthogonal. Case c) The ordered set {e01 , e02 , e03 } = {e1 − e2 , 3e1 − 3e2 , e3 } is not a basis because vectors are linear dependent: 3e01 − e02 = 0. 1 3 0 One can see it also studying the transition matrix. Transition matrix T = −1 −3 0 , (e01 , e02 , e03 ) = 0 0 1 (e1 , e2 , e3 )T . This matrix is degenerate, its rank ≤ 2, det T = 0. On the other hand second and third row of this matrix are linear dependent. Hence rank of the matrix T is equal to 2. This matrix is not orthogonal. Case d) 3 0 0 0 The transitionmatrix from the basis {e1 , e2 , e3 } to the ordered triple {e1 , e2 , e3 } = {e2 , e1 , e1 +e2 +λe3 } 0 1 1 is T = 1 0 1 , (e01 , e02 , e03 ) = (e1 , e2 , e3 )T 0 0 λ I-st case. λ 6= 0. The ordered set {e01 , e02 , e03 } is a basis because vectors are linear independent (see the exercise 3), This basis is not orthogonal, because the length vector is not equal to 1 ((e03 , e03 ) = |e03 |2 = 2+λ2 ). This matrix is not orthogonal, because the new basis is not orthonormal. II-nd case λ = 0. The ordered set {e01 , e02 , e03 } is not a basis because vectors are linear independent: 0 e1 + e02 − e03 = 0. The transition matrix T has rank less or equal to 2, because vectors are linear dependent. On the other hand vectors e01 , e02 are linear independent. Hence the rank of the matrix is equal to 2. 6† . Show that an arbitrary orthogonal transformation of two-dimensional Euclidean space can be considered as a composition of reflections. If the determinant of orthogonal transformation is equal to −1 then µ T = T̃ϕ = cos ϕ sin ϕ sin ϕ − cos ϕ ¶ One can see that this is reflection with respect to the axis which has an angle ϕ/2 with Ox axis. If the determinant of orthogonal transformation is equal to 1 then µ T = Tϕ = cos ϕ − sin ϕ sin ϕ cos ϕ ¶ µ = cos ϕ sin ϕ sin ϕ − cos ϕ ¶µ 1 0 0 −1 ¶ µ = T̃ϕ 1 0 0 −1 ¶ i.e. rotation on the angle ϕ is a composition of two reflections. 7† Prove the Cauchy–Bunyakovsky–Schwarz inequality (x, y)2 ≤ (x, x)(y, y) , where x, y are arbitrary two vectors and ( , ) is a scalar product in Euclidean space. Hint: For any two given vectors x, y consider the quadratic polynomial At2 + 2Bt + C where A = (x, x), B = (x, y), C = (y, y). Show that this polynomial has at most one real root and consider its discriminant. Pn Pn Consider quadratic polynomial P (t) = i=1 (txi + y i )2 = At2 +2Bt +C, where A = i=1 (xi )2 = (x, x), Pn P n B = i=1 (xi y i ) = (x, y), C = i=1 (y i )2 = (y, y). We see that equation P (t) = 0 has at most one root ( and this is the case if only vector x is collinear to the vector y). This means that discriminant of this equation is less or equal to zero. But discriminant of this equation is equal to 4B 2 − 4AC. Hence B 2 ≤ AC. It is just CBS inequality. ((x, y)2 = (x, x)(y, y), i.e. discriminant is equal to zero ⇔ vectors x, y are colinear. 4