Ph.D. Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2) EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified.

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1 PhD Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2 EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified Problem 1 [ points]: For which parameters λ R does the following system of equations have a unique solution? x 1 + x 3 x = 2 x 2 2x 3 = 3 x 1 + λx 3 = x 3 x = 2 Solution: A system of linear equations for variables has a unique solution if and only if the correspondent matrix of coefficients has full rank, i e rank Elementary row operations do not change the rank: λ λ λ λ (first we swapped the last two rows, then substracted a multiple of the first and third row resp from the last row The last matrix has rank if and only if λ 0 Problem 2 [ points]: Prove or give a counterexample: For any two 2-by-2 matrices A, B R 2 2, it holds that Det (A + B = Det (A + Det (B Solution: The equality is false( in general, as can( be easily seen by the choice of almost any matrices E g taking A = and B = yields Det (A + B = Det ( 1 0 = 1 0 = Det 0 1 Problem 3 [8 points]: Consider the matrix ( 2 1 A = 1 ( Det 0 0 ( (a Find the eigenvalues and the corresponding eigenvectors of A (b Is A diagonalizable? (c Find the Jordan decomposition A = P 1 JP Write all matrices in the expression explicitely Solution:

2 PhD Katarína Bellová Page 2 Mathematics 2 (10-PHY-BIPMA2 (a Eigenvalues can be characterized as roots of the characteristic polynomial: ( 2 λ 1 Det (A λi 2 = Det = (2 λ( λ + 1 = λ 2 6λ + 9 = (λ λ Hence λ = 3 is the only eigenvalue of A with algebraic multiplicity 2 Corresponding eigenvectors v = (v 1, v 2 solve (A λi 2 v = 0, where ( ( A λi 2 = = Hence, the only eigenvectors are the non-zero solutions of the corresponding homogeneous system of linear equations, which are all non-zero scalar multiples of (1, 1 We also see that the geometric multiplicity of λ = 3 is equal to 1 (as the dimension of the eigenspace (b Since the algebraic multiplicity of λ = 3 is different from its geometric multiplicity (2 1, matrix A is not diagonalizable (c The only possible Jordan form corresponding to a 2 2 matrix with a single eigenvalue λ = 3 with algebraic multiplicity 2 and geometric multiplicity 1 is the matrix ( The transition matrix P 1 (according to the somewhat suboptimal notation from the question can be chosen from any corresponding generalized eigenvectors: first column can be chosen for example as v 1 = (1, 1 t and the second column as any solution v 2 of the system (A λi v 2 = v 1 : Hence we can choose e g v 2 = (0, 1 t and 1v v 2 2 = 1, 1v v 2 2 = 1 P 1 = Computing the inverse as ( ( 1 0 gives P = and finally 1 1 ( 1 0 A = 1 1 ( ( ( ( Problem [6 points]: Let V be an inner product space over a field F (F = R or F = C

3 PhD Katarína Bellová Page 3 Mathematics 2 (10-PHY-BIPMA2 (a For a nonempty subset S V, define the orthogonal complement S (b Let U 1, U 2 be two vector subspaces of V Prove that (U 1 + U 2 = U 1 U 2 Solution: (a The orthogonal complement of S consists of all vectors from V which are orthogonal to all vectors from S: S = {v V : s, v = 0 s S} (b We will show two inclusions and to show the equality of the two sets Choose any vector v (U 1 + U 2, i e v is orthogonal to all vectors from U 1 + U 2 Since U 1 U 1 + U 2 ( 0 U 2, so any u U 1 and be written as u = u + 0, v is in particuar orthogonal to all vectors from U 1, i e v U1 In the same way, U 2 U 1 +U 2, so v U2 Hence v U1 U2 Since v (U 1 +U 2 was arbitrary, we showed (U 1 +U 2 U1 U2 On the other hand, choose any v U1 U2 Then v is orthogonal to all vectors from U 1 as well as all vectors from U 2 To prove that v (U 1 + U 2, we need to show that v is orthogonal to all vectors from U 1 + U 2 Let u (U 1 + U 2 Then u can be written as u = u 1 + u 2, where u 1 U 1, u 2 U 2 Using the linearity of the scalar product and v U1, v U2, we get u, v = u 1 + u 2, v = u 1, v + u 2, v = = 0 Since u (U 1 + U 2 was arbitrary, we see that v (U 1 + U 2 Since v U 1 U 2 was also arbitrary, we get the other inclusion U 1 U 2 (U 1 + U 2 Problem [ points]: Let (e 1, e 2 be an orthonormal basis of a vector space V over C Let T L(V, V be a linear operator such that T (e 1 = e 1 + e 2, T (e 2 = e 1 + e 2 (a Is T self-adjoint? (b Is T unitary? Solution: Matrix of T in basis (e 1, e 2 is A T = ( Since (e 1, e 2 is an orthonormal basis, matrix of the adjoint operator T is equal to A T = A t T = ( ( = (a Operator S is self-adjoint if and only if S = S Since for our T, A T operator is not self-adjoint A T, the

4 PhD Katarína Bellová Page Mathematics 2 (10-PHY-BIPMA2 (b Operator S is unitary if and only if S = S 1, or if and only if SS = id (in finitedimensional space Using the matrices of the operators, this is equvalent to A S A S = I Since indeed ( ( ( 1 0 =, 0 1 operator T is unitary Problem 6 [ points]: Consider the bilinear form B on R 2 R 2 whose Gram matrix in the basis ((1, 0, (0, 1 is ( 1 0 A B = 3 What is the Gram matrix of B in the basis ((2, 1, (1, 1? Solution: Transision matrix from ((1, 0, (0, 1 to ((2, 1, (1, 1 is equal to ( 2 1 P = 1 1 Then matrix of B in basis ((2, 1, (1, 1 is equal to ( ( ( Ã B = P t A B P = = Alternatively, one can compute the terms (ÃB ij as where ẽ i = (2, 1 t, ẽ j = (1, 1 (ÃB ij = B(ẽ i, ẽ j = ẽ t ia B ẽ j, ( Problem 7 [7 points]: Consider the following function f : R 2 R: { xy if (x, y (0, 0, x f(x, y = +y 0 if (x, y = (0, 0 (a Is f continuous on R 2? (b Do the partial derivatives f x, f y exist in the point (0, 0? (c Is f differentiable on R 2? Solution: (a If f was continuous at (0, 0, then we would have lim (x,y (0,0 xy x + y = lim f(x, y = f(0 = 0 (x,y (0,0 In particular, for any continuous curve (x, y = (α(t, β(t with (α(0, β(0 = (0, 0 (and (α(t, β(t (0, 0 for t 0, we would have lim t 0 α(tβ(t α(t + β(t = 0

5 PhD Katarína Bellová Page Mathematics 2 (10-PHY-BIPMA2 However, choosing (α(t, β(t = (t, t leads to lim t 0 α(tβ(t α(t + β(t = lim t 0 t 2 2t = lim t 0 1 2t 2 = +, so f cannot be continuous at 0, and hence is not continuous on R 2 (b Note that on the coordinate axes, f(x, 0 = x 0 = 0 x + 0 (x 0, f(0, y = 0 y = y (y 0 Hence f x (0, 0 = lim x 0 f(x, 0 f(0, 0 x f y (0, 0 = lim y 0 f(0, y f(0, 0 y = lim x x = lim y y = 0, = 0 and we see that both partial derivatives exist at the point (0, 0 (c Since f is not continuous on R 2, it cannot be differentiable on R 2 Problem 8 [ points]: Consider the following function f : R 2 R: f(x, y = cos x e 3x+y2 Compute D (1,2 f(0, 1, i e the directional derivative of f in direction (1, 2 at the point (0, 1 Solution: First compute the gradient: x f(0, 1 = sin xe 3x+y2 + cos xe 3x+y2 3 = 3e, (x,y=(0,1 y f(0, 1 = cos xe 3x+y2 2y = 2e, (x,y=(0,1 so f(0, 1 = (3e, 2e Now note that (1, 2 is not a unit vector Considering a unit vector in the same direction as (1, 2 ( (1, 2 (1, 2 1 l = = (1, = 2,, 2 we get D l (0, 1 = f(0, 1 l = (3e, 2e ( 1, 2 = 3e 1 + 2e 2 = 7e

6 PhD Katarína Bellová Page 6 Mathematics 2 (10-PHY-BIPMA2 Interpreting D (1,2 f(0, 1 as the derivative along the non-unit vector (1, 2 was also considered OK: D (1,2 f(0, 1 = f(0, 1 (1, 2 = 3e 1 + 2e 2 = 7e Problem 9 [ points]: Let w = xyz and consider the spherical coordinates (here r 0, θ [0, π], ϕ [0, 2π r > 0, θ (0, π, ϕ (0, 2π Solution: Using chain rule, x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ Find w ϕ w ϕ = w x x ϕ + w y y ϕ + w z z ϕ = yz r sin θ( sin ϕ + xz r sin θ cos ϕ + xy 0 in terms of r, θ, ϕ in a point where = r sin θ sin ϕ r cos θ r sin θ sin ϕ + r sin θ cos ϕ r cos θ r sin θ cos ϕ = r 3 sin 2 θ cos θ( sin 2 ϕ + cos 2 ϕ = 1 2 r3 sin θ sin(2θ cos(2ϕ Alternatively, one can express w = r 3 sin 2 θ cos θ sin ϕ cos ϕ and differentiate directly Problem 10 [6 points]: Find the maximum and minimum value of the function on the set f(x, y = x 2 + 2xy + 1 {(x, y R 2 : 2x 2 + y 2 9} Solution: First note that the set D = {(x, y R 2 : 2x 2 + y 2 9} is a bounded and closed subset of R 2, and hence the continuous function f(x, y = x 2 + 2xy + 1 attains its maximum and minimum on D The extrema can be attained either inside D = {(x, y R 2 : 2x 2 + y 2 < 9}, or on the boundary D = {(x, y R 2 : 2x 2 + y 2 = 9} Let us find all candidates If (x, y D is a point of extremum, then it is a critical point and we have 0 = f x (x, y = 2x + 2y, 0 = f y (x, y = 2x Second equality yields x = 0, and then first equality yields y = 0 The point (x, y = (0, 0 indeed lies in D and f(0, 0 = = 1 (1 If (x, y D is a point of extremum, we can use the method of Lagrange mulitipliers: setting g(x, y = 2x 2 + y 2 9, there must exist a λ R such that f(x, y = λ g(x, y

7 PhD Katarína Bellová Page 7 Mathematics 2 (10-PHY-BIPMA2 which together with the condition (x, y D gives From (3, we have x = λy Plugging this into (2, we get 2x + 2y = λ x, (2 2x = λ 2y, (3 2x 2 + y 2 = 9 ( 2λy + 2y = λ 2 y, y(2λ 2 λ 1 = 0 If y = 0, then x = λy = 0, violating ( Hence y 0 and necessarily (2λ 2 λ 1 = 0, which has the roots λ = 1 ± { = If λ = 1, then x = λy = y and plugging this into ( yields 3x 2 = 9, x = y = ± 3 (they need to have the same sign In these points, ( ( f 3, 3 = f 3, 3 = = 10 ( In the remaining case λ = 1, we get x = λy = 1 y, and plugging this into ( gives y2 + y2 = 9, yielding y = ± 6 and x = 1y = 6 At these points, 2 2 ( f 6 2, 6 = f ( 6 2, = = 7 2 (6 Comparing the values in (1, ( and (6, we see that f attains its maximum value of 10 at the points ( 3, 3 and ( 3, 3 and its minimum value of 7 at the points 2 ( 6, 6 and ( 6, Problem 11 [ points]: Find the general solution to the differential equation on a maximal possible interval xyy = x 2 + y 2 Solution: The eqeuation is homogeneous, so substitution y(x = xt(x (we can always assume y to have this form for x 0 should lead to an ODE with separated variables Indeed, plugging in y = t + xt yields x 2 t 2 + x 3 tt = x 2 + x 2 t 2, x 3 tt = x 2

8 PhD Katarína Bellová Page 8 Mathematics 2 (10-PHY-BIPMA2 If x 0, the equation can be further written as t dt dx = 1 x, t dt = 1 x dx, 1 2 t2 = ln x + C, t = ± 2(ln x + C, y = ±x 2(ln x + C Given C R, this defines two solutions which are defined for ln x + C 0, i e for x e C or x e C (note that in particular we never get to x = 0 Although the value y(±e C = 0 is finite, we cannot continue the solution up to this point, since it violates the original equation (y = 0, x 0, y does not exist Hence, y(x = ±x 2(ln x + C are solutions on maximal intervals (, e C and (e C, + Note that outside of x 0, these are the only possible solutions (in particular, y(x 0 for x 0 as noted above It remains to look at the case when x = 0 From the original equation we immediately get y(0 = 0 Assume that there is some solution y = y(x on an interval I containing 0 (e g on [0, a with y(0 = 0 As seen above, on interval ( a, a the solution must coincide 2 with some of the above solutions However, this is not possible, since none of the above solutions can be continued upto x = 0 Hence there are no solutions defined at x = 0