Course code: 8D020. Date: Wednesday April 11, Time: 09h00 12h00. Place: MA Read this first!

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1 EEXAMINATION MATHEMATICAL TECHNIQUES FO IMAGE ANALYSIS Course code: 8D00. Date: Wednesday April, 0. Time: 09h00 h00. Place: MA.46. ead this first! Write your name and student ID on each paper. The eam consists of 4 problems. The maimum credit for each item is indicated in the margin. Motivate your answers. The use of course notes is allowed. The use of problem companion opgaven- en tentamenbundel, calculator, laptop, or any other equipment, is not allowed. You may provide your answers in Dutch or English. Feel free to ask questions on linguistic matters or if you suspect an erroneous problem formulation. Good luck! 30. Inner Product Space Consider the set V {f : C f C and f k f } in which k V is a particular element of V for every and : C C C is a comple inner product on C. We take it for granted that C is a comple inner product space given the usual definitions of function addition and comple scalar multiplication. a. Show that the function k has the following properties: a. k y k y k ; Since k V we have, by definition of V, k y k y k. a. k y k y in which z denotes the comple conjugate of z C; Using the definition of V k y k y k k k y k y. first and last step and the definition of a comple inner product middle step we find a3. k 0 for all. By virtue of non-degeneracy of an inner product we have, for all, k k k 0. The following diagrams are abstract representations for k y and k y : y y ky ky

2 a4. Eplain what it means to say that these diagrams are mutually consistent. Assuming that the orientation of the graphics should not matter you can interpret the second graph as a 80 -rotated copy of the first graph, with free labels and y interchanged, i.e. as k y. But according to a this is identical to k y, indeed the interpretation given to the second graph. 0 b. Show that V is a comple vector space. Hint: Use the linear subspace theorem. It is given that C is a comple vector space, with V C. So we need to prove only closure. Let f, g V and λ, µ C, so in particular f k f and g k g for all. We have k λf + µg λ k f + µ k g λf + µg λf + µg, so λf + µg V. Here we have used the linearity property of a comple inner product, the definition of V, and the usual definition of linear function superposition. Below we take : C C C to be the standard comple inner product on C. 5 c. Eplain what this means by giving the eplicit formula for f g. The standard comple inner product for one-dimensional functions is given by f g f g d. 5 d. Eplain and prove the following diagrammatic equality: y Hint: The unlabeled central dot on the r.h.s. represents an integration dummy. The r.h.s. diagram symbolizes k y k z k yz dz. To see that this identity is correct, consider the l.h.s. diagram. By definition this diagram equals k y a k k y c k z k yz dz a k z k k z k y dz. The integrand on the r.h.s. consists of two factors, which can be diagrammatically represented as two oriented line elements with a common dummy label z, once with an incoming arrow and once with an outgoing arrow. By graphically contracting this dummy into a single inner node one obtains the diagram on the r.h.s. 35. Algebra Eam March 8, 005, Problem In this problem we consider the set G def endowed with certain internal and eternal operators. We identify an element θ G with its column representation in : θ def θ θ in which θ, θ the components of θ. To begin with we interpret G as the linear space over by introducing vector addition and

3 scalar multiplication, in the usual way. The vector sum of η, θ G is written as η + θ, and the scalar multiple of θ G and λ as λ θ. 5 a. Eplain what is meant by the usual way by indicating eplicitly how η + θ and λ θ are defined in terms of their components. For arbitrary λ, µ and η, θ G we define λ η + µ θ def λ η + µ θ. λ η + µ θ We furthermore introduce an algebraic operation, which we shall refer to as multiplication. The product of η, θ G is simply written as η θ, for which we agree that, in terms of components, η θ def η θ η θ + η θ G. 5 b. Prove that G, endowed with the aforementioned multiplication operation, constitutes an algebra. Proceed as follows without proof we take it for granted that G is a linear space, cf. part a: b. Prove that η, θ, γ G η θ γ η θ γ. Epanding in terms of components we get η θ γ def η θ γ def η θ + η θ γ def η θ γ. η θ γ η θ γ + η θ + η θ γ η θ γ η θ γ + θ γ + η θ γ def η θ γ η θ γ + θ γ The triviality marked with eploits associativity of ordinary multiplication on. b. Prove that η, θ, γ G η θ + γ η θ + η γ. η θ + γ η θ + γ η θ + γ η θ + γ η θ + γ + η θ + γ η θ + η θ + η θ η γ η θ + η γ. η γ + η γ In distributivity of ordinary multiplication on has been used to eliminate parentheses in the components and also of the definition of vector addition to split the column vector into two terms. b3. Prove that η, θ, γ G η + θ γ η γ + θ γ. You can construct the proof in terms of components analogous to the solution for b. Alternatively one can make use of commutativity, which will be proven under d below, and of the distributivity property in part b : η + θ γ γ η + θ γ η + γ θ η γ + θ γ. b4. Prove that η, θ G, λ λη θ λ η θ η λ θ. 3

4 We first prove the first equality: λη θ def λ η θ η θ + η θ λ η θ λη θ + η θ λ η θ λ η θ + λ η θ λ η θ λ η θ + λ η θ def λ η θ. In we have used the definition of scalar multiplication, in we have used associativity of multiplication on to shift parentheses. In the first and last step the definition of multiplication on G has been used. The second equality can be proven in a similar fashion, or with the help of commutativity part d. 5 c. Show that, moreover, there eists a unit element G not to be confused with the number, and give its column representation in. Call the components of G e respectively e. Let G be arbitrary, and suppose e e e e + e def for all,, in which in the final step the definition of the identity element has been used, then it follows by necessity that e and e 0. Conclusion: def. 0 One still needs to verify whether for all G. We could do this by componentwise analysis as previously, or by eploiting once more commutativity, with a modest amount of foresight, cf. part d : for all G. 5 d. Is multiplication on G commutative? If so, prove, if not, give a counter eample. Suppose, y G, then y y y y y + y y y. y + y We now consider the subset G 0 G, defined by G 0 {θ G θ 0}. With θ we mean θ θ. 5 e. Give an eplicit characterization of G 0 by indicating what the column representation in of an arbitrary element θ G 0 looks like. In terms of components we have θ θ def 0. θ θ 0 In the last step we have used the fact that θ G 0. This is equivalent to the condition θ 0, i.e. an arbitrary element θ G 0 has the form 0 θ θ with θ arbitrary. Finally we introduce on G a degenerate, non-negative, symmetric, real valued, bilinear form. For η, θ G this is indicated by η θ. In terms of the components of η and θ we define this as follows: η θ η θ. Caveat: The adjective degenerate indicates that does not define an inner product. 4

5 5 f. Eplain the adjective degenerate by eplaining why does not define an inner product. For a non-degenerate inner product we have the condition that θ θ 0 iff θ 0 G. In the case of the definition above, however, we have θ θ 0 iff θ 0, i.e. irrespective of the value of θ. Thus there are non-trivial elements θ G viz. all elements with θ 0 and θ 0 for which θ θ 0 degeneracy. We now consider the subset G G, defined by G {θ G θ θ }. 5 g. Prove that G constitutes a group with respect to multiplication. Proceed as follows: g. Show that, if η, θ G then η θ G closure. You need to show that the subset G G is closed under multiplication. Suppose η, θ G, then η η θ θ η θ + η θ so η θ η θ η θ η θ η θ η η θ θ. In the definition of the bilinear form η θ has been used, in that of the first component of the algebraic product η θ G. In the last step the fact that η, θ G has been used. Since η θ η θ it follows that η θ G. g. Show that η, θ, γ G η θ γ η θ γ associativity. Multiplication on G is associative cf. b. Since G G it is also associative within G. g3. Show that the unit element of part c satisfies G. θ G iff θ G and θ θ. For the identity element it has already been shown in c that G. Using the components of the identity element and the definition of the bilinear form it immediately follows that. Conclusion: G. g4. Show that, given θ G, there eists an inverse θ G, such that θ θ θ θ G. Give the column representation of θ in in terms of the components of θ. Given θ G arbitrary. Write θ η for notational convenience. We must have η θ G. If this holds then θ η holds automatically by virtue of commutativity, recall part d. In other words, η θ η θ + η θ 0 which can also be written as θ 0 η. θ θ η 0 Inversion yields θ def η θ 0 inv θ 0 θ η θ θ 0 θ θ θ 0 θ θ θ θ θ. Note that this is well defined, since θ 0 for all θ G. θ θ θ for all θ G. The last step, indicated with a, eploits the fact that 5

6 35 3. Fourier Transformation and Distribution Theory Consider the generalized function f defining a tempered distribution T f S, given by f, y u, y δy m, in which δ denotes the one-dimensional Dirac function, u : C is a given function with well-defined Fourier transform û : C, and m is a parameter. That is, for φ S, T f φ f, y φ, y ddy. Note that the support of f the part of the, y-domain where f, y may not vanish is effectively the line given by l : y m. For this reason we define the function u m : C by u m u, m. In this problem the two-dimensional Fourier transform is defined as ˆfω, ν e iω iνy f, y ddy. 0 a. Show that ˆfω, ν û m ω + mν. Working out the definitions we have ˆfω, ν e iω iνy f, y ddy e iω iνy u, y δy m ddy e iω+mν u m d û mω+mν. b. Sketch the graph of l in the, y-plane. The graph of l in the, y-plane is a straight line through the origin with an inclination angle α relative to the -ais, such that m tan α. b. Epress the angle α by which l intersects the -ais in terms of the parameter m. ecall b: α arctan m. b3. Sketch the family of lines in the ω, ν-plane on which ˆfω, ν assumes constant values. For constant c C and generic function ˆf we have ˆfω, ν c, i.e. û mω + mν c, along lines in the ω, ν-plane for which ω + mν k for any constant k. b4. Under which angle does the normal vector to this family intersect the ω-ais? The common normal of the family of lines given in b3 is, up to an arbitrary non-zero constant, equal to, m, so that it makes the same angle α arctan m with the ω-ais as the line l does with the -ais. 6

7 Below we consider the case u, y Ae +y, for some amplitude A > 0. In the following problem you may use the following standard integral, valid for all ξ, η : e ξ+iη dξ π. 0 c. Compute û m ω. Plugging in the definition of the function u m we obtain û mω e iω u m d A e iω e +m d. The r.h.s. can be rewritten as A e ω 4+m e +m + iω +m d A + m e ω 4+m ξ+ iω +m e dξ A π ω + m e 4+m. In we have substituted + m ξ, in we have applied the given standard integral. 5 d. Suppose the function u m is normalized such that u m d. Determine A. This means that û m0, so A + m / π. 7