Problem Sheet 2: Eigenvalues and eigenvectors and their use in solving linear ODEs

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1 Problem Sheet 2: Eigenvalues an eigenvectors an their use in solving linear ODEs If you fin any typos/errors in this problem sheet please The material in this problem sheet is not examinable It is intene for the more mathematicallyincline stuents who want to obtain a more thorough unerstaning of eigenvalues an eigenvectors an their use in solving linear ODEs Even though in the course we only eal with moels of imension 1 an 2, here we consier vectors, matrices an linear moels of arbitrary (but finite) imension n The reason why is because we believe that, when iscussing the material in this problem sheet, nothing is gaine in terms of simplicity an comprehensibility, quite the contrary, by limiting the imensions of the vector/matrices we eal with to be of a fixe low imension However, any course assessment will only require you to be able to apply the concepts iscusse below to vectors, matrices an linear moels of imension 1 or 2 1 An eigenvector v of an n n matrix of real numbers A is efine as a non-trivial vector 1 of complex numbers such that Av = λv where λ is a complex number calle an eigenvalue If real, eigenvalues an eigenvectors have very simple geometric interpretations For instance, an eigenvector v is a vector such that A maps it onto itself, in other wors, Av lies on the same line as v, scale (up if λ > 1 or own if λ < 1) an/or rotate by 18 o (if λ < ) Further information on linear algebra can be foun in (short camcasts which are excellent if you have never seen any linear algebra before in your life) (bit more avance, great lecturer, first 7 lectures cover useful linear algebra concepts with engineering-type applications) Linear algebra one right by Shelon Axler (it is a nice, brief an rigorous introuction to the subject that oesn t rely on eterminants for proofs, which in our opinion is a goo thing) Any other of the numerous textbooks on the subject (a) We can represent any vector x R n as a linear combination of the eigenvectors of A if an only if A has n eigenvectors that form a linearly inepenent set vectors 2 In the wors, if the previous is true, an letting v 1, v 2,, v n enote n linearly inepenent eigenvectors, then for any given x R n we can fin some complex numbers c 1, c 2,, c n such that x = c 1 v 1 + c 2 v c n v n Use the above to fin an expression for the vector c = [c 1, c 2,, c n ] T in terms of the matrix of linearly inepenent eigenvectors V = [v 1, v 2,, v n ] an in terms of x (Note that v 1, v 2,, v n are column vectors corresponing to the 1st, 2n,, an nth column of the matrix V, respectively) Hint: The inverse of a matrix exists if an only if its columns form a linearly inepenent set of vectors In aition, if it exists it is unique (b) Eigenvectors an eigenvalues are useful because they provie us a very straightforwar way of etermining how A transforms any given vector x (ie, what Ax looks like) Show that Ax = c 1 λ 1 v 1 + c 2 λ 2 v c n λ n v n, where λ i enotes the eigenvalue such that Av i = λ i v i with i {1, 2,, n} λ i is sai to be the eigenvalue corresponing to, or associate with, eigenvector v i 1 A vector is sai to be non-trivial if at least one of its elements is non-zero 2 A set of vectors {v 1, v 2,, v n} is sai to be {[ linearly ] [ ]} inepenent {[ ] [ ]} if none of the vectors is a linear combination of the other vectors, for example,,, are linearly inepenent, while {[ ] [ ]} {[ ] [ ]} {[ ] [ ]} ,,,,, are not (they are sai to be linearly epenent sets of vectors)

2 (c) Using the answers to the previous two parts, show that Ax = V ΛV 1 x, where Λ is an n n iagonal matrix such that Λ ii = λ i Expressing A as V ΛV 1 is calle iagonalising A If it is possible to iagonalise A, then A is sai to beiagonalisable It is possible to o this if an only if one can fin n eigenvectors that form a set of linearly inepenent vectors Otherwise we can t carry out the step we i in part (a); we can t express any arbitrary x R n as a linear combination of the eigenvectors of A () All the above is very well, but to put it to practice one must first be able to fin the eigenvectors an eigenvalues of a given matrix A A property of eterminants is that, for any given matrix A, there exists a non-trivial vector x such Ax = if an only if et(a) = By the efinition of eigenvectors, Av = λv Av λv = (A Iλ)v = et(a Iλ) = for any eigenvalue λ Note that et(a Iλ) is a polynomial in λ, also note that this implies that there is at most n eigenvalues (can you see why?, Hint: think about the orer of the polynomial et(a Iλ)) Thus, by solving for the roots of et(a Iλ) we fin the eigenvalues Practice this by fining the eigenvalues of the following matrices i) [ ], ii) [ ], iii) [ ], iv) [ ] 2 2 (e) Once we have the eigenvalues then we simply use (A λi)v = to figure out the eigenvectors For the above four matrices fin all linearly inepenent eigenvectors 3 (f) Which of the four matrices are iagonalisable? 2 In this exercise we erive the solution of arbitrary (but finite) imensional linear moels efine by ẋ(t) = Ax(t), x() = x, (1) where A is n n an iagonalisable In other wors we fin a function x(t) that obeys two rules: I ẋ(t) = Ax(t) for all t, ie, that at any given time, it s time-erivative equals A times itself an II x() = x, ie, its value at time is x In aition, with no extra effort we show that there is only one such solution x(t), ie, that the solution x(t) is unique 4 (a) Show that t eλt = Λe Λt = e Λt Λ, where Λ is as in exercise 1 an e λ 1t e Λt e λ 2t := e λnt Hint: If D 1 an D 2 are iagonal n n matrices, then D 1 D 2 = D 2 D 1 (b) Pre-multiply ẋ(t) = Ax(t) by (V e Λt V 1 ) an show that x(t) = V e Λt V 1 c where c R n is a vector of constants Hint 1: Remember integrating factors? Hint 2: If M, N an Q(t) are n n matrices such that Q(t) varies in time but M, N o not, then t (MQ(t)N) = M t (Q(t))N In aition, if v(t) R n varies with time, then t (Q(t)v(t)) = t (Q(t))v(t) + Q(t) t (v(t))5 (c) Show that e Λ = I, where I is the ientity matrix () Use part (c) an rule II to show that c = x 3 This is slang, linear inepenence is a property of a sets of vectors not of vectors; to say that a vector v is linearly inepenent has no meaning However, it is common practice to say that a bunch of vectors are linearly inepenent if the set of those vectors is linearly inepenent 4 It is well known that there only exists one solution, that is why we write we will erive the solution to (1) instea of we will erive a solution to (1) Inee, generally, it is best to say the unique solution instea of the solution as it removes all possible ambiguity 5 This is one of the generalisations of the prouct rule in multivariable calculus Page 2

3 (e) Is x(t) = V e Λt V 1 x the unique solution to (1)? 3 (More avance) In this exercise we take a ifferent approach to show that x(t) = V e Λt V 1 x, (2) as efine in exercise 2, is the unique solution to (1) In particular, we first show that, at most, (1) has a single solution Then, separately, we show that (2) is a solution to (1) by verifying that it obeys rules I an II (see the escription of exercise 2) (a) The Gronwall-Bellman inequality implies that if a continuous function y : [, + ) R satisfies y(t) α + where α R an β are constants, then βy(τ)τ y(t) αe βt (3) Assume that the solution is not unique, ie, that we have two functions, x(t) an z(t) both of which satisfy (1) Consier the norm ifference between x(t) an z(t), ie, δ(t) = x(t) z(t) Use (3) to show that δ, that is that δ(t) = for all t, an thus argue that the solution is inee unique (b) Next, show that (2) satisfies rule I, ie, that ẋ(t) = Ax(t) Hint: Use exercise 2(a) an the secon hint given in exercise 2(b) (c) Finish by showing that (2) satisfies rule II, ie, that x() = x Hint: Use exercise 2(c) Page 3

4 Solutions 1 (a) The fact that the eigenvectors are linearly inepenent, in other wors, that the columns of V are linearly inepenent, implies that the inverse of V exists, hence x = c 1 v 1 + c 2 v c n v n = V c c = V 1 x (b) This follows simply from linearity, A(b 1 x 1 + b 2 x 2 ) = b 1 Ax 1 + b 2 Ax 2 for any matrix A, vectors x 1, x 2 an scalars b 1, b 2 From (a), we have that x, written in terms of the linearly inepenent eigenvectors of A, is given by x = c 1 v 1 + c 2 v c n v n Therefore, we have: Ax = A(c 1 v 1 +c 2 v 2 + +c n v n ) = c i are scalars c 1Av 1 +c 2 Av 2 + +c n Av n = c 1 λ 1 v 1 +c 2 λ 2 v 2 + +c n λ n v n (c) Using (b) an (a), we have: Ax = (using (b)) c 1λ 1 v 1 + c 2 λ 2 v c n λ n v n = V λ 1 c 1 λ 2 c 2 = V Λc = V ΛV 1 x (using (a)) λ n c n () By the funamental theorem of algebra, a polynomial of orer n has at most n complex roots Hence any matrix of imension n has at most n eigenvalues i) λ 1 = 2, λ 1 = 1 A little trick is that the eigenvalues of triangular matrices are the elements on the main iagonal of the matrix ii) λ 1 =, λ 2 = 3 iii) λ 1 = 1, λ 2 = 1 (repeate eigenvalue at 1) iv) λ 1 = 2, λ 2 = 2 (repeate eigenvalue at 2) (e) i) v 1 = [1, ] T, v 2 = [1, 1] T, where v 1 enotes the eigenvector that correspons to λ 1 Note that av 1 an av 2 are also vali eigenvectors for any non-zero real number a However, the biggest set of linearly inepenent eigenvectors has two elements v 1 an v 2 or any scale versions of them ii) v 1 = [1, 1] T, v 2 = [1, 1/2] T iii) v 1 = [1, 1] T iv) v 1 = [1, ] T, v 2 = [, 1] T There is one eigenvalue but two eigenvectors! This is fine, however you can t have the contrary; there is at least one eigenvector per eigenvalue (otherwise the efinition of eigenvector/eigenvalue oesn t make sense) (f) Matrices i), ii), an iv) are the only ones that have 2 linearly inepenent eigenvectors, hence they are the only ones that are iagonalisable 2 (a) From the hint, we know that Λe Λt = e Λt Λ Thus, we only nee to show t eλt = Λe Λt (or, equivalently, t eλt = e Λt Λ): t eλ 1t λ 1 e λ 1t t eλt t = eλ 2t = λ 2 e λ 2t = ΛeΛt t eλnt λ n e λnt Page 4

5 (b) Starting from rule I ẋ(t) = Ax ẋ(t) Ax(t) = (V e Λt V 1 )(ẋ(t) Ax(t)) = Notice that the fact that (V e Λt V 1 ) is invertible (with inverse (V e Λt V 1 ) 1 = V e Λt V 1 ) is responsible for the secon if an only if, ie, the secon Next, But, A = V ΛV 1, so (V e Λt V 1 )(ẋ(t) Ax(t)) = V e Λt V 1 ẋ(t) V e Λt V 1 Ax(t) V e Λt V 1 Ax(t) = V e Λt V 1 V ΛV 1 x(t) = V e Λt ΛV 1 x(t) = V e Λt ( Λ)V 1 x(t) From part (a) we have that t e( Λ)t = e Λt ( Λ) So, V e Λt ( Λ)V 1 x(t) = V t (e( Λ)t )V 1 x(t) Using the secon hint we have that V t e( Λ)t V 1 = t (V e( Λ)t V 1 ) an thus V e Λt V 1 ẋ(t) V e Λt V 1 Ax(t) = V e Λt V 1 t (x(t)) + t (V e( Λ)t V 1 )x(t) = t (V e( Λ)t V 1 x(t)) ẋ(t) = Ax t (V e( Λ)t V 1 x(t)) = t (V e( Λ)t V 1 x(t))t = V e ( Λ)t V 1 x(t) = c where c R n is a vector of integration constants Then, pre-multiplying the above with V e (Λ)t V 1 we get that x(t) = V e Λt V 1 c (c) e λ 1 e 1 e Λ e λ 2 = = e = 1 = I e λn e 1 () x() = V e Λ V 1 c = V IV 1 x = V V 1 c = Ic But, x() = x, hence, c = x (e) In parts (a) () we prove that x(t) = V e Λt V 1 x is a solution to (1) In aition, given that throughout the whole erivation we only employe if an only ifs,, we also prove ( for free! ) that x(t) = V e Λt V 1 x is the only solution to (1) 3 (a) By assumption, we have that ẋ(t) = Ax(t), x() = x, ż(t) = Az(t), z() = x an x z, ie, that there exists a t such that x(t) z(t) Note that x(t) = x() + ẋ(τ)τ = x + Ax(τ)τ, z(t) = z() + Hence, using both linearity of integration an the properties of norms, ( δ(t) = x(t) z(t) = x + Ax(τ)τ x + Az(τ)τ) = A(x(τ) z(τ)) τ Using α = an β = A an applying (3) we get A x(τ) z(τ) τ = δ(t) αe βt = e A t =, t ż(τ)τ = x + Az(τ)τ A(x(τ) z(τ))τ A δ(t)τ Hence, δ, so x z, which contraicts our initial assumption that x(t) an z(t) are ifferent (at least for one value of t) Thus, there are not multiple solutions to (1), there is at most one Page 5

6 (b) ẋ(t) = t x(t) = t (V eλt V 1 x ) = V t (eλt )V 1 x = V Λe Λt V 1 x But A = V ΛV 1, hence AV = V Λ Plugging this into the above we get ẋ(t) = AV e Λt V 1 x = Ax(t) (c) x() = V e Λ V 1 x = V IV 1 x = V V 1 x = Ix = x Thus we can conclue that the unique solution to (1) is given by (2) Page 6