HW # 2 Solutions. The Ubiquitous Curling and Hockey NBC Television Schedule. March 4, 2010

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1 HW # 2 Solutions The Ubiquitous Curling and Hockey NBC Television Schedule March 4, 2010 Hi everyone. NBC here. I just got done airing another 47.2 hours of either curling or hockey. We understand that Corey likes both of those sports, but we both understand that he also likes basically every other olympic sport, and simultaneously refuse to air anything other that curling and hockey. Even on the internet. Enjoy these solutions! Again, Corey is posting those solutions which he grades or finds interesting and nontrivial enough to post. He wouldn t post stuff he specifically covered already in class. Have fun! (a) By definition, W = span{v, T (v), T 2 (v),...}. Let w W. Then for some scalars a i, we have w = a i T i (v), where by convention, we express T 0 (v) = v. Then T (w) = a i T i+1 (v) span{v, T (v), T 2 (v),...}. So it follows that W is T invariant. (b) Let Z be a T invariant subspace, and suppose v Z. We must show that W V, where W is defined above (in part (a)). This can be accomplished by showing that each of the elements of the spanning set {v, T (v), T 2 (v),...} of W is an element of Z. But Z is T invariant, so T (v) Z, and by repeating the argument, one has (inductively) T i (v) Z for all i. Thus {v, T (v), T 2 (v),...} Z, so W Z The proof follows very closely to that of Theorem 5.5, page 261. Proceeding inductively, we see that if v 1 W (that s the sum of the first one elements of the set {v 1,..., v k }, then v 1 W (that s each of the summands from the sum above). Now suppose that the sum of the first k 1 elements from the set {v 1,..., v k } being in W is equivalent to the assertion v i W for i = 1,..., k 1. Now consider v v k W. Clearly, if v i W, then this sum is in W. Conversely, applying T λ k I, where v i E λi for i = 1,..., k, we have (T λ k I)(v v k ) = (λ 1 λ k )v (λ k 1 λ k )v k 1 W, so by the induction hypothesis, each of (λ i λ k )v i W, so that v i W for i = 1,..., k 1. Now, v = v 1 + +v k 1 +v k W means that v k = v v 1 v k 1 W We first show that if W is a T invariant subspace, and W V = E λ, then W = (E λ W ). Let W λ = E λ W. Then first we show that W λ = W. Let w W λ. Then w = w λ, where the sum is taken over all distinct eigenvalues λ, and w λ W λ = E λ W. So, each w λ W, and so w W. Conversely, suppose 1

2 w W V = E λ. So w, then, must be the sum of eigenvectors corresponding to distinct eigenvalues w = λ w λ, w λ E λ. By the above problem , we have each w λ W. So w λ W λ, and w ( W λ. Now we show that η λ η) W W λ = {0}. If η λ w η W λ, then using the T invariant subspace W λ and problem , we have each w η W λ, which could only be the case if w η = 0, since E η E λ = {0} when η λ. So, W = W λ. Now, we can find a basis β λ for W λ for each λ and proceed. The space W λ is T invariant, and T Wλ = λi. So the basis β λ diagonalizes T Wλ, and the union β = λ β λ over all eigenvalues λ of T yields a basis for W = W λ that diagonalizes T W (a) We suppose that T and U are diagonalizable operators, and that UT = T U. We show that there exists one basis β for V for which both [U] β and [T ] β are diagonal. We note that V = λ E λ, where the E λ are the eigenspaces of T, and the direct sum is taken over all eigenvalues λ of T. We also note that if, for each eigenvalue λ of T, β λ can be found which simultaneously diagonalizes both T and U on E λ, then the basis β = β λ is a desired basis for V that diagonalizes both T and U on V, where the union is taken over all eigenvalues λ of T. So we set out to find a basis β λ for E λ diagonalizing both T and U. First we show that U is E λ invariant. After that, we observe that the restriction U λ of U to E λ is diagonalizable since it is the restriction of a diagonalizable operator to an invariant subspace (see # 24). Last, we ll notice that every basis of E λ diagonalizes T on E λ since T λ, the restriction of T to E λ, is a scalar multiple of the identity: T λ = λi. Thus, a desired basis β λ for E λ is one which diagonalizes U λ, since such a basis would necessarily also diagonalize T λ = λi. So showing the U is E λ invariant is our last duty. Suppose v E λ. We must show that Uv E λ. To check this, we have to verify that T (Uv) = λ(uv). But by our commutativity hypothesis, T (Uv) = (T U)v = (UT )v = U(λv) = λ(uv) The proof that this is an inner product is actually independent of A, provided that A satisfies a condition known as being Hermitian, which for the most part means that A = A (here, A is the adjoint of the matrix A). The first part follows from direct observation about the algebra of appropriately sized matrices: x + y, z = (x + y)az = xaz + yaz = x, z + y, z. The next property also follows quite easily The third property follows since A = A: cx, z = (cx)az = c(xaz ) = c x, z. 2

3 x, z = x, z T = xaz T = z T A T x T = zax = z, x. Finally, we do the multiplication, with i = 1, v 1 = a + bi and v 2 = c + di so that v 2 = c di. ( ) ( ) ( ) 1 i v1 v1 v 2 = ( ) ( ) v v i 2 v 1 v 1 + i v i v v 2 = v v i(v 1 v 2 v 1 v 2 ) = a 2 + b 2 + 2c 2 + 2d 2 + 2ad 2bc = d 2 + c 2 + (a + d) 2 + (b c) 2 > 0 when either of v 1 or v 2 is We compute by hand, using a different dummy variable in the second slot. Keep in mind that the hypotheses state that {v 1,..., v n } is an orthogonal set, so that v i, v j = 0 if i j: i a iv i 2 = i a iv i, j a jv j = i,j a iā j v i, v j = i a iā i v i, v i = a i 2 v i Here are some thoughts on this problem, as it s sort of a big deal. Here s a proof: Proof. If every operator in C has only one eigenvalue, then since they re all diagonalizable, each one is represented by a scalar matrix for any choice of basis. (That means I couldn t screw up each of them being diagonal if I tried. This is a big point.) Namely, they re simultaneously diagonalizable. Notice also that if the dimension of the vector space is 1, then the same statement is true: any basis diagonalizes every element in C. We induct on the dimension of V, and the base case has already been verified. Suppose that on any vector space of dimension less than n, that whenever we have a family of commuting diagonalizable operators on such a vector space, that they are simultaneously diagonalizable. Since the result is true for any vector space V of dimension n where every operator has only one eigenvalue, we assume that there is at least one operator T with more than one eigenvalue. Suppose λ is an eigenvalue of T, and that Λ is the set of eigenvalues of T. Note that since Λ > 1 by hypothesis, we have 1 dim E λ < n. Since T is diagonalizable, we have V = η Λ E η, and that the vector space η Λ,η λ E η has a dimension dim( η Λ,η λ E η ) = n dim(e λ ) {1,..., n 1}. 3

4 Namely, if there were a commuting collection of linear operators for each of these spaces, the induction hypothesis would then apply to the vector spaces E λ and W := η Λ,η λ E η. To do this, we must find such a collection C λ for E λ and C W for W. We establish this presently. Suppose U C is an arbitrary element of C. We show that E λ and W are both U invariant. We begin by noting that if η is any eigenvalue of T, then E η is U invariant. This fact follows precisely as in the last part of the argument in number (a). So, U is E λ invariant (η = λ). For every eigenvalue η λ, U : E η E η, and thus U : W W since W is the direct sum of eigenspaces. So E λ and W are U-invariant. If U C, then its restrictions U λ to E λ and U W to W are both still diagonalizable. Furthermore, if we set C λ = {U λ U C} and C W = {U W U C}, then C λ and C W is a set of commuting diagonalizable operators. So, the induction hypothesis applies to both, and we have bases β λ of E λ and β W of W which simultaneously diagonalize the operators in C λ and C W respectively. Since V = E λ W, the collection β = β λ β W is a basis for V. Since U = U λ U W, each of the operators remains simultaneously diagonalized. So, here s the deal. With most induction proofs I feel as though if you gave me n, and enough time, that the proof is instructive enough to actually produce what is being asked for. You know, the idea that you could do one step, put your work aside and focus on what s left, then repeat until you find yourself finished and yearning to see something other than curling or hockey on TV. In this case, it didn t seem as though I could do that until I gave it some thought. So, here s a description of what I might do if given a collection C on a vector space of dimension n. I would first choose any operator not already diagonal (i.e., choose any basis, and see if any operators are not scalar matrices.) If every one of them is, then I m done! Back to watching curling and/or hockey on NBC. If not, then choose any one which is not. This means that it must have more than one eigenvalue. Choose any one of its eigenvalues λ and get to work. Everything we accomplish below (i.e., finding a suitable basis for E λ ) can then be applied to each eigenvalue and eigenspace, and we ll be done. Suppose for just a moment that C contained 3 operators: T, U and Ũ. Number 25 shows that we may find a basis for E λ that diagonalizes U, since E λ is U invariant. But what about Ũ? Like before, E λ is Ũ invariant. Here s where thinking about it too much leads one to not many conclusions, but remembering something simple helps a lot. Rather than formulae and proofs (since I ve already given a valid proof above), let me speak a little less formally about this situation, and you ll see that you could indeed diagonalize all of T, U, and Ũ on E λ. Suppose {v 1,..., v k } is a basis for E λ diagonalizing both T and U. What if one of the v i is not an eigenvector for Ũ? This makes v 1 somewhat of a problem. Then say i = 1 and Ũv 1 = av 1 + bv 2 for 4

5 simplicity. Then UŨv 1 = aλ 1 v 1 + bλ 2 v 2 = λ 1 av 1 + λ 1 bv 2 = ŨUv 1. Thus, λ 1 b = λ 2 b, and since v 1 was not an eigenvector, b 0. So λ 1 = λ 2. Furthermore, if v 2 isn t an eigenvector for Ũ, then Ũv 2 spreads out amongst possibly other parts of the space, but the same argument would give us λ 2 is equal to any of the other corresponding eigenvectors across which Ũv 2 might be, same with the rest of any other problem vectors that associate with one another in this manner. That is, for each problem vector, there is a set of some other problem vectors that all correspond to the same eigenvalue. Say that eigenvalue is η (of U). Then on W = E λ E η (that s λ, the eigenvalue for T, and η, the eigenvalue for U), the restriction of U to W is U W = ηi, a scalar multiple of the identity. So, all I have to do is collect all of the problem vectors associated to one eigenvalue into one pot that s W and proceed. On W, both T and U are multiples of the identity, so try as I might, there s no way that any basis for W could possibly screw up the diagonal look of T W and U W (see (a), and further, it s easy to see that W is both T and U invariant). It s also easy to check in the same manner as before that W is Ũ invariant since all of T, U and Ũ commute. So Ũ W may be diagonalized on W, and so, on W, all of T, U, and Ũ are simultaneously diagonalizable. It s possible that there were other problem vectors that correspond to another eigenvalue of U. Those are in a seperate spot, i.e., none of the problem vectors corresponding to one eigenvalue could also be problem vectors corresponding to any other eigenvalue. So, for each eigenvalue of U, we adjust each of the problem spots as described above, and since E λ = Eη U since U Eλ is diagonalizable, where the direct sum is taken over all eigenvalues η of U, and the spaces Eη U are the eigenspaces of U corresponding to the eigenvalue η, this produces a basis for E λ which simultaneously diagonailzes all of T, U and Ũ. Now choose a different eigenvalue of T and do the same thing. The process eventually stops since dim(v ) <. So what if there is another operator Û? Then we go back to our first problem in W, diagonalize T, U, and Ũ, and do the same thing to Û that we did to Ũ before, but this time with reference to Ũ instead of U. There is no need to pay any attention to T and U as we search for a suitable basis for W, since any basis would diagonalize T and U on W. Again, since dim(v ) <, as we consider more and more operators, the successive intersections of eigenspaces will eventually lead to a bunch of spaces of dimension 1 (which would be easy to find such a basis on each such space anything works), or there would be some subspace (of possibly higher dimension) that is an eigenspace (or subspace of an eigenspace) for all of the operators in C, and all of the operators are multiples of the identity on such a space. (Actually, the situation that we end up with a bunch of 1 dimensional spaces is the same situation!) We patch all the bases together, and move eigenvalue by eigenvalue. Whew! So, I now believe that for any collection of diagonalizable commuting operators, that I could simultaneously diagonalize them. The bonus in this discussion is that this reveals the inductive step that was described above: if you could do it for E λ, then you could do it for the 5

6 whole space. Then, within E λ, you ask yourself the same question and get down to W. Then ask yourself the same question and repeat until your fingers fall off from typing, or the olympics end and you long for both curling and hockey. 6