Math 32a Fall 2003 R. Palais Third Assignment Answers Here are a few exercises concerning adjoints of linear maps. We recall that if V and W are inner-product spaces and T : V → W is a linear map then there is a uniquely determined map T ∗ : W → V , called the adjoint of T , satisfying hT v, wi = hv, T ∗wi for all v ∈ V and w ∈ W . Exercise 1. Show that (T ∗)∗ = T . Answer By definition of adjoint linear maps, hw, (T ∗)∗vi = hT ∗w, vi = hw, T vi for all v ∈ V and w ∈ W . Therefore (T ∗)∗ = T . Exercise 2. Recall that if Tij is an m × n matrix (i.e., m rows and n columns) and Sji an n × m matrix, then Sji is called the transpose of Tij if Tij = Sji for 1 ≤ i ≤ m, 1 ≤ j ≤ n. Show that if we choose orthonormal bases for V and W , then the matrix of T ∗ relative to these bases is the transpose of the matrix of T relative to the same bases. Answer Let v1, v2,... and w1, w2,... be orthonormal bases for V and W . Then hT vi, wji = P ∗ D P ∗ E ∗ ∗ h k Tkivk, wji = Tji and hvi,T wji = vi, k Tkjwk = Tij. Since hT vi, wji = hvi,T wji, ∗ we get that Tij = Tji. Exercise 3. Show that ker(T ) and im(T ∗) are orthogonal complements in V , and similarly, im(T ) and ker(T ∗) are each other’s orthogonal complements in W . (Note that by Exercise 1, you only have to prove one of these.) Answer Let v ∈ ker(T ), so T v = 0, and let u ∈ im(T ∗), so u = T ∗w for some w ∈ W . Then hv, ui = hv, T ∗wi = hT v, wi = h0, wi = 0, so v and u are orthogonal. Exercise 4. Show that a linear operator T on V is in the orthogonal group O(V ) if and only if TT ∗ = I (where I denotes the identity map of V ) or equivalently, if and only if T ∗ = T −1. Answer If T ∈ O(V ), then T preserves the inner product. Then for any v, w ∈ V , hv, wi = hT v, T wi = hv, T ∗(T w)i. Therefore T ∗T = I, or T ∗ = T −1. Conversely, if T ∗ = T −1, then for any v, w ∈ V , hv, wi = hv, T ∗(T w)i = hT v, T wi, so T preserves the inner product, and therefore is in O(V ). If T : V → V is a linear operator on V , then T ∗ is also a linear operator on V , so it makes sense to compare them and in particular ask if they are equal. Definition. A linear operator on an inner-product space V is called self-adjoint if T ∗ = T , i.e., if hT v1, v2i = hv1, T v2i for all v1, v2 ∈ V . Note that by Exercise 3 above, self-adjoint operators are characterized by the fact that 1 Math 32a Fall 2003 R. Palais their matrices with respect to an orthonormal basis are symmetric. Exercise 5. Show that if W is a linear subspace of the inner-product space V , then the orthogonal projection P of V on W is a self-adjoint operator on V . Answer Let u, v ∈ V , and decompose each into a vector in W and a vector perpendicular to W : u = uW + uW ⊥ , v = vW + vW ⊥ . Then hP u, vi = huW , vi = huW , vW + vW ⊥ i = huW , vW i, and hu, P vi = hu, vW i = huW + uW ⊥ , vW i = huW , vW i. Therefore hP u, vi = hu, P vi, so P is self-adjoined. Definition. If T is a linear operator on V , then a linear subspace U ⊆ V is called a T -invariant subspace if T (U) ⊆ U, i.e., if u ∈ U implies T u ∈ U. Remark. Note that if U is a T -invariant subspace of V , then T can be regarded as a linear operator on U by restriction, and clearly if T is self-adjoint, so is its restriction. Exercise 6. Show that if T : V → V is a self-adjoint operator, and U ⊆ V is a T -invariant subspace of V , the U ⊥ is also a T -invariant subspace of V . Answer Let u ∈ U. Since U is a T -invariant subspace, T u ∈ U. Then for any v ∈ U ⊥, 0 = hT u, vi = hu, T vi, so T v ∈ U ⊥. Thus U ⊥ is a T -invariant subspace of V . We next recall for convenience the definitions relating to eigenvalues and eigenvectors. We assume that T is a linear operator on V . If λ is a real number, then we define the linear subspace Eλ(T ) of V to be the set of v ∈ V such that T v = λv. In other words, if I denotes the identity map of V , then Eλ(T ) = ker(T − λI). Of course the zero vector is always in Eλ(T ) . If Eλ(T ) contains a non-zero vector, then we say that λ is an eigenvalue of T and that Eλ(T ) is the λ- eigenspace of T . A non-zero vector in Eλ(T ) is called an eigenvector of T belonging to the eigenvector λ. The set of all eigenvalues of T is called the spectrum of T (the name comes from quantum mechanics) and it is denoted by Spec(T ). Exercise 7. Show that a linear operator T on V has a diagonal matrix in a particular basis for V if and only if each element of the basis is an eienvector of T , and that then Spec(T ) consists of the diagonal elements of the matrix. Answer Suppose that T has a diagonal matrix in a particular basis v1, v2,... of V , so the diagonal entries are Tii = λi, and the rest are 0. Then T vi = λivi, so the vi are eigenvectors of T , and the λi are in the spectrum Spec(T ). Moreover, the λi describe all P P the elements of Spec(T ), since, for w = i aivi, T w = i aiλivi = λw if and only if all the λi are equal to λ. Conversely, suppose that the spectrum of T is Spec(T ), and that its eigenvectors v1, v2,... form a basis of V . Then we have that T vi = λivi for some number λi, which shows that j Tij = λiδi , so T has a diagonal matrix. 2 Math 32a Fall 2003 R. Palais . Exercise 8. If T is a self-adjoint linear operator on an inner-product space V and λ1, λ2 are distinct real numbers, show that Eλ1 (T ) and Eλ2 (T ) are orthogonal subspaces of V . In other words, eigenvectors of T that belong to different eigenvalues are orthogonal. (Hint: Let vi ∈ Eλi (T ), i = 1, 2. You must show that hv1, v2i = 0. Start with the fact that hT v1, v2i = hv1, T v2i.) Answer Let v1 ∈ Eλ1 (T ) and v2 ∈ Eλ2 (T ). Then hT v1, v2i = hλ1v1, v2i = λ1 hv1, v2i and hv1, T v2i = hv1, λ2v2i = λ2 hv1, v2i. Since T is self-adjoint, hT v1, v2i = hv1, T v2i, so λ1 hv1, v2i = λ2 hv1, v2i, Since λ1 is not equal to λ2, we have that hv1, v2i = 0, so the eigenvectors of T that belong to different eigenvalues are orthogonal. 3.