LECTURE VI: SELF-ADJOINT AND UNITARY OPERATORS MAT 204 - FALL 2006 PRINCETON UNIVERSITY ALFONSO SORRENTINO 1 Adjoint of a linear operator Note: In these notes, V will denote a n-dimensional euclidean vector space and we will denote the inner product by, Denition Let (V,, ) be a n-dimensional euclidean vector space and T : V V a linear operator We will call the adjoint of T, the linear operator S : V V such that: T (u), v = u, S(v), for all u, v V Proposition 1 Let (V,, ) be a n-dimensional euclidean vector space and T : V V a linear operator The adjoint of T exists and is unique Moreover, if E denotes an orthonormal basis of V (with respect to, ) and T has matrix B with respect to E (ie, T (E) = EB), then the adjoint of T is the linear operator S : V V, that has matrix B T with respect to E For this reason, the adjoint of T is sometimes called the transpose operator of T and denoted T T Proof Let E an arbitrary basis of V, let T (E) = EB, with B M n (R) and let A be the matrix of, (wrt E) Therefore: u, v = Ex, Ey = x T Ay for all u = Ex, v = Ey V [observe that A GL n (R), since, is an inner product] Let us denote with S : V V an arbitrary linear operator with matrix C M n (R) (wrt E), ie, S(E) = EC For any u, v V we have: Hence: T (u), v = T (Ex), Ey = EBx, Ey = (Bx) T Ay = x T (B T A)y u, S(v) = Ex, S(Ey) = Ex, ECy = x T A(Cy) = x T (AC)y S is the adjoint of T B T A = AC C = A 1 B T A Therefore, the adjoint of T exists and is unique (in fact, its matrix wrt E is uniquely determined by A and B) Moreover, if E is orthonormal, then A = I n and C = B T Denition Let (V,, ) be a n-dimensional euclidean vector space and T : V V a linear operator T is said to be self-adjoint [or symmetric] if T = T T ; ie, T (u), v = u, T (v), for all u, v V Remark Let T : V V be a linear operator and E an orthonormal basis of V If T (E) = EB, then it follows from the previous proposition that: T is self-adjoint B = B T 1
2 ALFONSO SORRENTINO Hence, self-adjoint operators on V (with dim V = n) are in 1 1 correspondence with symmetric matrices in M n (R) In particular, they form a vector subspace of End(V ), that is isomorphic to the vector subspace of symmetric matrices in M n (R) Notice that if E is NOT orthonormal, then a self-adjoint operator might not have a symmetric matrix (wrt that basis); viceversa, an operator with a symmetric matrix (wrt a non-orthonormal basis) might not be self-adjoint Example Let V be a two dimensional euclidean vector space, with inner product, dened by ( ) 2 1 A = 1 1 with respect to a xed basis E Let T : V V a linear operator, dened by T (E) = EB, where ( ) 1 2 B = 2 1 Let us verify that T is not self-adjoint (although B is symmetric) and nd the matrix of its adjoint operator S (wrt E) Observe that, if T were self-adjoint, we would have from which T (Ex), Ey = Ex, T (Ey), for all Ex, Ey V ; x T (B T A)y = x T (AB)y for all x, y M n,1 (R) Therefore, we should have B T A = AB, but: ( ) B T 0 1 A = 3 1 The adjoint of T is dened by: ( 0 3 1 1 S(E) = EC with C = A 1 B T A = ) ( 3 0 0 1 ) 2 Spectral theorem for self-adjoint operators The main theorem of this section states that if T is a self-adjoint operator on an eucliden vector space, then there exists an orthonormal basis of V formed by eigenvectors of T Theorem 1 (Spectral theorem for self-adjoint operators) Let V be a n- dimensional euclidean vector space and T : V V a self-adjoint linear operator Then, there exists an orthonormal basis of V formed by eigenvectors of T The proof of this theorem is based on the following preliminary results Proposition 2 Let A M n (R) be a symmetric matrix Then, its characteristic polynomial P = P A has n real roots (each counted with its algebraic multiplicity); hence, P can be factorized in the product of n linear polynomials in R[x] Proof From the Fundamental Theorem of Algebra, it follows that P can be always linearly factorized in C[x] We need only to show that each root λ C of P is indeed a real number; ie, λ R Let T : C n C n be the linear operator with matrix A, with respect to the canonical basis E of C n Since λ is an eigenvalue of T, then there exists a non-zero vector z such that T (z) = λz, ie, (1) Az = λz
LECTURE VI: SELF-ADJOINT AND UNITARY OPERATORS 3 Let us remember some denitions and simple results about the complex numbers For any α = x + iy C, we dene its conjugate as α = x iy In particular, the following holds: αα = x 2 + y 2 [it is called norm of α ]; αα 0 and αα = 0 α = 0 ; α = α ; α = α α R ; α + β = α + β and αβ = αβ Let us consider a non-zero vector z = z 1 z n C n One can easily verify that z T z = z 1 z n C n and its conjugate z = n z i z i > 0 [at least one z i is dierent from zero] ; i=1 ( ) z T Az = z T Az = z T Az [since A is real] If we multiply (1) by z T (on the left), we get and consequently z T Az = z T λz = λ(z T z), λ = 1 z T z (zt Az) In order to show that λ R, it suces to verify that z T Az R or equivalently: In fact, using that A is symmetric: (z T Az) = z T Az) (z T Az) = z T Az = (z T Az) T = z T Az Proposition 3 Let T be a self-adjoint operator on a n-dimensional euclidean vector space V and u an eigenvector Then, T (u ) u ; ie, T let the subspace u xed Proof Let T (u) = λu For any v u [ie, v, u = 0] we have: Therefore, T (v) u ; T (v), u = v, T (u) = v, λu = λ v, u = 0 We can now prove the Spectral Theorem Proof [Spectral Theorem] We proceed by induction on n = dim (V ) If dim V = 1, then the assertion is evident (it is sucient to choose any basis E = (e 1 ) with e 1 = 1) Suppose that n 2 and assume that the theorem is true for self-adjoint operators on euclidean vector spaces of dimension n 1 According to proposition 2, T has at least one real eigenvalue λ and call e 1 one of the corresponding eigenvectors We can assume that e 1 = 1 (otherwise, we just divide this vector by its norm)
4 ALFONSO SORRENTINO One can observe that e 1 is an euclidean vector subspace of V of dimension n 1 (see Lecture V, 2) From proposition 3, T (e 1 ) e 1, therefore we can restrict T to the subspace e 1 We obtain a new linear operator T : e 1 e 1, that is still self-adjoint (since it acts like T on the vectors in e 1 ) Using the inductive hypothesis, T has an orthonormal basis {e 2,, e n }, formed by eigenvectors Since e 1 e i (for all i = 1,, n), then the vectors e 1,, e n are pairwise orthogonal and therefore linearly independent These vectors form the desired basis Remark Let T be a self-adjoint operator and F an orthonormal basis of eigenvectors of T We want to point out that, with respect to this basis, T is represented by a diagonal matrix: λ 1 0 0 0 λ 2 0 D =, 0 0 λ n where λ 1,, λ n are the eigenvalues of T They are not necessarily distinct: each of them appears h λi times on the diagonal (where h λi is its algebraic multiplicity = geometric multiplicity ) Proposition 4 Let T : V V be a self-adjoint operator If u, v are eigenvectors corresponding to distinct eigenvalues, then they are orthogonal Therefore, the eigenspaces of T are pairwise orthogonal Proof Let T (u) = λu and T (v) = µv, with λ, µ R and λ µ We have: T (u), v = λu, v = λ u, v and u, T (v) = u, µv = µ u, v Since T (u), v = u, T (v), then λ u, v = µ u, v and therefore (since λ µ) u, v = 0 It follows also that E λ Eµ and E µ Eλ (where E λ and E µ are the associated eigenspaces) Remark From the previous proposition, it follows that in order to compute an orthonormal basis F of eigenvectors of a self-adjoint operators, it is enough to nd the basis of each eigenspace E λ and orthonormalize it (using Gram-Schmidt, for instance) The union of such bases provides the desired one Example In R 4 with the canonical inner product, consider the linear operator dened (wrt the canonical basis of R 4 ) by: 1 0 0 0 A = 0 1 0 0 0 0 0 1 0 0 1 0 Determine an orthonormal basis F of eigenvectors of T and write the matrix of T with respect to F Solution: T has characteristic polynomial P = (x 1) 2 (x + 1) 2 and therefore its specturm is Λ(T ) = {1, 1}, with multiplicities h 1 = 2 and h 1 = 2 The eigenspace E 1 has equations: { 2x2 = 0 x 3 + x 4 = 0
LECTURE VI: SELF-ADJOINT AND UNITARY OPERATORS 5 and therefore: E 1 = (1, 0, 0, 0), (0, 0, 1, 1) This basis is already orthogonal, but we need to normalize the vectors, dividing by their norm: ( ) 1 1 E 1 = (1, 0, 0, 0), 0, 0,, 2 2 The eigenspace E 1 has equations: { 2x1 = 0 x 3 + x 4 = 0 and therefore: E 1 = (0, 1, 0, 0), (0, 0, 1, 1) This basis is already orthogonal, but we need to normalize the vectors, dividing by their norm: ( 1 E 1 = (0, 1, 0, 0), 0, 0,, 1 ) 2 2 Concluding, an orthonormal basis F for T is given by: 1 0 0 0 0 0 1 0 F = EC, where C = 1 0 2 1 0 2 1 0 2 0 1 2 The matrix of T with respect to this basis is: T (F) = FD, where D = C T AC = O 4(R) 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 3 Unitary operators Denition Let (V,, ) be a n-dimensional euclidean vector space and T : V V a linear operator We will say that is unitary if: T (u), T (v) = u, v, for all u, v V [ie, T preserves the inner product, in V ] Proposition 5 Let T be a linear operator on (V,, ) We have: Proof T is unitary T is invertible and T 1 = T T (= ) It is sucient to verify that T T T = Id, that is equivalent to T T (T (v)) = v for all v V In fact, from the denition above and that of adjoint of T, one can conclude: therefore, u, v = T (u), T (v) = u, T T (T (v)) ; u, v T T (T (v)) = 0 for all u V We can deduce from this (using the non-degeneracy of the inner product) that T T (T (v)) = v, for any v V ( =) We have that, for any u, v V : T (u), T (v) = u, T T (T (v)) = u, T 1 (T (v)) = u, v
6 ALFONSO SORRENTINO Let us try to deduce some information about the matrices of these unitary operators Let E be a basis for (V,, ) and suppose that Ex, Ey = x T Ay If T is a linear operator on V, such that T (E) = EB, then: T is unitary T (Ex), T (Ey) = Ex, Ey, for all Ex, Ey V EBx, EBy = Ex, Ey, for all Ex, Ey V (Bx) T A(By) = x T Ay, for all x, y M n,1 (R) x T (B T AB)y = x T Ay, for all x, y M n,1 (R) B T AB = A Corollary 1 Let (V,, ) be a n-dimensional euclidean vector space, with an orthonormal basis E and let T : V V a linear operator, such that T (E) = EB Then, T is unitary B O n (R) [ie, B is an orthogonal matrix] Proof For what observed above: T is unitary if and only if B T AB = A Since E is orthonormal, then A = I n and we get: T is unitary B T B = I n B O n (R) Remark i) Let T be a unitary operator If E is an orthonormal basis, then also T (E) is an orthonormal basis (see Lecture V, prop 8) Therefore, unitary operators send orthonormal bases into orthonormal bases ii) If T is a unitary operator, then det T = ±1 In fact, if T (E) = EB, with E orthonormal, then B O n (R) and det T = det B = ±1 In particular, unitary operators with det T = 1 are called special unitary operators (or rotations) of V iii) If T is unitary, then its spectrum Λ T {1, 1} In fact, if T (u) = λu, then: u, u = T (u), T (u) = λu, λu = λ 2 u, u ; therefore λ 2 = 1, that implies λ = ±1 iv) Let us point out that, in general, a unitary linear operator might not be diagonalizable Consider, for instance, R 2 with the canonical inner product and the operator that is dened (wrt the canonical basis) by ( ) R π 0 1 = SO 2 1 0 2 (R) O 2 (R) This is a unitary operator and it has no eigenvalues, hence it is not diagonalizable v) If T is unitary and Λ T = {1, 1}, then the eigenspaces E 1 and E 1 are orthogonal to each other In fact, if T (u) = u and T (v) = v, then: u, v = T (u), T (v) = u, v = u, v ; this implies that u, v = 0 and therefore u, v = 0 vi) If T is unitary and u is one of its eigenvectors, then T (u ) u In fact, for any v u (remember that λ = ±1): T (v), u = 1 λ T (v), λu = 1 λ T (v), T (u) = 1 v, u = 0 ; λ hence, T (v) u Department of Mathematics, Princeton University E-mail address: asorrent@mathprincetonedu