Isomorphisms of inner product spaces Aim lecture: We examine the notion of isomorphisms for inner product spaces. In this lecture, we let F = R or C. Let V ; W be inner product spaces with inner products denoted (·|·). Prop-Defn A linear map T : V −! W preserves inner products if for all v; v0 2 V we have (T vjT v0) = (vjv0). In this case T is injective. If T is also bijective, then we say that T is an isomorphism of inner product spaces. In this case T −1 also preserves n inner products. A co-ordinate system C : F −! V is said to be orthonormal if C is an isomorphism of inner product spaces. Proof. Suppose T preserves inner products & v 2 ker T . Then 0 = (T vjT v) = (vjv) so v = 0 which shows T to be injective. It is an easy ex to see T −1 preserves inner products. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 1 / 8 Orthonormal co-ordinate systems Prop n Let C = (v1 ::: vn): F −! V be linear. Then C preserves inner products iff B = fv1;:::; vng is orthonormal. In particular, C is an orthonormal co-ordinate system iff B is an orthonormal basis for V . Proof. Suppose first that B is orthonormal & let T 0 0 0 T n v = (β1; : : : ; βn) ; v = (β1; : : : ; βn) 2 F . Then 0 X X 0 X 0 X 0 0 (CvjCv ) = ( βi vi j βj vj ) = βi βj (vi jvj ) = βi βi = (vjv ) i j i;j i Thus C preserves inner products. The converse is an easy ex reversing the above computation. E.g. Consider the R-space V = Span(sin x; cos x) with inner product R π (f jg) = −π f (t)g(t)dt. Then the co-ord system p1 2 C = π (sin x cos x): R −! V is orthonormal. Rem Every fin dim inner product space has an orthonormal co-ord system by Gram-Schmidt. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 2 / 8 Isometries Defn Let T : V −! W be a (not necessarily linear) fn. We say T is an isometry of V if it preserves \distances" in the sense that kT v − T v0k = kv − v0k for all v; v0 2 V . n n n E.g. 1 Pick v0 2 R . The translation by v0 map T : R −! R : v 7! v + v0 is an isometry which is not linear unless v0 = 0. Why? 3 3 E.g. 2 Reflection about a plane in R is an isometry of R which is linear if the plane is a subspace. 3 3 E.g. 3 Rotation about a line in R is an isometry of R which is linear if the line is a subspace. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 3 / 8 Linear isometries Prop Suppose that the inner product space V is over R 1 0 1 0 2 2 0 2 0 (vjv ) = 2 kv + v k − kvk − kv k for all v; v 2 V . 2 A linear map T : V −! W preserves inner products iff T is a linear isometry. Proof. 1) Just calculate. 2) If T preserves inner products then kT v − T v0k2 = (T (v − v0)jT (v − v0)) = (v − v0jv − v0) = kv − v0k2 so T is an isometry. Conversely, if T is a linear isometry, then kT vk = kT v − T 0k = kv − 0k = kvk for all v 2 V so T preserves inner products by 1). Rem The result 2) above is true over C but you need a more complicated version of 1). Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 4 / 8 Unitary operators Prop-Defn Let U : V −! W be linear. The following condns on U are equiv. 1 U is an isomorphism of inner product spaces. ∗ ∗ ∗ ∗ −1 2 U exists & U U = id V ; UU = id W (i.e. U = U ). We say U : V −! V is unitary or a unitary operator if U∗ = U−1. A complex ∗ −1 matrix A 2 Mnn(C) is unitary if A = A . A real matrix A 2 Mnn(C) is orthogonal if AT = A−1. In both the complex & real case, a matrix is unitary, n n resp orthogonal iff the columns are an orthonormal basis of C , resp R . Proof. We need only show 1) () 2). Suppose first that U is an isomorphism of inner product spaces. Then 2) will follow if we can show U−1 is the adjoint. But for all v 2 V ; w 2 W we have (U−1wjv) = (UU−1wjUv) = (wjUv) as required. Suppose now that U−1 is an adjoint of U. We need to show 1) holds. Indeed, Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 5 / 8 The unitary group Let Un denote the set of all unitary matrices in Mnn(C) Prop-Defn 0 0 1 If U; U 2 Un then UU 2 Un. −1 2 If U 2 Un then U 2 Un. In particular, Un is a group when endowed with matrix multiplication. It is called the unitary group & has group identity In. Proof. This follows from the following easy ex Lemma If T : V −! W ; S : W −! X are linear maps between inner product spaces preserving inner products, then S ◦ T also preserves inner products. Similary, the set of all orthogonal matrices in Mnn(R) forms a group under matrix multn called the orthogonal group On. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 6 / 8 Unitarily similar matrices Defn Two matrices A; B 2 Mnn(C) are unitarily similar if there is a unitary matrix −1 ∗ U 2 Mnn(C) such that A = U BU. In this case A = U BU. Two matrices A; B 2 Mnn(R) are orthogonally similar if there is an orthogonal matrix −1 T U 2 Mnn(R) such that A = U BU. In this case A = U BU. If B 2 Mnn(C)& A is some matrix representing it wrt some orthonormal change of co-ord systems, then A & B are unitarily similar. Prop Let U : W −! W be linear & T : V −! W be an isomorphism of inner product spaces. Then (T ∗UT )∗ = T ∗U∗T . In particular, if U is unitary, so is T ∗UT : V −! V . Proof. Indeed we just compute Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 7 / 8 Basic properties of unitary matrices Prop Let U 2 Un. 1 j det Uj = 1. 2 If U is orthogonal then det(U) = ±1 3 The e-values of U also have modulus 1. Proof. 1) & 2) Just note that T 1 = det I = det(UU∗) = det(U) det(U ) = det(U) det(U) = det(U)det(U) = j det(U)j2: 3) Let v be an e-vector with e-value λ. Then λ(vjv) = (vjλv) = (vjUv) = (U∗vjv) = (U−1vjv) = (λ−1vjv) = λ−1(vjv): Hence λ = λ−1 & jλj2 = 1 which gives the result. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 8 / 8.