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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 ∈ V we have (T v|T v0 ) = (v|v0 ). 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 inner products. A co-ordinate system C : Fn −→ V is said to be orthonormal if C is an isomorphism of inner product spaces. Proof. Suppose T preserves inner products & v ∈ ker T . Then 0 = (T v|T v) = (v|v) 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 Let C = (v1 . . . vn ) : Fn −→ V be linear. Then C preserves inner products iff B = {v1 , . . . , vn } 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 v = (β1 , . . . , βn )T , v0 = (β10 , . . . , βn0 )T ∈ Fn . Then X X X X (C v|C v0 ) = ( βi vi | βj0 vj ) = βi βj0 (vi |vj ) = βi βi0 = (v|v0 ) 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 Rπ (f |g ) = −π f (t)g (t)dt. Then the co-ord system C = √1π (sin x cos x) : R2 −→ 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 v0 k = kv − v0 k for all v, v0 ∈ V . E.g. 1 Pick v0 ∈ Rn . The translation by v0 map T : Rn −→ Rn : v 7→ v + v0 is an isometry which is not linear unless v0 = 0. Why? E.g. 2 Reflection about a plane in R3 is an isometry of R3 which is linear if the plane is a subspace. E.g. 3 Rotation about a line in R3 is an isometry of R3 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 (v|v0 ) = 12 kv + v0 k2 − kvk2 − kv0 k2 for all v, v0 ∈ 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 v0 k2 = (T (v − v0 )|T (v − v0 )) = (v − v0 |v − v0 ) = kv − v0 k2 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 ∈ 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. 2 U ∗ exists & U ∗ U = id V , UU ∗ = id W (i.e. U ∗ = U −1 ). We say U : V −→ V is unitary or a unitary operator if U ∗ = U −1 . A complex matrix A ∈ Mnn (C) is unitary if A∗ = A−1 . A real matrix A ∈ Mnn (C) is orthogonal if AT = A−1 . In both the complex & real case, a matrix is unitary, resp orthogonal iff the columns are an orthonormal basis of Cn , resp Rn . 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 ∈ V , w ∈ W we have (U −1 w|v) = (UU −1 w|Uv) = (w|Uv) 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 1 If U, U 0 ∈ Un then UU 0 ∈ Un . 2 If U ∈ Un then U −1 ∈ 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 ∈ Mnn (C) are unitarily similar if there is a unitary matrix U ∈ Mnn (C) such that A = U −1 BU. In this case A = U ∗ BU. Two matrices A, B ∈ Mnn (R) are orthogonally similar if there is an orthogonal matrix U ∈ Mnn (R) such that A = U −1 BU. In this case A = U T BU. If B ∈ 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 ∈ Un . 1 | det U| = 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) = | det(U)|2 . 3) Let v be an e-vector with e-value λ. Then λ(v|v) = (v|λv) = (v|Uv) = (U ∗ v|v) = (U −1 v|v) = (λ−1 v|v) = λ−1 (v|v). Hence λ = λ−1 & |λ|2 = 1 which gives the result. Daniel Chan (UNSW) Lecture 38: Unitary operators Semester 2 2012 8/8