Lecture 38: Unitary Operators Semester 2 2012 1 / 8 Orthonormal Co-Ordinate Systems

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 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 ∈ 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 n Let C = (v1 ... vn): F −→ 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 T 0 0 0 T n v = (β1, . . . , βn) , v = (β1, . . . , βn) ∈ F . Then 0 X X 0 X 0 X 0 0 (Cv|Cv ) = ( βi vi | βj vj ) = βi βj (vi |vj ) = βi βi = (v|v ) 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 |g) = −π f (t)g(t)dt. Then the co-ord system √1 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 ∈ V .

n n n E.g. 1 Pick v0 ∈ 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 (v|v ) = 2 kv + v k − kvk − kv k for all v, v ∈ 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)|T (v − v0)) = (v − v0|v − 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 ∈ 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 ∈ Mnn(C) is unitary if A = A . A real matrix A ∈ 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 ∈ V , w ∈ W we have (U−1w|v) = (UU−1w|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 0 0 1 If U, U ∈ Un then UU ∈ Un. −1 2 If U ∈ Un then U ∈ 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 −1 ∗ U ∈ Mnn(C) such that A = U BU. In this case A = U BU. Two matrices A, B ∈ Mnn(R) are orthogonally similar if there is an orthogonal matrix −1 T U ∈ Mnn(R) such that A = U BU. In this case A = U 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−1v|v) = (λ−1v|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