Download Lecture 38: Unitary operators

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.
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Lecture 38: Unitary operators
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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.
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Lecture 38: Unitary operators
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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.
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Lecture 38: Unitary operators
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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
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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,
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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 .
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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
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Lecture 38: Unitary operators
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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
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