Lecture 19: Isometries, Positive operators, Polar
and singular value decompositions; Unitary
matrices and classical groups; Previews (1)
Travis Schedler
Thurs, Nov 18, 2010 (version: Wed, Nov 17, 2:15 PM)
Goals (2)
• Isometries
• Positive operators
• Polar and singular value decompositions
• Unitary matrices and classical matrix groups
• Preview of rest of course!
Characterization of isometries (3)
Correction: An isometry V → W need not be invertible even if V and W
are finite-dimensional: e.g., if V = 0 the zero map is an isometry! However, it
must be injective so invertible if dim V = dim W < ∞.
Proposition 0.1. If T ∗ exists (e.g., if V is finite-dimensional), then T ∈
L(V, W ) is an isometry iff T ∗ T = IV .
Proof.
• First, T is an isometry iff hT ∗ T v, vi = hv, vi for all v.
• Next, T ∗ T and hence S := T ∗ T − I is always self-adjoint.
• T is an isometry iff hSv, vi = 0 for all v.
• Now, apply the skipped result from last time below.
Proposition 0.2 (Proposition 7.4). If T is self-adjoint, then hT v, vi = 0 for
all v iff T = 0.
1
Proof of the proposition (4)
Proposition 0.3 (Proposition 7.4). If T is self-adjoint, then hT v, vi = 0 for
all v iff T = 0.
Proof.
• Assume hT v, vi = 0 for all v. Then hT (v + u), v + ui − hT (v), vi −
hT (u), ui = 0 for all v, u.
• Thus, hT (v), ui + hT (u), vi = 0 for all u, v.
• When T is self-adjoint, this says 2<hT (v), ui = 0 for all u, v.
• Plugging in iu for u, also 2=hT (v), ui = 0 for all u, v. Thus T (v) = 0 for
all v, i.e., T = 0.
Note: From another skipped result on last week’s slides, when V is finitedimensional, hT v, vi = 0 for all v iff T = 0 or F = R and T is anti-self-adjoint.
Positive operators (5)
Definition 1. A positive operator T is a self-adjoint operator such that hT v, vi ≥
0 for all v.
In view of the spectral theorem, a self-adjoint operator is positive iff its
eigenvalues are nonnegative (part of Theorem 7.27).
Theorem 2 (Remainder of Theorem 7.27).
T = S ∗ S is positive.
(i) Every operator of the form
(ii) Every positive operator admits a positive square root.
Proof.
• (i) First, T ∗ = (S ∗ S) = S ∗ S is self-adjoint.
• Next, hT v, vi = hSv, Svi ≥ 0 for all v.
√
• (ii) For any orthonormal eigenbasis of T , let T be the operator with the
same orthonormal eigenbasis, but with the nonnegative square root of the
eigenvalues.
Polar decomposition (6)
After the spectral theorem, the second-most important theorem of Chapters
6 and 7 is:
Theorem
3 (Polar decomposition: Theorem 7.41). Every T ∈ L(V ) equals
√
S T ∗ T for some isometry S.
Main difficulty: T need not be invertible!
√
Lemma 4. For all v ∈ V , ||T v|| = || T ∗ T v||.
√
√
√
√
√
Proof. || T ∗ T v||2 = h T ∗ T v, T ∗ T vi = h( T ∗ T )∗ T ∗ T v, vi = hT ∗ T v, vi =
hT v, T vi = ||T v||2 .
√
√
∼
Corollary:
null(T ) = null( T ∗ T ). We may√
thus define S1 : range( T ∗ T ) →
range(T )
√
∗
∗
by S1 ( T T√v) = T v. Thus, for all v ∈ V , S1 T T v = T v. Also, ||S1 u|| = ||u||
for all u(= T ∗ T v) by the lemma. So S1 is an isometry.
2
Completion of proof (7)
• We only have to extend S1 to an isometry on all of V .
√
√
• Note that range( T ∗ T ) ⊕ range( T ∗ T )⊥ = V = range(T ) ⊕ range(T )⊥ .
√
∼
• Thus, the extensions of S1 : range( T ∗ T ) →
range(T
√ ) to an∼isometry S⊥:
∼
range(T )
V → V are exactly S = S1 ⊕ S2 , where S2 : range( T ∗ T )⊥ →
is an isometry.
• Since these are inner product spaces of the same dimension, there always
exists an isometry, by taking an orthonormal basis to an orthonormal
basis.
Recall here that T1 ⊕ T2 on U1 ⊕ U2 means (T1 ⊕ T2 )(u1 + u2 ) = T1 (u1 ) + T2 (u2 ),
∀u1 ∈ U1 , u2 ∈ U2 .
Singular value decomposition (SVD) (8)
Let V be finite dimensional and T ∈ L(V ).
Theorem 5. There exist orthonormal bases (e1 , . . . , en ) and (f1 , . . . , fn ) and
nonnegative values si ≥ 0 such that T ei = si fi .
The si are called the singular values.
Proof.
• Let (e1 , . . . , en ) be an orthonormal eigenbasis of the positive
Let s1 , . . . , sn be the nonnegative eigenvalues.
√
• Using polar decomposition, T = S T ∗ T for S an isometry.
√
T ∗T .
• Let fi := Sei . Then T ei = si fi .
Corollary: M(ei ),(fi ) (T ) = D, where D is diagonal with entries si ≥ 0.
Improvement on normal form: (ei ), (fi ) orthonormal! Note: To compute, first
find √
eigenvalues s2i of T ∗ T , then orthonormal eigenbasis ei . This yields fi , S,
and T ∗ T !
Unitary matrices and A = U1 DU2−1 decomposition (9)
Corollary: If A = M(vi ) (T ) for any orthonormal basis, then A = U1 DU2−1
for the change-of-basis matrices U1 , U2 . Moreover, the columns of U1 and U2
form orthonormal bases of Mat(n, 1, F): such matrices are called unitary.
Proposition 0.4. The following are equivalent for U ∈ Mat(n, n, F):
• The columns form an orthonormal basis;
• U U ∗ = I = U ∗U ;
• U = M(ei ) (S) for some isometry S ∈ L(V ) and some orthonormal basis
(ei ) of V .
3
Proof.
• The first two are immediately equivalent.
• In general, M(ei ) (S) has columns equal to Sei in the basis (ei ), so hSei , Sej i =
the dot product of the i-th and j-th column of M(ei ) (S).
Example ofSVD (A= U1 DU2−1 decomposition)! (10)
10 −2
Let A =
. Compute the decomposition A = U1 DU2−1 .
−5 11
125 −75
• First step: Find eigenvalues and eigenbasis of A∗ A =
.
−75 125
• Char. poly of A = x2 − (tr A)x + det A = x2 − 250x + 10000 = (x −
200)(x − 50). Eigenvalues: s21 = 200, s22 = 50.
1
−75 −75
.
• Nullspace of A∗ A − 200I =
: spanned by e1 := √12
−1
−75 −75
1
• Same computation for e2 : get e2 := √12
. Alternative: e2 determined
1
up to scaling so that e1 ⊥ e2 .
Computation continued
(11)
1
1
10 −2
1
1
2
2
√
√
, e2 = 2
.
Recall: A =
, s1 = 200, s2 = 50, e1 = 2
−1
1
−5 11
12
3/5
−1
1
• Now, f1 = s1 Ae1 = 20
=
.
−16
−4/5
8
4/5
−1
1
• Similarly, f2 = s2 Ae2 = 10
=
. Caution: f2 ⊥ f1 , but this
6
3/5
only determines f2 up to scaling by absolute value one (here ±1)!
√
3/5 4/5
s1 0
10 2 √
0
• Now, U1 = (f1 f2 ) =
, D =
=
,
−4/5 3/5
0 s2
0
5 2
1 1
and U2 = (e1 e2 ) = √12
. Check: A = U1 DU2−1 !
−1 1
Classical matrix groups (12)
Definition 6. A group is a set G with an associative operation G × G → G
with an identity and inverses (not necessarily commutative).
Definition 7. GL(n, F) ⊆ Mat(n, n, F) is the group of invertible matrices “general linear group,” under multiplication. SL(n, F) ⊆ Mat(n, n, F) is the subgroup of matrices of det= 1 “special linear group.”
(We haven’t defined determinant yet. But det(AB) = det(A) det(B) so
SL(n, F) is closed under mult.)
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Definition 8. O(n, R) = the group of orthogonal (=unitary) matrices: O such
that Ot O = I = OOt . U(n, C) = the group of unitary matrices: U such that
U ∗U = I = U U ∗.
Note: Also have O(n, F): matrices such that Ot O = I = OOt for any F.
Hence U(n, C) 6= O(n, C)!
Finite subgroups; platonic solids (13)
• O(n, R) = group generated by reflections (and rotations)!
• Big question: What are the finite subgroups G < O(n, R)?
(
cos θ − sin θ
• In case n = 2: Just “cyclic” groups Cm =
: θ =
sin θ
cos θ
)
2πk/m; 1 ≤ k ≤ m , and “dihedral” groups Cm ∪ T Cm , where T is a
reflection.
• In case n = 3: Groups of rotations only are: cyclic and dihedral groups
on the plane, and platonic solid rotation groups: groups of rotation fixing
the platonic solids!
• The other groups are G ∪ T G where G is as above, and T is a single
reflection.
• This gives a classification of the platonic solids: the only three finite rotation groups in O(3, R) which don’t fix a plane!
Classical groups of V ; bilinear and quadratic forms (14)
Similarly:
• GL(V ), SL(V ) ⊆ L(V ) are the groups of invertible, det= 1 linear transformations (V = vector space);
• O(V ), U(V ) are the groups of isometries in the cases F = R, C (V = inner
product space);
• For general F, we can define groups O(V ) if we generalize inner product
spaces:
– A bilinear form h−, −i : V × V → F is a map which is additive and
homogeneous (in both slots!).
– It is symmetric if hu, vi = hv, ui.
– It is nondegenerate if hu, vi = 0 for all u ∈ V implies v = 0, and also
hv, ui = 0 for all u ∈ V implies v = 0.
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Definition 9. Let V have a (nondegenerate symmetric) bilinear form. Then
O(V ) ⊆ L(V ) = {T : hu, vi = hT u, T vi for all u, v ∈ V }.
Quadratic form: q(v) := hv, vi, satisfying q(λv) = λ2 q(v), parallelogram identity; uniquely determines h−, −i if 1/2 ∈ F.
Preview of rest of course: C case (15)
• Characteristic polynomial: write upper-triangular matrix A = M(T ). If
the diagonal entries are λ1 , . . . , λN , then char. poly := (x−λ1 ) · · · (x−λN ).
N
• Theorem 1: This does not depend on basis.
Call
Pit χT (x) = x +
N −1
a
+ · · · + a0 . Definition: tr(T ) = −aN −1 = λi ; det(T ) = a0 =
QN −1 x
λi .
• Theorem 2 (Cayley-Hamilton): χT (T ) = 0.
• Theorem 3: tr(T ) = tr(A) = sum of diagonal entries. Corollary: tr(T ) +
tr(S) = tr(T + S) (sum of sums of eigenvalues = sum of eigenvalues of
sum!)
• Theorem 4: a0 = det(A) = an explicit formula we will study, satisfying
det(AB) = det(A) det(B).
– Case F = R: | det(A)| = volume of A(unit n-cube).
Corollary: det(T S) = det(T ) det(S). Corollary: χT (T ) = det(M(T ) −
xI). So the ai are polynomial functions in entries of A !
Arbitrary F (16)
Definition: χT (x) = det(M(T ) − xI).
• Theorem 1’: Does not depend on basis.
• Theorem 2’ (C-H): χT (T ) = 0.
• Corollary: The roots of χT are exactly the eigenvalues.
Jordan canonical form (17)
Finally, back to F = C. The following only requires Theorem 1:
Theorem 10. For

 F = C, in some basis, T has block diagonal form with blocks
λ 1


..
..


.
.

.


.
. . 1

λ
More generally (unfortunately we don’t have time to prove):
6
• For F = R, we can say the same thing if we also allow λ to be
and then 1 becomes I (alternatively,
).
1
a
b
−b
,
a
• For general F, we can say the same if we allow λ to be amatrix with
irreducible characteristic polynomial, and replace 1 by
.
1
7