Compact Operators in Hilbert Space
Hart Smith
Department of Mathematics
University of Washington, Seattle
Math 526/556, Spring 2015
Hilbert-Schmidt integral kernels on L2 (A)
If K (x, y ) ∈ L2 (A × A), and H = L2 (A), define TK ∈ B(H) by
Z
TK g(x) = K (x, y ) g(y ) dy .
Then hf , TK gi = hTK f , gi iff
Z
Z Z
Z
K (x, y ) g(y ) dy dx =
f (x)
K (y , x)f (x) dx g(y ) dy
TK is self-adjoint iff K (x, y ) = K (y , x) for a.a. x, y ∈ A × A.
TK is compact since it’s a Hilbert-Schmidt operator.
Self-adjoint Hilbert-Schmidt integral kernels on L2 (A)
Corollary: assume K (x, y ) = K (y , x)
∃ orthonormal set {φj } ⊂ L2 (A) of eigenvectors for TK :
Z
K (x, y ) φj (y ) dy = λj φj (x) , λj 6= 0 ,
such that if {ψk } is an orthonormal basis for ker(TK ), then
{φj } ∪ {ψk } is an orthonormal basis for L2 (A).
It follows that
{φj (x)φk (y )} ∪ {φj (x)ψk (x)} ∪ {ψj (x)φk (y )} ∪ {ψj (x)ψk (y )}
is an orthonormal basis for L2 (A × A). Expand K (x, y ) in basis:
K (x, y ) =
∞
X
j=1
λj φj (x) φj (y )
Square root of T ∗ T
Suppose T is compact on H (not necessarily self-adjoint).
T ∗ T is compact, self-adjoint on H, and hT ∗ Tx, xi = kTxk2 ≥ 0
Lemma
There exists a unique compact, non-negative, self-adjoint
operator S such that S 2 = T ∗ T .
Proof. Let N0 = ker(T ∗ T ) = ker(T ) , E = N0⊥ , so H = E ⊕ N0 .
T ∗ T : E → E is compact, 1-1, so in some orthonormal basis
T ∗ T = diag(µ1 , . . . , µ1 , µ2 , . . . , µ2 , µ3 , . . .)
where µ1 > µ2 > µ3 > · · · > 0 and µj → 0 unless finitely many.
Square root of T ∗ T
Define S : E → E by the matrix
√
√
√
√
√
S = diag µ1 , . . . , µ1 , µ2 , . . . , µ2 , µ3 , . . .
Extend S to H = E ⊕ N0 by Sy = 0 if y ∈ N0 = ker(T ).
S 2 x = T ∗ Tx if x ∈ E, and S 2 y = T ∗ Ty = 0 if y ∈ N0 .
Any z ∈ H of form x + y ∈ E ⊕ N0 , so S 2 = T ∗ T on H.
If S 2 = T ∗ T is as in the statement, then S commutes with
S 2 = T ∗ T , so each Nµj , as well as N0 , is invariant for S.
p
• S self-adjoint, non-negative, ⇒ S = µj on Nµj .
• Similarly, S = 0 on N0 , so S unique.
Note:
hSx, Sxi = hS 2 x, xi = hT ∗ Tx, xi = hTx, Txi
therefore: kSxk = kTxk for all x ∈ H.
Polar Decomposition for compact T on H.
Theorem
Suppose T is compact on a Hilbert space H. Let E = (ker T )⊥ .
One can write T = US, where U : E → H is a norm preserving
map, S is self-adjoint non-negative, and kSxk = kTxk ∀ x ∈ H.
Proof. Let S 2 = T ∗ T , with S ≥ 0 self-adjoint, so kSxk = kTxk.
S : E → E is self-adjoint, 1-1, so dense range: S(E) = E.
Define U : S(E) → H by U(Sx) = Tx for x ∈ E.
Since kUy k = ky k and S(E) = E, U extends uniquely to a
map E → H with kUy k = ky k for all y ∈ E. Indeed, U is
an isometry of E onto T (E).
Since U(Sx) = Tx for all x ∈ E, and U(Sx) = Tx = 0 for
x ∈ E ⊥ = ker T , then U(Sx) = Tx for all x ∈ H.
Remark: U and S are unique. The condition kSxk = kTxk
forces S : ker T → {0}, and S = S ∗ then implies S : E → E.
Then T ∗ T = SU ∗ US = S 2 , so S is the unique non-negative
self-adjoint square root of T ∗ T .
Example:
T (x1 , x2 , x3 , . . .) = (0, x1 , 21 x2 , 31 x3 , . . .) , T = SR ◦ diag(1, 12 , 13 , . . .)
Corollary
If T ∈ B(H) is compact, there exists a sequence {TN } ⊂ B(H)
of finite-rank operators such that limN→∞ kT − TN kB(H) = 0.
Proof. Let SN be the truncation of S to N√µ1 ⊕ · · · ⊕ N√µN .
√
Then kS − SN k ≤ µN+1 → 0. Let TN = USN , then
kT − TN k ≤ kUk kS − SN k → 0 .
Singular value decomposition
Definition
If µ1 > µ2 > µ3 . . . are the eigenvalues of T ∗ T , the numbers
√
ρk = µk are called the singular values of T .
Remarks.
kT k = maxk ρk .
T is Hilbert-Schmidt iff
∞
X
ρ2k mult(ρk ) < ∞ , and
k =1
kT k2HS =
∞
X
ρ2k mult(ρk ) .
k =1
Proof. Take {xn } = eigenvector basis for T ∗ T . Then
∞
X
n=1
2
kTxn k =
∞
X
n=1
∗
hT Txn , xn i =
∞
X
k =1
ρ2k mult(ρk ) .
Singular value decomposition
Singular Value Decomposition Theorem
If T ∈ B(H) is compact with singular values {ρk }, there exists
an orthonormal basis {xn } for (ker T )⊥ , and an orthonormal
basis {yn } for T (H) , such that
∞
X
Tx =
ρk (n) hxn , xi yn .
j=0
Proof. Let {xn } be O.N. basis of E = (ker T )⊥ in which
S = diag(ρ1 , . . . , ρ1 , ρ2 , . . . , ρ2 , ρ3 , . . . )
Since U : E → T (H) is isometry, then {yn } = {Uxn } is an
P
O.N. basis for T (H). Then Txn = ρk (n) yn , and x = n hxn , xixn
with convergence in H, so the result holds by continuity.