Let H 1 and H 2 be Hilbert spaces, let T ∈ B ( H 1 , H 2 ). Suppose that Ker ⁡ T is finite-dimensional and that Im ⁡ T is closed in H 2 . Prove that Ker ⁡ ( T + K ) is finite-dimensional for each K ∈ K ( H 1 , H 2 ).1. Define Hilbert space as direct sum of two complemented subspaces2. For compact operator use Hilbert Schmidt decompositionMain idea is to prove that intersection of Spectrum of such $T and compact K is finite

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Let       H    1   and       H    2   be Hilbert spaces, let   T  ∈  B  (      H    1    ,      H    2    ). Suppose that   Ker  ⁡  T is finite-dimensional and that   Im  ⁡  T is closed in       H    2  . Prove that   Ker  ⁡  (  T  +  K  ) is finite-dimensional for each   K  ∈  K  (      H    1    ,      H    2    ).1. Define Hilbert space as direct sum of two complemented subspaces2. For compact operator use Hilbert Schmidt decompositionMain idea is to prove that intersection of Spectrum of such $T and compact K is finite

Stevinivm · Accepted Answer

By assumptions the operator   T  :     ker  ⁡      T    ⊥    ⟶      I    m    T is invertible. Therefore for any   w  ⊥  ker  ⁡  T we have  ‖  T  w  ‖  ≥  c  ‖  w  ‖for a constant   c  &amp;gt;  0.Assume by contradiction, that   ker  ⁡  (  T  +  K  ) is infinite dimensional. Then there exists an infinite orthonormal system       v    n   in   ker  ⁡  (  T  +  K  )  . As   K is compact, then   K      v    n    →  0. Thus   T      v    n    →  0. We have      v    n    =      w    n    +      u    n    ,        w    n    ∈  ker  ⁡  T  ,         u    n    ⊥  ker  ⁡  Tand  ‖  T      v    n    ‖  =  ‖  T      u    n    ‖  ≥  c  ‖      u    n    ‖hence       u    n    →  0. The sequence       w    n   has an accumulation point, therefore the sequence       v    n   has an accumulation point as well. This leads to a contradiction.

Let H 1 </msub> and H 2 </msub> be Hilbert spaces, let T

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