5
$\begingroup$

Can someone tell me if and why $\text{Im}\,f+\ker f=R$ holds for a selfadjoint operator $f:R\rightarrow R$, where $R$ is a finite dimensional inner product space?

Can someone give me an example of an operator in an infinite dimensional space for which that equation is not true ?

  • 1
    It is holds for any finite dimensional space and any linear operator.2012-08-19
  • 1
    Ok, but could you give me an idea why it holds ?2012-08-19
  • 1
    Represent your linear operator as a matrix. Then you immediately get $\mathrm{dim}\;\mathrm{Ker}f+\mathrm{dim}\;\mathrm{Im}f=\mathrm{dim} R$2012-08-19
  • 3
    @Norbert, the dimensions work, but is that enough for the spaces to be the same?2012-08-19
  • 7
    Here is the idea: The dimensions fit and the kernel is a subspace of the orthogonal of the image (if $Sz=0$, then, by self-adjointness, $\langle Sx,z\rangle=\langle x,Sz\rangle=0$ for every $x$), hence the kernel **is** the orthogonal of the image, QED.2012-08-19
  • 0
    But why do the dimensions fit ? I can't see how I would obtain that, if I represent $f$ by a matrix (which I know is hermitian, if the basis is orthogonal).2012-08-19
  • 0
    @Norbert: This is wrong, only the dimension holds.2012-08-19
  • 0
    My statement is wrong if we consider inner direct sum of this spaces, otherwise this correct.2012-08-19
  • 2
    @Norbert: so I guess what you wanted to say is that your statement is correct for *selfadjoint* operators, not for *any* linear operator, didn't you?2012-08-19
  • 0
    @AgustíRoig Nevermind2012-08-19

1 Answers 1