Let $C$ be a self-adjoint, compact operator from a Hilbert space $H$ onto itself. Let the spectral decomposition of $C$ be $C = S^*S$, where $S^*$ is the adjoint of $S$. Assuming $C$ be invertible, is the following true?

$$S(C)^{-1}S^* \text{ is the identity operator.}$$

In vector spaces, I believe it holds but I am not sure about the case with general operators.

  • $\begingroup$ What do you mean by the case of general operator; this operator is from a Hilbert space into itself so it is a vector space ! $\endgroup$ – The_lost Jan 30 '18 at 6:41
  • $\begingroup$ I meant finite-dimensional vector spaces. Sorry for the confusion. $\endgroup$ – Enayat Jan 30 '18 at 22:11

One way of approaching this exercise is to note that the fact that there is an operator on $H$ that is both compact and invertible tells you quite a lot about $H$.

Recall that if $K$ is a compact operator on a Hilbert space and $B$ is a bounded operator on the same Hilbert space then $BK$ is compact.

It follows that if $C$ is a compact operator on a Hilbert space, and $C$ is invertible, then the identity operator $I$ on that same Hilbert space is compact. (Proof: $C^{-1}$ is a bounded operator, so take $B = C^{-1}$ and $K = C$ in the previous recollection to deduce that $C^{-1} C = I$ is compact.)

It is also true, however, that if the identity operator on a Hilbert space is compact, then the Hilbert space must be finite dimensional. See for example the information in this answer: Invertibility of compact operators in infinite-dimensional Banach spaces.

Once you know that the Hilbert space $H$ on which $C$ acts is finite dimensional, the fact that $C = S^* S$ is invertible implies that $S$ is invertible (because in a finite dimensional Hilbert space, a product of linear operators can only be invertible if each factor is itself invertible). But this means $C^{-1}$ can be expressed as the product of invertible operators: $C = (S^* S)^{-1} = S^{-1} (S^*)^{-1}$.

But then $$ SC^{-1} S^{*} = S (S^{*} S)^{-1} S^{*} = S (S^{-1} (S^{*})^{-1}) S^* = (SS^{-1})((S^*)^{-1} S^*) = I \cdot I = I,$$ as desired.

  • $\begingroup$ Thanks for the answer. I have a question. Suppose $H=R^m$, but $S:\ L^2(X,\rho) \rightarrow R^m$ such that $C = S^*S$. Here $X$ is some finite dimensional vector space and $\rho$ a measure on it. Does it still hold? $\endgroup$ – Enayat Jan 30 '18 at 22:17

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