I am looking at Arveson's book, an invitation to $C^*$ algebras. There, it is explained p. 21 ($C^*$ algebras of compact operators) that any representation of a compact $C^*$ algebra can be decomposed into direct sum of multiples of irreducible representations. Then this is applied to the $C^*$ algebra generated by a compact operator. My problem is that I don't understand what this means already for a finite dimensional Hilbert space. That is, to what topic in linear algebra is this (the decomposition of the $C^*$ algebra generated by a compact operator) related to? To compare for instance, I understand the relation between non cyclicity and multiple eigenvalues for self adjoint operators.

Thank you for any help


1 Answer 1


In my experience, the topic of subalgebras of $M_n(\mathbb C)$ is not part of the usual linear algebra curriculum.

What you need to understand first is the form that finite-dimensional C$^*$-algebras have. A finite-dimensional C$^*$-algebra $A$ is always a finite direct sum $$\bigoplus_{k=1}^m M_{n(k)}(\mathbb C).$$ The "blocks" can be identified via the minimal projections of the centre of $A$.

And because C$^*$-algebras of compact operators always have minimal projections, the same game can be played.

  • 2
    $\begingroup$ I'd like to add, if i may, that there's a section on this in Goodearl's Real and Complex C*-algebras. $\endgroup$ Jan 15, 2016 at 0:06
  • $\begingroup$ First, sorry for late answer and thanks for your answer; I needed to think well about what you said and my question. If I understand you point : to understand the case of compact operators, one needs to have in mind rather this example (akin to Wedderburn, that I knew) rather than the * algebra generated by a single operator, may it be in finite dimensional Hilbert space. But nevertheless, is there nothing to be said when A is the algebra generated by an operator T in finite dimension? Eg what are the minimal projections, how is the spectrum of the algebra related to the eigenvalues of T etc $\endgroup$
    – John
    Jan 21, 2016 at 18:13
  • 1
    $\begingroup$ There is absolutely no relation whatsoever between the eigenvalues of $T$ and the structure of $C^*(T)$. For instance, let $t,s\in\mathbb C$ be any two distinct nonzero complex numbers. Let $$ T_1=\begin{bmatrix}s&0\\0& t\end{bmatrix},\ \ \ T_2=\begin{bmatrix}s&1\\ 0& t\end{bmatrix}. $$ Then $$C^*(T_1)=\mathbb C\oplus\mathbb C,\ \ \ C^*(T_2)=M_2(\mathbb C). $$ $\endgroup$ Jan 22, 2016 at 0:19
  • $\begingroup$ Ah that's right, thank you. Well that surely kills any hope to find any relation ! $\endgroup$
    – John
    Jan 22, 2016 at 15:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .