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if $S$ be a self-adjoint bounded operator and $\sigma(S)=\left\{\lambda_{1}, \dots, \lambda_{n}\right\} .$ show that there exist projections operator $P_{1}, \dots, P_{n} \in B(H)$ that $\sum_{j=1}^{n} P_{j}=I$ and $P_i P_j=0$ if $j \neq k$ and $S=\sum_{j=1}^{n} \lambda_{j} P_{j}$. I can see how we can define projection operator for each term in spectrum but I can't show that they are complete in the sense $\sum_{j=1}^{n} P_{j}=I$. actually its strange for me that in general self-adjoint bounded operator with finite spectrum have a complete set of projection operator.

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For $S$ as described, one has $p(S)=0$ where $$ p(\lambda)=\prod_{k=1}^{n}(\lambda-\lambda_k). $$ This follows from
$\;\;$ (a) $p(S)^*=p(S)$, and
$\;\;$ (b) $\sigma(p(S))=p(\sigma(S))=\{0\}$,
which forces the self-adjoint operator $p(S)$ to be the $0$ operator.

The Lagrange polynomials $$ p_k(\lambda)=\frac{\prod_{j\ne k}(\lambda-\lambda_j)}{\prod_{j\ne k}(\lambda_k-\lambda_j)}. $$ satisfy $p_{k}(\lambda_j)=\delta_{k,j}$, which gives $$ p_1(\lambda)+p_2(\lambda)+\cdots+p_n(\lambda)=1. $$ This forces $p_1(S)+p_2(S)+\cdots+p_n(S)=I$. And $$ p_k(S)p_k(S)=p_k(S) \\ p_k(S)p_l(S)=(p_kp_l)(S)=0,\;\; k\ne l. $$ This is because $p_k(\lambda)^2=p_k(\lambda)$ for $\lambda\in\sigma(S)$, and because $p_k(\lambda)p_l(\lambda)=0$ for $\lambda\in\sigma(S)$ whenever $k\ne l$. These have the desired properties.

Finally, \begin{align} S &= S(p_1(S)+p_2(S)+\cdots+p_n(S)) \\ &= \lambda_1 p_1(S)+\lambda_2 p_2(S)+\cdots+\lambda_n p_n(S) \end{align} The projections you want are $P_k = p_k(S)$.

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  • $\begingroup$ (+1) While our answers are more or less the same, yours is a little bit more explicit. $\endgroup$ – QuantumSpace Jan 4 at 9:36
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When you want to prove something about self-adjoint operators on a Hilbert space, your first instinct should be to think about the continuous functional calculus.

Recall that the functional calculus is an isometric $*$-morphism $$\phi: C(\sigma(S)) \to B(H): f \mapsto f(S)$$ that sends the inclusion $z: \sigma(S) \hookrightarrow \mathbb{C}$ to $S$ and $1$ to $\text{id}_H$.

Since the spectrum is finite, it is discrete and the characteristic functions $f_i = \chi_{\{\lambda_i\}}$ are continuous on the spectrum. Let $P_i:= f_i(S) \in B(H)$. Now, note that on $\sigma(S)$, the following identities hold: $$z = \sum_i \lambda_i f_i$$ $$f_i f_j = 0, \quad i \neq j$$ $$f_i^2 = f_i = \overline{f_i}$$

Hence, since the functional calculus is a unital $*$-morphisms that maps $z$ to $S$, we find that these identities are also satisfied for the images, and this is what you want.


Remark: The same is true (with the same proof) when we replace $B(H)$ with a unital $C^*$-algebra.

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  • $\begingroup$ Thanks, I take the general idea of the proof but I have some problems. I dont exactly understand what you mean by "functional calculus is an isometric *-morphism" and what you write below it. Is there any simpler argument or where I can read more about this or you discuss more in the answer about it! $\endgroup$ – a.p Jan 3 at 21:09
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    $\begingroup$ The spectrum $\sigma(S)$ is a compact topological space, and you can consider continuous functions on it. What I use is that there is a map $C(\sigma(S)) \to B(H)$ which has nice properties: (1) It maps the inclusion $\sigma(S) \hookrightarrow \mathbb{C}$ to $S$, (2) It maps $1$ to $\text{id}_H$, (3) It is a morphism of algebras (it is a linear map that preserves multiplication) and (4) It preserves adjoints: thus $\overline{f}$ is mapped to the adjoint of the image of $f$. See also here: en.wikipedia.org/wiki/Continuous_functional_calculus $\endgroup$ – QuantumSpace Jan 3 at 21:14
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    $\begingroup$ I cannoto quite think of a simpler argument. The argument I used is pretty routine once you get used to these things, but if you have any additional questions I'll gladly help. $\endgroup$ – QuantumSpace Jan 3 at 21:18

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