# Bounded Self-adjoint operator with finite spectrum

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.

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 $$P_k = p_k(S)$$.

• (+1) While our answers are more or less the same, yours is a little bit more explicit. – QuantumSpace Jan 4 at 9:36

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.

• 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! – a.p Jan 3 at 21:09
• 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 – QuantumSpace Jan 3 at 21:14
• 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. – QuantumSpace Jan 3 at 21:18