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Welcome everybody :)

I'm working together with a little group preparing for the upcoming exams in "Mathematical Methods of Physics." There's one tricky task involving some linear algebra, namely the generalization of the Spectral Decomposition of Self-Adjoint Operators.

The task is the following:

Let $A_1,A_2,…,A_s$ be a set of pairwise commuting self-adjoint operators on the finite dimensional complex inner product space $V$. Show the Spectral Theorem for Commuting Self-Adjoint Operators:

There are orthogonal projections $P_1,P_2,…,P_r$ such that:

  • $P_jP_k = δ_{jk}P_k$
  • $\mathbb I_V = P_1 + P_2 + \cdots + P_r$
  • Each $A_i$ can be written as $\sum_{j=1}^r \lambda_{ij}P_j$ for some scalars $\lambda_{ij}$ (which will necessarily be eigenvalues of $A_i$)
  • If the $A_i$ are written as self-adjoint matrices acting on $\mathbb{C}^{\mathrm{dim} V}$, then there is a unitary matrix $U$ such that all $U^{-1}A_iU$ are diagonal

For the task one has to prove first that the statement above holds for just one operator, lets say $A_1$. This is the induction basis (s = 1). Since we have already proven this theorem we can safely use it here without proving it again. So I'm really interested in the induction step (s → s+1).

Is the following argumentation correct?

Let $P_j$ be the orthogonal projections onto the simultaneous eigenspaces $E_j ≔ \text {Im } P_j$ for $A_1,…,A_s.$
Thus $A_r = \sum_{j=1}^n \lambda_{rj}P_j$ for $1\leq r \leq s$. Now for $v \in E_j$, $A_r A_{s+1}(v) = A_{s+1}A_r (v) = \lambda_{rj}A_{s+1}(v)$ and thus $A_{s+1}$ restricts to a self adjoint operator $A_{s+1}|_{E_j} : E_j \to E_j$.

Let $A_{s+1}|_{E_j} = \sum_{k=1}^{m_j}μ_{jk}Q_{jk}$ be its spectral decomposition.

Set $P_{jk} ≔ Q_{jk} \circ P_j$ and consider it as a map $V \to V$.

Then $\sum_{k=1}^{m_j}Q_{jk} = \mathbb I ·E_j \ ⇒ \ P_j = \sum_{k=1}^{m_j}P_{jk}$.

From $P_j P_{j'} = δ_{jj'}P_j ∧ Q_{jk}Q_{jk'} = δ_{kk'}Q_{jk}$ one deduces $P_{jk}P_{j'k'} = δ_{jj'}δ_{kk'}P_{jk}$.

Therefore $\sum_{j=1}^{n}P_j = \mathbb I_V \ ⇒ \ \sum_{1≤j≤n,1≤k≤m_j}P_{jk} = \mathbb I_V$.
Further, $A_r = \sum_{1≤j≤n,1≤k≤m_j}\lambda_{r,jk}P_{jk}$ for $1≤r≤s+1$ and some $\lambda_{r,jk}$.

For $v,w \in V$, $\sum_{j=1}^n P_j = \mathbb I_V$ implies that:

$$\langle P_{jk}v,w\rangle = \langle Q_{jk}P_j v, \sum_{j=1}^n P_j w\rangle = \langle Q_{jk}P_j v, P_j w\rangle = \langle P_j v,Q_{jk}P_j w\rangle = \langle P_j v, P_{jk} w\rangle = \langle v,P_{jk} w\rangle$$

i.e. the $P_{jk}$ are self-adjoint.

I would really appreciate if some smart guys would help me a bit. Thank you in advance!



EDIT: For the induction basis one can safely uses the Spectral Decomposition of Self-Adjoint Operators:

Let $(V,\langle·,·\rangle)$ be a complex inner product space of dimension $n∈ℤ_{>0}$ and $A:V→V$ be a self adjoint operator. Let $\lambda_j$ and $E_j, j=1,…,r$ be the distinct eigenvalues and the corresponding eigenspaces of $A$. Furthermore set $P_j = \prod_{k≠j}\frac{A-\lambda_k \mathbb I_n}{\lambda_j - \lambda_k}, j=1,…,r$.

  • $\lambda_j∈ℝ$ and the $E_j$ are pairwise orthogonal.
  • $P_j$ are orthogonal projections commuting with $A$ that satisfy $P_jP_k = δ_jk P_k$, $\text{Im}P_j = E_j$, $\text{Tr} P_j = \text{dim}E_j =m_j$, $\text{ker}P_j = \oplus_{k≠j}E_k$.
  • $\mathbb I_n = \sum_{j=1}^r P_j$, $V=E_1 \oplus \cdots \oplus E_r$
  • $A = \sum_{j=1}^r \lambda_jP_j$ (spectral decomposition of A)
  • Let $m_j^- = \sum_{i=1}^{j-1}m_i$ and $v_{m_j^-+1},\cdots,v_{m_j^-+m_j}$ be an orthonormal basis for $E_j$. Then $v_1,\cdots,v_n$ is an orthonromal basis for $V$. The matrix of $A$ in this basis is $A = \oplus_{j=1}^r \lambda_j \mathbb I_{m_j}$.
  • Fix a basis of $V$. Let $A$ and $P_j$ denote the matrices of $A$ and $P_j$ in this basis. Let $U$ be the matrix whose columns are the coordinates of the $v_j$ in this basis. Then the matrices $U^{-1}AU = \oplus_{j=1}^r \lambda_j \mathbb I_{m_j}$ and $U^{-1}P_jU = 0_{m_j^-} \oplus \mathbb I_{m_j} \oplus 0_{m_j^+}$ are diagonal. Here $m_j^+ = \sum_{i=j+1}^r m_i$ and $0_m$ is the $m×m$ zero matrix.

share|improve this question
A [related problem](math.stackexchange.com/questions/33883/spectral-decomposition-of-a-norm‌​al-matrix/252181#252181) for normal matrices. –  Mhenni Benghorbal Jul 13 '13 at 22:05
Your inference that $A_iA_{s+1}(v) = \lambda_{ij}A_{s+1}(v)$ for $v \in E_j$ implies that $A_{s+1}$ restricted to $E_j$ is a well-defined operator $E_j \to E_j$ is incorrect. –  Amit Kumar Gupta Jul 13 '13 at 23:12
You are right! It's corrected now with $A_r A_{s+1}(v)$ –  Red Pencil Jul 14 '13 at 11:24
Welcome? Really? I'm sorry if I sound like a douche, but I hate it when people say "welcome" when they're actually new. And it's not exactly a language barrier thing: I've heard people do the exact same thing in my mother tongue (Polish) and it doesn't put me off any less. –  tomasz Jul 14 '13 at 11:45
I wanted to use Hi,folks but strangely this first phrase never appeared. Better not to say Hello at all I switched to the one written above. Even then I think It's like a friendly start into a new question posted from a user, if he's new or not shouldn't make any difference. Anyway I have now posted also the induction base to get better into the notation $m_j$, and the role of $P_j$ in comparison to the later used $P_{jk}$ containing it's own little projections with the $Q_{jk} \circ P_j$. I hope this helps a bit :) –  Red Pencil Jul 14 '13 at 12:11

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