# Is the spectral decomposition of self-adjoint compact operators unique?

Given a compact, self-adjoint operator $T$ on a Hilbert space $H$, then there is the Spectral Theorem which says that $T=\sum_i \lambda_i P_i$ where the sum is over the number of eigenvalues of $T$, which are countable and $P_i$ are finite-rank operators.

Now, an almost immediate corollary is that there exist $\{\mu_n\} \subset \mathbb{R}, \{e_n\}$ orthonormal base for $N$(ker$T$) (the orthogonal space of the ker) such that for every $h\in H$, $Th= \sum_n \mu_n (h,e_n)e_n$.

Now, my question is: these two representations of $T$ are unique under change of orthonormal basis? The first one actuallu doesn't require the choice of a basis so I think that it is unique, but the second one?

Also is there a relationship between the second representation of $T$ and its eigenvalues? I.e. can one recover the eigenvalues from that given representation?

• What does "unique under change of orthonormal basis" mean? That string of words doesn't actually make sense... Sep 12, 2018 at 21:48
• Also, you will certainly have no uniqueness whatsoever for the first representation unless you make additional assumptions about the operators $P_i$ (indeed, writing $Q_i=\lambda_i P_i$, all your statement says is that we can write $T$ as a sum of some finite-rank operators $Q_i$). The usual statement is that the $P_i$ are pairwise orthogonal finite-rank projections, not just finite-rank operators. Sep 12, 2018 at 21:54

I don't know what "unique under change of orthonormal basis" means but here is what we can say about uniqueness. If $$T$$ is a compact self-adjoint operator, then there exists a unique set of pairs $$\{(\lambda_i,P_i)\}_{i}$$ such that each $$\lambda_i$$ is a nonzero scalar, the $$\lambda_i$$ are all distinct, each $$P_i$$ is a nonzero projection operator, the $$P_i$$ are pairwise orthogonal, and $$T=\sum_i \lambda_iP_i.$$ Moreover, the $$\lambda_i$$ are exactly the nonzero eigenvalues of $$T$$, $$P_i$$ is the projection onto the eigenspace of $$T$$ with eigenvalue $$\lambda_i$$, and each $$P_i$$ has finite rank.
For the second representation $$Th=\sum_n \mu_n (h,e_n)e_n,$$ the only uniqueness statement we can say is that coefficients $$\mu_n$$ are uniquely determined (up to permutation) by $$T$$. Indeed, these coefficients $$\mu_n$$ are none other than the eigenvalues of $$T$$, listed with multiplicity. This formula should not look mysterious at all: if we were in a finite-dimensional space and $$\{e_n\}$$ was the standard (finite) basis, this would literally just be saying that $$T$$ is the diagonal matrix with the $$\mu_n$$ as the diagonal entries. So this formula is saying that there exists an orthonormal basis in which $$T$$ is a diagonal matrix. As usual for diagonal matrices, the diagonal entries are just the eigenvalues (with multiplicity). The orthonormal basis $$\{e_n\}$$ with respect to which $$T$$ is diagonal is not unique, just as a diagonalizable matrix does not have a unique basis of eigenvectors.
To relate the two representations, note that if $$e_{i_1},\dots,e_{i_k}$$ is an orthonormal basis for the eigenspace of $$T$$ with eigenvalue $$\lambda_i$$, then the projection $$P_i$$ onto that eigenspace is just $$P_ih=\sum_{j=1}^k (h,e_{i_j})e_{i_j}.$$ The second representation is obtained from the first representation by just picking an orthonormal basis for each eigenspace in this way and then using this formula for $$P_i$$. The non-uniqueness of the second representation comes from the fact that there are many different orthonormal bases for each eigenspace.