3
$\begingroup$

If $H$ is a complex Hilbert space and $T:H\to H$ is a bounded linear operator, we say that $T$ is diagonalizable if there exists an orthonormal basis of $H$ formed by eigenvectors of $T$ ($0\neq{x}\in H$ is a eigenvector if there exists $\lambda \in \mathbb{C}$ with $T(x)=\lambda x$).

When $H$ is finite dimensional, it is known that if $T$ is normal then it is diagonalizable. However, by the spectral theorem for normal compact operator, in infinite dimensional Hilbert spaces we need to make an extra hypothesis: $T$ is compact. So, if I have understood correctly, there must exist bounded linear operator which are normal but not diagonalizable. Could someone give me an example of this?, because I have not found such operator.

$\endgroup$
3
$\begingroup$

There are many counterexamples. Before I give one I want to give some context. Recall that for any normal matrix $A$ there exists a basis of eigenvectors. Let $\sigma(A)=\{\lambda_1,\ldots,\lambda_n\}$ be the eigenvalues with eigenspaces $\{V_{\lambda_1}, \ldots, V_{\lambda_n}\}$. Let $P_k$ be the orthogonal projection onto $V_{\lambda_k}$. Rewriting the spectral theorem for finite dimensional spaces a bit shows that $$ A = \sum_{k=1}^n \lambda_k P_k. $$ Note that sums can be written as integrals over atomic measures and we can loosely write this as $$ A = \int_{\sigma(A)} \lambda dP(\lambda). $$ More precisely, we do not integrate w.r.t. a measure in the usual sense. But integrate w.r.t a "projection-valued measure". I won't go to deep however.

Why did I write the spectral theorem for finite dimensional operators in such a convoluted way? Well there is a general spectral theorem that says that for any normal bounded operator $A$ on a Hilbert space $\mathcal{H}$ there exists a so called spectral measure $P$ such that we can write $$ A = \int_{\sigma(A)} \lambda(A) dP(\lambda). $$ However, while previously, the set $\sigma(A)$ consisted of the eigenvalues of $A$ (i.e. the values $\lambda \in \mathbb{C}$ such that $A - \lambda 1$ is not injective), we now need to take the following definition: $$ \sigma(A) = \{ \lambda \in \mathbb{C} \mid A-\lambda 1 \text{ is not invertible}. \} $$ In the finite dimensional case (and even when $A$ is compact) this boils down to $\sigma(A)$ being the eigenvalues of $A$. However in general, $A-\lambda 1$ might not be invertible while there exists no non-zero vector $v \in \mathcal{H}$ such that $A v =\lambda v$. While the generalized spectrum $\sigma(A)$ may be defined for any bounded operator (and having some really nice properties popping-up from complex analysis), it can be split up (for normal operators) into a discrete part (the real eigenvalues) and a continuous part. The continuous part makes general normal operators so different than the finite dimensional ones. This also makes the domain of the integral above over a non-discrete set.

Now to give a counterexample, take the operator $$A: L^2([0,1]) \rightarrow L^2([0,1]), A(f)(x) = x f(x) \quad \text{(multiplication by }x). $$ $A$ has no eigenvalues (for every $z \in \mathbb{C}$, there is no non-zero function $f$ with $A(f) = z f$, this would mean that $x = z$ for all $x$ where $f(x)$ is non-zero, so $f$ is zero a.e.). In particular, there is no basis of eigenvectors. However the spectrum $\sigma(A)$ can be shown to be equal to $[0,1]$ (the image of the function $x$ on $[0,1])$.

Edit: To show that $\sigma(A) = [0,1]$ for the example above, assume that $\lambda \notin \sigma(A) $. This means that there exists some $T \in \mathcal{B}(L^2([0,1])$ such that $$T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = \text{Id}. $$ Put $g = T(1) \in L^2([0,1])$. It follows that$(x-\lambda 1) g(x) = 1$ for all $x \in [0,1]$ a.e. or written differently that $$ g(x) = \frac{1}{x-\lambda} \quad \forall x \in [0,1] \text{ a.e.} $$ However $g$ is not square integrable if $\lambda \in [0,1]$. This shows that $[0,1] \subset \sigma(A)$. For other inclusion, it is easy to see that whenever $\lambda \notin [0,1]$, then multiplication by the function $g(x) = \frac{1}{x-\lambda}$ is an inverse to $A-\lambda 1$. Because $g$ is uniformly bounded, the multiplication operator by $g$ is a bounded operator.

As a side note, $A$ from above is not a compact operator, as you can tell from the fact that is has a continuous spectrum (compact operators are diagonisable in the classical sense (having a countable orthonormal basis of eigenvectors), hence they have a discrete spectrum).

$\endgroup$
  • $\begingroup$ Really good answer. Thanks a lot! $\endgroup$ – Loye94 Mar 26 '20 at 19:09
  • $\begingroup$ Sorry, but I am reading the answer again and I would like to know a prooof of that $\sigma(A)=[0,1]$. I don't see how to prove it. $\endgroup$ – Loye94 Mar 27 '20 at 16:45
  • $\begingroup$ No problem, see the edit for an answer to your question. Let me now if you have any further questions. $\endgroup$ – abcdef Mar 27 '20 at 18:05
  • $\begingroup$ Why do you write $T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = 1 $ and not $T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = Id $? $\endgroup$ – Loye94 Mar 27 '20 at 19:25
  • $\begingroup$ Oh yeah, you are right! There are to many units to manage my notation. I edited it. $\endgroup$ – abcdef Mar 27 '20 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.