# Non diagonalizable normal, linear and bounded operator

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.

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).

• Really good answer. Thanks a lot! Mar 26, 2020 at 19:09
• 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. Mar 27, 2020 at 16:45
• No problem, see the edit for an answer to your question. Let me now if you have any further questions. Mar 27, 2020 at 18:05
• Why do you write $T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = 1$ and not $T \circ (A-\lambda \text{Id}) = (A-\lambda \text{Id}) \circ T = Id$? Mar 27, 2020 at 19:25
• Oh yeah, you are right! There are to many units to manage my notation. I edited it. Mar 27, 2020 at 19:37