Proof. We show the first equality, since the other ones follow analogously.

Since $M$ is reducing for $T$, $Q_MTP_M = 0$ and therefore,

\[ Q_MDP_M = Q_M(T-u\otimes v)P_M = Q_M (u\otimes v)P_M. \]

In addition, for every $x \in H$

\[ (Q_M u \otimes v)P_M x = \langle{P_M x,v}\rangle Q_M u = \langle{x,P_M v}\rangle Q_M u = (Q_M u \otimes P_Mv)x, \]

which finally leads to the desired expression.

Proof. First, observe that $Q_M= I - P_M$ and $P_M$ both commute with $T$ and $T^{*}$. Now, since $u$ is a separating vector for ${\{{T}\}'}$ and $v$ for ${\{{T^{*}}\}'}$ (see [Reference Foias, Jung, Ko and Pearcy19, theorem 1.4]) it follows that the vectors $P_M u,\, P_M v,\, Q_M u,\, Q_M v$ are non-zero. In particular, one deduces that $u,\,v \notin M \cup M^{\perp }$.

On the other hand, since $P_M T = T P_M$ then

\[ D P_M - P_M D = u \otimes P_M v - P_M u \otimes v. \]

Let $n \in \mathbb {N}$ and assume, for the moment, that $e_n \in M$. Then,

(2.1)\begin{equation} 0 = D P_M e_n - P_M D e_n = \langle{e_n,v}\rangle u - \langle{e_n,v}\rangle P_M u. \end{equation}

Note that $\langle {e_n,\,v}\rangle \neq 0$ since $\alpha _n\beta _n \neq 0$, so $u = P_Mu$ and therefore $u \in M$, which is a contradiction. Hence $e_n\not \in M$.

The case $e_n \in M^{\perp }$ is analogous.

Proof. Let us argue by contradiction assuming $P_M v = 0$. Then, by lemma 2.2, we have

\[ Q_M DP_M ={-}Q_M u\otimes P_Mv = 0, \]

where $Q_M=I-P_M$. Then $0 = (I-P_M)DP_M = DP_M-P_M D P_M$ or, equivalently, $DP_M = P_M DP_M$. Hence the closed subspace $M := P_M(H)$ is a non-trivial closed invariant subspace for $D$ (see [Reference Radjavi and Rosenthal32, theorem 0.1], for instance).

Moreover, $D$ is a completely normal operator (see [Reference Wermer39, theorem 3] or [Reference Radjavi and Rosenthal32, chapter 1]). Therefore, every invariant subspace of $D$ is reducing and it is spanned by a subset of eigenvectors. Accordingly, $DP_M = P_M D$ and $M = \overline {\textrm {span}\; \{e_n : n \in \Lambda \}},$ where $\Lambda$ is a proper subset of the natural numbers $\mathbb {N}$. Observe that, in particular, $DP_M = P_M D$ implies that $P_M u = 0$.

On the other hand, since $P_M$ is the orthogonal projection onto $M=\overline {\textrm {span}\; \{e_n : n \in \Lambda \}}$, trivially $P_M$ is a diagonal operator with respect to $(e_n)_{n\geq 1}$ (since $P_M e_n = e_n$ for every $n \in \Lambda$ and $P_M e_n = 0$ for every $n \notin \Lambda$). Having into account that $P_M u = P_M v = 0$, we deduce

\[ 0 =\sum_{n \in \Lambda } \alpha_n e_n = \sum_{n \in \Lambda} \beta_ne_n. \]

Hence, for every $n \in \Lambda$ we have $\alpha _n = \beta _n = 0$, which is a contradiction unless $\Lambda = \emptyset$. But, in this latter case, it would follow that $P_M = 0$ which is also absurd since $P_M$ is a non-trivial projection. Therefore, $P_M v \neq 0$ as the statement asserts.

The proof of the statement $u\notin \ker P_M \cup \ker (I-P_M)$ is analogous, just considering $T^{\ast }$.

Finally, in order to show that $e_n \notin \ker P_M \cup \ker (I-P_M)$ for all $n \in \mathbb {N}$, we may argue as in lemma 2.3 considering (2.1) if $e_n$ would be in $M$ (note that, by assumption, $\alpha _n$ and $\beta _n$ are not simultaneously zero).

Proof of theorem 2.1. Proof of theorem 2.1

Assume that $T = D + u\otimes v \in \mathcal {L}(H)$ where $D$ is a diagonal operator with respect to an orthonormal basis $(e_n)_{n\geq 1}$ and $u =\sum _n \alpha _n e_n$, $v=\sum _n \beta _n e_n$ two nonzero vectors in $H$.

Clearly, if $T$ is a normal operator, it has non-trivial reducing subspaces. In addition, if there exists $n_0 \in \mathbb {N}$ such that $\alpha _{n_0} = \beta _{n_0} = 0$, then the subspace generated by the basis vector $e_{n_0}$ is reducing for $T$ as pointed out in the preliminary section. In both cases, $T$ has a non-trivial reducing subspace. For the converse, assume that $T$ has a non-trivial reducing subspace and let us show that $T$ is normal or there exists $n_0 \in \mathbb {N}$ such that $\alpha _{n_0} = \beta _{n_0} = 0$.

Case 1: $D$ is a self-adjoint operator.

Assume that $T=D + u\otimes v$ has a non-trivial reducing subspace $M\subset H$ and that $\alpha _n$ and $\beta _n$ are not simultaneously zero. Let $P_M$ be the non-trivial orthogonal projection onto $M$ and $Q_M = I - P_M$. By lemma 2.2, both relations

\[ Q_M u \otimes P_M v ={-}Q_M DP_M = Q_M v \otimes P_Mu \]

and

\[ P_M u \otimes Q_M v ={-}P_M D Q_M = P_M v \otimes Q_M u \]

hold. Moreover, lemma 2.4 ensures that the vectors $P_M u,\, Q_M u,\, P_M v$ and $Q_M v$ are non-zero.

A little computation shows that

\[ \langle{P_M u,P_M v}\rangle Q_M u = \|P_M u\|^{2}Q_M v, \]

and since $\left \lvert \left \lvert {P_M u}\right \rvert \right \rvert > 0$, we deduce

\[ Q_M v = \frac{ \langle{P_M u,P_M v}\rangle }{ \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} }Q_M u. \]

In a similar way, we have

\[ P_M v = \frac{\langle{Q_M u,Q_M v}\rangle}{\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2}}P_M u = \frac{\langle{P_M v,P_M u}\rangle}{\left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}}P_M u. \]

Now, let us write $\alpha = {\langle {P_M v,\,P_M u}\rangle }/{\left \lvert \left \lvert {P_M u}\right \rvert \right \rvert ^{2}}$, so $P_M v = \alpha P_M u$ and $Q_M v = \overline {\alpha } Q_M u$. Hence,

\[ v = P_M v + Q_M v = \alpha\, P_M u + \overline{\alpha}\, Q_M u. \]

Notice that it is enough to show that $\alpha \, \in \mathbb {R}$, so $v = \alpha \, u$ and therefore $T$ would be a self-adjoint operator, since it would be the sum of two self-adjoint operators.

Observe that

\[ \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2} = \left\lvert\left\lvert{P_M v}\right\rvert\right\rvert^{2} + \left\lvert\left\lvert{Q_M v}\right\rvert\right\rvert^{2} = |\alpha|^{2} \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} + |\alpha|^{2} \left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} = |\alpha|^{2} \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}. \]

Moreover,

\begin{align*} \langle{u,v}\rangle & = \langle{P_M u,P_M v}\rangle + \langle{Q_M u,Q_M v}\rangle \\ & = \langle{P_M u,\alpha P_M u}\rangle + \langle{Q_M u, \overline{\alpha} Q_M u}\rangle \\ & =\overline{\alpha} \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} + \alpha \left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} \\ & =(\textrm{Re}\; (\alpha) - i \textrm{Im}\; (\alpha))\left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} + (\textrm{Re}\; (\alpha) + i\textrm{Im}\, (\alpha))\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} \\ & =\textrm{Re}\; (\alpha)(\left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} + \left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2}) + i\textrm{Im}\, (\alpha)(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}) \\ & = \textrm{Re}\; (\alpha)\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} + i\textrm{Im}\, (\alpha)(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}). \end{align*}

Note that $\alpha u - v \in M^{\perp },$ since $P_M(\alpha u - v) = 0.$ Thus, $Q_M (\alpha u - v) = \alpha u -v$.

Similarly, $\overline {\alpha }u-v \in M$. Furthermore, $(\alpha P_M + \overline {\alpha } Q_M)u = v$. Since $P_M$ and $Q_M$ commute with $T$, it follows that

\[ (\alpha P_M + \overline{\alpha} Q_M)T = T(\alpha P_M + \overline{\alpha} Q_M). \]

Now,

\[ T(\alpha P_M + \overline{\alpha} Q_M)u = Tv = Dv + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}u = Dv + |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}u \]

and

\begin{align*} (\alpha P_M + \overline{\alpha} Q_M)Tu & = (\alpha P_M + \overline{\alpha} Q_M)(Du + \langle{u,v}\rangle u) \\ & = \alpha P_M Du + \overline{\alpha}Q_M Du + \langle{u,v}\rangle (\alpha P_Mu + \overline{\alpha} Q_Mu) \\ & = \alpha P_M Du + \overline{\alpha}Q_M Du + \langle{u,v}\rangle v \end{align*}

Thus,

\[ Dv + |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}u = \alpha P_M Du + \overline{\alpha}Q_M Du + \langle{u,v}\rangle v. \]

So,

\[ P_M Dv + Q_M Dv + |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}u = \alpha P_M Du + \overline{\alpha} Q_M Du + \langle{u,v}\rangle v. \]

Upon applying lemma 2.2 it follows that

\begin{align*} |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}u - \langle{u,v}\rangle v & = P_M D(\alpha u -v) + Q_M D(\overline{\alpha} - v)\\ & = P_M DQ_M (\alpha u -v) + Q_M DP_M ( \overline{\alpha}u -v) \\ & = \alpha( P_M u \otimes Q_M u)(v-\alpha u) + \alpha(P_M u \otimes Q_M u)(v - \overline{\alpha}u) \\ & =\alpha \langle{v-\alpha u,u}\rangle P_M u + \alpha \langle{v - \overline{\alpha}u,u}\rangle Q_M u \\ & = (\alpha\langle{v,u}\rangle - \alpha^{2} \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2})P_M u + ( \alpha\langle{v,u}\rangle - |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2})Q_M u. \end{align*}

Moreover, $v = \alpha P_M u + \overline {\alpha }Q_M u$, so

\[ |\alpha|^{2} \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}u - \langle{u,v}\rangle v = (|\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - \alpha\langle{u,v}\rangle)P_M u + (|\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - \overline{\alpha}\langle{u,v}\rangle)Q_M u. \]

Consequently,

\[ |\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - \alpha\langle{u,v}\rangle = \alpha \langle{v,u}\rangle - \alpha^{2} \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} \]

and

\[ |\alpha^{2}|\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - \overline{\alpha} \langle{u,v}\rangle = \alpha\langle{v,u}\rangle - |\alpha|^{2} \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}. \]

From the second equality we have

(2.2)\begin{align} 2|\alpha|^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} & = \alpha\langle{v,u}\rangle+ \overline{\alpha}\langle{u,v}\rangle \nonumber\\ & = 2\textrm{Re}\, (\alpha\langle{v,u}\rangle). \end{align}

Let us write $\alpha = a+bi$, with $a,\, b \in \mathbb {R}$ real numbers. The aim is to show that $b=0$.

Recall that

\[ \langle{v,u}\rangle = \overline{\langle{u,v}\rangle} = a\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - ib(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}). \]

Then,

\begin{align*} \alpha \langle{v,u}\rangle & =(a+bi)( a\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - ib(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2})\\ & = a^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} - aib(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2})\\ & \quad + aib\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} + b^{2}(\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}). \end{align*}

So, $\textrm {Re}\; (a\langle {v,\,u}\rangle ) = a^{2}\left \lvert \left \lvert {u}\right \rvert \right \rvert ^{2} + b^{2} (\left \lvert \left \lvert {Q_M u}\right \rvert \right \rvert ^{2} - \left \lvert \left \lvert {P_M u}\right \rvert \right \rvert ^{2}).$ From equation (2.2) we have

\[ (a^{2}+b^{2})\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} =a^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} + b^{2} (\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}), \]

and therefore,

\[ b^{2}\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2} = b^{2} (\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2} - \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}). \]

If $b\neq 0$, this identity yields that $\left \lvert \left \lvert {P_M u}\right \rvert \right \rvert = 0$, what contradicts lemma 2.4. Accordingly, $b=0$ and then $\alpha$ is a real number as we wish to prove in order to show that $T$ is a normal operator.

Finally, if $T$ has a reducing subspace and it is not a normal operator, we deduce that there exists $n \in \mathbb {N}$ such that $\alpha _n = \beta _n = 0$, so the proof in case 1 is complete.

Case 2: $D$ is not a self-adjoint operator.

Assume first that the set of eigenvalues $\Lambda (D)=(\lambda _n)_{n\geq 1}$ is contained in a line $\Gamma$ in the complex plane parallel to the real axis. Let $h\in \mathbb {R}$ such that $\Gamma =\{x+ih: x\in \mathbb {R}\}$.

Observe that the linear bounded operator $D - hiI$ is a diagonal, self-adjoint operator of uniform multiplicity one. Now, observe that $T$ has the same reducing subspaces that $T - hiI$ has. Then, by case 1, it follows that $T-hi I$ has a non-trivial reducing subspace if and only if either $T-hi I$ is a normal operator or there exists $n\in \mathbb {N}$ such that $\alpha _n=\beta _n= 0.$ Accordingly, the same conclusion holds for $T$.

Finally, if $\Lambda (D)$ is contained in a line $\Gamma$ that intersects the real axis, let $\theta \in [0,\,\pi )$ denote the angle formed by $\Gamma$ and the real axis measured in the positive direction. Then, $\widetilde {D}= {\rm e}^{-i\theta }D$ satisfies that its set of eigenvalues $\Lambda (\widetilde {D})$ is contained in a line parallel to the real axis. The final statement follows upon applying case 1 to the operator ${\rm e}^{-i\theta }T$. This concludes the proof of theorem 2.1.

Proof. Let $D$ be a self-adjoint diagonal operator of uniform multiplicity one. Let us consider $u$ and $v$ two non-zero vectors in $H$ with non-zero components and being not real-proportional. Then the operator $T= D+ u\otimes v$ is strongly decomposable since $T^{\ast } - T$ is a rank two operator (and hence belongs to the Schatten class $S_p(H)$ for every $1\leq p < \infty$). Nevertheless, $T$ is not normal and by theorem 2.1, it has no non-trivial reducing subspaces.

Proof. First, we consider the case that the diagonal operator $D$ is a unitary operator and therefore, its spectrum is contained in the unit circle $\mathbb {T}$.

Case 1: $D$ is a unitary operator.

Assume, on the contrary, that $M$ is a non-trivial reducing subspace for $T=D+u\otimes u$ and denote, as usual, $P_M$ the orthogonal projection onto $M$. We may assume, without loss of generality, that $M$ is infinite dimensional (otherwise we would argue with $M^{\perp }$ since it would be an infinite-dimensional reducing subspace).

Let $x$ be a non-zero vector in $M$. Clearly, $Tx = Dx+\langle {x,\,u}\rangle u \in M$, and since $M$ is reducing, $T^{\ast } x = D^{\ast } x + \langle {x,\,u}\rangle u$ is also in $M$. Accordingly, $(D-D^{\ast })x \in M$ for every $x \in M$, or equivalently, $M$ is an infinite-dimensional non-trivial closed invariant subspace for $D-D^{\ast }$.

Note that $D-D^{\ast }$ is a non-trivial diagonal operator with the spectrum contained in the imaginary axis of the complex plane. Indeed, if $\Lambda (D)=(\lambda _n)_{n\geq 1}$ denotes the set of eigenvalues of $D$, then $\Lambda (D-D^{\ast })=(2 i\textrm{Im}\, (\lambda _n))_{n\geq 1}$. In particular, by [Reference Radjavi and Rosenthal32, theorems 1.23 and 1.25] $D-D^{\ast }$ is a completely normal operator and therefore, $M$ is spanned by eigenvectors of $D-D^{\ast }$.

On the other hand, since $D$ has uniform multiplicity one then the eigenvalues of $D-D^{\ast }$ has multiplicity at most two. Actually, if $2 i\textrm{Im}\, (\lambda _n)$ has multiplicity two, then there exists $m \in \mathbb {N}$ such that $\textrm {Im}\; (\lambda _n) = \textrm {Im}\; (\lambda _m)$. For each pair of indexes of this form, let us denote by $n$ the smaller index and by $n'$ the greater one. That is, let us denote by $\Omega$ the set of pairs of indexes

\[ \Omega= \{(n, n')\in \mathbb{N}\times \mathbb{N}:\; n< n' \mbox{ and } \textrm{Im}\; (\lambda_n) = \textrm{Im}\; (\lambda_{n'}) \mbox{ where } \lambda_n, \lambda_{n'} \in \Lambda(D) \}. \]

Doing so, we may consider a disjoint partition of the natural numbers

\[ \mathbb{N} = N_1 \cup N_2 \cup N_3, \]

where $N_1\subset \mathbb {N}$ consists of the indexes of the eigenvalues with multiplicity two that are the smaller index of its corresponding pair, $N_2$ consists of the indexes of the eigenvalues with multiplicity two that are the larger index of its corresponding pair and $N_3$ the indexes of the eigenvalues with multiplicity one. In other words, the set $\Omega =N_1\times N_2$.

Then, the eigenvectors of $D-D^{*}$ are

\[ \{e_n : n\in N_3\} \mbox{ and } \{\lambda e_n + \tau e_{n'} : (n,n')\in \Omega,\, \lambda, \tau \in \mathbb{C}\}. \]

With the characterization of the eigenvectors of $D-D^{*}$, our goal now will be identify those eigenvectors spanning $M$, that is, determine the proper subset $\Lambda$ of $\mathbb {N}$ such that

\[ M=\overline{\textrm{span}\; \{e_n : n \in N_3\cap \Lambda \}+\{\lambda e_n + \tau e_{n'} : n\in \Lambda\cap N_1,\, (n,n')\in \Omega,\, \lambda, \tau \in \mathbb{C}\}}. \]

By lemma 2.4 for every $n \in \mathbb {N}$ we have $e_n \notin M\cup M^{\perp }$, so $e_n \notin M$ for every $n \in N_3.$ Assume for the moment that $N_3\cap \Lambda$ is not empty and let $n_0\in N_3\cap \Lambda$. Denote by

\[ e_{n_0}=e_{n_0}^{M}\oplus e_{n_0}^{M^{{\perp}}} \]

the orthogonal decomposition of $e_{n_0}$ with respect to $H=M\oplus M^{\perp }$. Note that, in particular, $e_{n_0}^{M}$ is non-zero because otherwise $e_{n_0}\in M^{\perp }$.

Having in mind that $(D-D^{*}) P_M = P_M (D-D^{*})$ because $M$ is reducing for $D-D^{*}$, we deduce upon applying it to $e_{n_0}$ that

\[ (D-D^{*}) P_M e_{n_0}= (D-D^{*}) e_{n_0}^{M}= (\lambda_{n_0}- \overline{\lambda_{n_0}}) e_{n_0}^{M}. \]

That is, $2 i\textrm{Im}\, (\lambda _{n_0})$ is an eigenvalue of $D-D^{*}$ with eigenspace $\textrm {span}\;\{e_{n_0},\, e_{n_0}^{M}\}$. Hence, $2 i\textrm{Im}\, (\lambda _{n_0})$ is of multiplicity 2, but this contradicts the fact that $n_0\in N_3$. Therefore, $N_3\cap \Lambda$ is empty. Accordingly, we deduce that $\Lambda$ must be a non-void subset $\Lambda$ of $N_1$ (being possible $\Lambda =N_1$), and

(3.1)\begin{equation} M = \overline{\textrm{span}\; \{\lambda e_n + \tau e_{n'}:\; n\in \Lambda\subseteq N_1,\, (n,n')\in \Omega \mbox{ and } \lambda, \tau \in \mathbb{C} \}}. \end{equation}

Moreover, we observe that for every eigenvector $\lambda e_n + \tau e_{n'} \in M$, the coefficients $\lambda$ and $\tau$ are non-zero (otherwise we would have $e_n$ or $e_{n'} \in M$). Accordingly, every eigenvector $\lambda e_n + \tau e_{n'} \in M$ is a multiple of $e_n + ({\tau }/{\lambda })e_{n'}$. Since $e_{n'}\notin M$, we deduce that there exists coefficients $\tau _n \in \mathbb {C}$ such that

(3.2)\begin{equation} M = \overline{\textrm{span}\; \{e_n + \tau_n e_{n'}:\; n\in \Lambda\subseteq N_1,\, (n,n')\in \Omega \}}. \end{equation}

On the other hand, since $M$ is a non-trivial closed invariant subspace under $T$, we have that $T(e_n + \tau _n e_{n'}) \in M$ for every $n\in \Lambda \subseteq N_1$ with $(n,\,n')\in \Omega$. If we consider $u =\sum _n \alpha _n e_n$ where $\alpha _n=\langle {u,\,e_n}\rangle \neq 0$ for every $n \in \mathbb {N}$ by hypotheses, then

(3.3)\begin{equation} \begin{aligned} T(e_n + \tau_n e_{n'}) & = \lambda_ne_n - \overline{\lambda_n}\tau_n e_{n'} + \langle{e_n + \tau_n e_{n'},u}\rangle u \\ & = \lambda_ne_n - \overline{\lambda_n}\tau_n e_{n'} + (\overline{\alpha_n} + \tau_n \overline{\alpha_{n'}})u \in M \end{aligned} \end{equation}

for every $n\in \Lambda \subseteq N_1$ with $(n,\,n')\in \Omega$.

Now, let us prove that $\textrm {Re}\; (\lambda _{n_0})=0$ for at least one positive integer $n_0 \in N_1$.

First, by means of the orthogonality relations of the basis elements $\{e_n\}_{n\in \mathbb {N}}$, it follows that for every vector $x \in M$

(3.4)\begin{equation} \tau_m \langle{x,e_m}\rangle = \langle{x,e_{m'}}\rangle \qquad \mbox{for every} \ m\in \Lambda\subseteq N_1 \mbox{ with } (m,m')\in \Omega. \end{equation}

Let $n_0$ be any positive integer in $\Lambda$ (recall that $\Lambda$ is not empty). If

(3.5)\begin{equation} \overline{\alpha_{n_0}} + \tau_{n_0} \overline{\alpha_{n_0'}} = 0, \end{equation}

then by (3.3) the vector

\[ \textbf{a}= \lambda_{n_0} e_{n_0} - \overline{\lambda_{n_0}}\tau_{n_0} e_{n_0'} \]

is in $M$ which, by means of (3.4) particularized in the index $n_0$, satisfies

\[ \tau_{n_0} \langle{\textbf{a},e_{n_0}}\rangle=\langle{\textbf{a},e_{n_0'}}\rangle. \]

That is,

\[ \tau_{n_0} \lambda_{n_0}={-} \overline{\lambda_{n_0}}\tau_{n_0}. \]

Having in mind that $\tau _n\neq 0$ for every $n\in \mathbb {N}$, we deduce that $\lambda _{n_0}=- \overline {\lambda _{n_0}}$ as far as (3.5) holds, or equivalently $\textrm {Re}\; (\lambda _{n_0}) = 0$ whenever (3.5) is satisfied.

Assume, now, that (3.5) is not satisfied, that is

(3.6)\begin{equation} \overline{\alpha_{n_0}} + \tau_{n_0} \overline{\alpha_{n_0'}} \neq 0. \end{equation}

Let us consider $m_0\in \Lambda$ with $m_0\neq n_0$. Observe that this is possible since, in particular, $M$ is given by (3.2) and the dimension of $M$ is infinite.

Now, by (3.3), the vector

\[ \textbf{b}= \lambda_{n_0} e_{n_0} - \overline{\lambda_{n_0}}\tau_{n_0} e_{n_0'}+(\overline{\alpha_{n_0}} + \tau_{n_0} \overline{\alpha_{n_0'}})u \]

is in $M$. We argue as before upon considering (3.4) at $\textbf {b}$ and $e_{m_0}$, that is, $\tau _{m_0} \langle {\textbf {b},\,e_{m_0}}\rangle = \langle {\textbf {b},\,e_{m_0'}}\rangle$. Then

\[ \tau_{m_0} (\overline{\alpha_{n_0}} + \tau_{n_0} \overline{\alpha_{n_0'}})\alpha_{m_0} = (\overline{\alpha_{n_0}} + \tau_{n_0} \overline{\alpha_{n_0'}})\alpha_{m_0'}. \]

Since $\overline {\alpha _{n_0}} + \tau _{n_0} \overline {\alpha _{n_0'}} \neq 0$ (i.e. our assumption (3.6)), we deduce that

\[ \tau_{m_0} = \frac{\alpha_{m_0'}}{\alpha_{m_0}}. \]

Therefore, having in mind that $e_{m_0} + \tau _{m_0} e_{m_0'}\in M$, we deduce that

\[ \textbf{c}= T(e_{m_0} + \tau_{m_0} e_{m_0'}) = \lambda_{m_0} e_{m_0} - \overline{\lambda_{m_0}}\frac{\alpha_{m_0'}}{\alpha_{m_0}}e_{m_0'} + \left (\overline{\alpha_{m_0}} + \frac{|\alpha_{m_0'}|^{2}}{\alpha_{m_0}} \right ) u \]

is also in $M$. Once again applying (3.4), we have $\tau _{m_0} \langle {\textbf {c},\,e_{m_0}}\rangle = \langle {\textbf {c},\,e_{m_0'}}\rangle$ and therefore

\[ \frac{\alpha_{m_0'}}{\alpha_{m_0}}\left (\lambda_{m_0} + |\alpha_{m_0}|^{2} + |\alpha_{m_0'}|^{2} \right ) ={-}\frac{\alpha_{m_0'}}{\alpha_{m_0}}\overline{\lambda_{m_0}} + \overline{\alpha_{m_0}}\alpha_{m_0'} + \frac{|\alpha_{m_0'}|^{2}}{\alpha_{m_0}}\alpha_{m_0'}. \]

Multiplying by ${\alpha _{m_0}}/{\alpha _{m_0'}}$ we obtain

\[ \lambda_{m_0} + |\alpha_{m_0}|^{2} + |\alpha_{m_0'}|^{2} ={-} \overline{\lambda_{m_0}} + |\alpha_{m_0}|^{2} + |\alpha_{m_0'}|^{2}, \]

from where, clearly, $\textrm {Re}\; (\lambda _{m_0}) = 0$.

Consequently, independently if (3.5) is or not satisfied, there exists at least a positive integer $n_0\in N_1$ such that $\textrm {Re}\; (\lambda _{n_0})=0$, as we wished to show.

In order to finish the proof of case 1, let us show that the existence of such $n_0 \in N_1$ yields the desired contradiction. Since $D$ is unitary, its spectrum is contained in the unit circle and therefore, such $\lambda _{n_0}$ is either $i$ or $-i$. Assume $\lambda _{n_0}=i$ (the other case is analogous). If $n_0\in N_1$, by definition, there exist $n_0'\in \mathbb {N}$ with $n_0< n_0'$ and $\lambda _{n_0'}$ an eigenvalue of $D$, $\lambda _{n_0'} \in \Lambda (D)$, such that $\textrm {Im}\; (\lambda _{n_0})=\textrm {Im}\; (\lambda _{n_0'})$. Hence, $\textrm {Im}\; (\lambda _{n_0'})=1$, and therefore $\lambda _{n_0'}$ must be also $i$. Accordingly, $i$ is an eigenvalue of $D$ of multiplicity 2, which contradicts the hypotheses that the uniform multiplicity of $D$ is one.

Case 2: $D$ is not a unitary operator.

Now, assume $D$ is a diagonal operator such that $\sigma (D)$ is contained in a circle with center $\alpha$ and radius $r>0$. As in case 1, we argue by contradiction assuming $T = D+u \otimes u$ has a non-trivial reducing subspace $M$. Since $M$ is reducing for $T$, it is also reducing for $({1}/{r})(T - \alpha I) = ({1}/{r})(D- \alpha I) + ({1}/{r}) u\otimes u,$ which is the contradiction.

This concludes the proof that statement (a) of theorem 3.1 holds.

Now, we will show statement (b) of theorem 3.1. As before, the key argument will be to prove the result when $D$ is a unitary diagonal operator.

Hence, suppose $D$ is a unitary and $u \in H$ such that $\langle {u,\,e_n}\rangle \neq 0$ for every $n \in \mathbb {N}$ and $\langle {u,\,Du}\rangle \neq 0.$ In order to show that there exists $v \in H$ such that $T = D+u\otimes v\in \mathcal {L}(H)$ is not normal but has non-trivial reducing subspaces, we take

\[ v={-} \frac{1}{\langle{u,Du}\rangle}D^{*2} u. \]

The goal will be showing that the subspace $M:=\textrm {span}\;\{D^{*}u\}$ reduces the operator $T = D + u\otimes v$.

First, note that

\[ \langle{D^{*2} u,e_n}\rangle = \langle{u,D^{2}e_n}\rangle = \overline{\lambda_n}^{2}\langle{u,e_n}\rangle \neq 0 \]

since $|\lambda _n| = 1$. Then, $\langle {v,\,e_n}\rangle \neq 0$ for every $n \in \mathbb {N}$.

Now, observe that

\[ \langle{D v,u}\rangle = \left\langle{- \frac{1}{\langle{u,Du}\rangle}D^{*}u,u } \right\rangle={-} \frac{1}{\langle{u,Du}\rangle} \langle{D^{*}u,u}\rangle ={-}1. \]

So, we have

\[ TD^{*}u = (D+u\otimes v)D^{*}u = u + \langle{D^{*}u,v}\rangle u = u + \langle{u,D v}\rangle u = u-u = 0. \]

Moreover,

\[ T^{*}D^{*}u = (D^{*}+v\otimes u)D^{*}u = D^{*2} u + \langle{u,Du}\rangle v = D^{*2} u - D^{*2}u = 0. \]

So, $D^{*}u$ is a non-zero vector which turns out to be an eigenvector associated with the eigenvalue $0$ for $T$ and $T^{*}$.

Finally, let us show that $T$ is not a normal operator. Assume, on the contrary, that $T$ is normal. Having into account condition (ii$'$) in Ionascu theorem 1.1, we deduce the existence of $\gamma \in \mathbb {C}$ such that $Dv = \gamma u$. Then, since $Dv = - ({1}/{\langle {u,\,Du}\rangle })D^{*}u$, we have

\[ D^{*}u ={-}\gamma \langle{u,Du}\rangle u. \]

That is, $u$ is an eigenvector for $D^{*}$ associated with the eigenvalue $-\gamma \langle {u,\,Du}\rangle$. Observe that $\gamma \neq 0$ since $Dv \neq 0$ and $\langle {u,\,Du}\rangle \neq 0$ by hypothesis. Therefore, there exists $n_0 \in \mathbb {N}$ such that $u\in \textrm {span}\;\{e_{n_0}\}$, and this contradicts the fact that $\langle {u,\,e_n}\rangle \neq 0$ for every $n\in \mathbb {N}$. Hence $T$ is not a normal operator.

Now, let us prove the general case. Assume $D$ is not unitary but $\sigma (D)$ is contained in a circle with center $\alpha$ and radius $r>0$. Then the operator $\widetilde {D}= ({1}/{r})(D - \alpha I)$ is a diagonal unitary operator satisfying $\langle {u,\,\widetilde {D} u}\rangle \neq 0$ (since $\langle {u,\, \widetilde {D} u}\rangle = 0$, would imply $\langle {u,\, (D-\alpha I) u}\rangle = 0,$ and therefore $\langle {u,\,Du}\rangle = \overline {\alpha }\left \lvert \left \lvert {u}\right \rvert \right \rvert ^{2},$ which contradicts our hypotheses). Hence, from the unitary case, we ensure the existence of $\widetilde {v} \in H$ such that

\[ \widetilde{T} = \widetilde{D} + u\otimes \widetilde{v} \]

is not normal but has a non-trivial reducing subspace $M$. Thus,

\[ r \widetilde{T}+\alpha I = D + r(u\otimes \widetilde{v})=D + u\otimes (r\widetilde{v}) \]

is a non-normal operator with $M$ as a reducing subspace. Accordingly, $v= r \widetilde {v}$ completes the argument and the proof of theorem 3.1.

Proof. First, we claim that every non-zero reducing subspace of $T$ contains at least one eigenvector. In order to show the claim, let $M$ be a non-zero reducing subspace. Since the eigenvectors of $T$ span $H$, there exists an eigenvector $x$ not orthogonal to $M$. Let $Tx=\lambda x$ and write $x=x_1+x_2$ with respect to the orthogonal decomposition $H=M\oplus M^{\perp }$. Hence, from

\[ Tx=T (x_1+x_2)= \lambda (x_1+ x_2), \]

it follows

\[ Tx_1-\lambda x_1= x_2-T x_2 \]

and since $M$ is reducing, $Tx_1-\lambda x_1 \in M$ and $x_2-T x_2 \in M^{\perp }$. Thus, $Tx_1=\lambda x_1$ and $M$ contains at least one eigenvector of $T$ as claimed.

Now, let us prove the statement of the theorem. Take $M$ a non-zero reducing subspace for $T$ and denote by $N$ the closed subspace of $M$ generated by all the eigenvectors of $T$ in $M$. Note that $N$ is a non-zero reducing subspace because $T$ is normal. We claim that $N=M$.

Indeed, arguing by contradiction, suppose that $N\neq M$. Then $N' = M\cap N^{\perp }$ is a non-zero invariant subspace for $T$ which is reducing (since it is the intersection of two reducing subspaces). Accordingly, $N'$ contains at least one eigenvector $z\neq 0$ of $T$. Thus $z\in N^{\perp }$. But, since $z\in M$ and is an eigenvector of $T$, $z\in N$ by definition. But this implies that $z=0$, which is a contradiction. Accordingly, $N=M$ which yields the statement.

Proof of proposition 4.1. Proof of proposition 4.1

Let $P_M: H \rightarrow M$ be the orthogonal projection onto $M$, which clearly commutes with $T$ since $M$ is reducing. The goal is to show that either $P_M$ is the identity or the zero operator. Assume $P_M$ is not the zero operator, $Q_M = I-P_M$ and suppose first that $u,\,v \in M$. Hence, $Q_M u = Q_M v = 0$.

By lemma 2.2 we have $Q_M D P_M = Q_M D^{*} P_M = 0$. Then, $DP_M = P_M D P_M$ and $D^{*}P_M = P_M D^{*} P_M$, so it follows that $M$ is an invariant subspace for $D$ and $D^{*}$, that is, $M$ is a reducing subspace for $D$. Thus, according to Wermer's theorem, $M$ is spanned by a set of eigenvectors of $D$ and, therefore, there exists $\Lambda \subset \mathbb {N}$ such that

\[ M = \overline{\textrm{span}\; \{e_n : n \in \Lambda\}}. \]

Note that, if $n \in \Lambda$ we have $Q_M e_n = 0$ and if $n \in \mathbb {N}\setminus \Lambda$ then $Q_M e_n = e_n.$

Now, since $u =\sum _n \alpha _n e_n$ and $v=\sum _n \beta _n e_n$ and both $Q_M u = Q_M v = 0$, we deduce

\[ \sum_{n \notin \Lambda} \alpha_n e_n = \sum_{n\notin \Lambda }\beta_n e_n = 0. \]

Hence, $\alpha _n = \beta _n = 0$ for every $n \in \mathbb {N}\setminus \Lambda$ which contradicts the hypothesis that $\alpha _n$ and $\beta _n$ are not simultaneously zero unless $\Lambda = \mathbb {N}$ and, $P_M$ the identity operator as we wished to prove.

The case $u,\,v \in M^{\perp }$ is analogous.

Proof of lemma 4.4. Proof of lemma 4.4

Let us begin by proving statement (i). Since $\{u,\,v,\, D^{*}u,\,Dv\}$ are linearly independent, Ionascu theorem implies that $T$ is not normal (observe that statement (i) in Ionascu theorem asks $u$ and $v$ be linearly dependent vectors while (ii) asks $Dv$ and $u$ be linearly dependent). Since $T\mid _M$ is normal, we deduce that $M\subsetneq H.$

On the other hand, a straightforward computation shows

(4.1)\begin{equation} [T,T^{*}] = Dv\otimes u + u \otimes (Dv + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}u) - D^{*}u\otimes v - v \otimes (D^{*}u + \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}v). \end{equation}

By hypotheses, $T\mid _M$ is normal so $[T,\,T^{*}]x = 0$ for every $x \in M$. Since $u,\,v,\,D^{*}u,\, Dv$ are linearly independent, lemma 4.5 yields that

\[ \langle{x,u}\rangle=\langle{x,v}\rangle = 0 \]

for every $x \in M$. Hence, $u,\,v \in M^{\perp }$ and therefore, by proposition 4.1, $M$ is trivial. Now, $T$ is not normal and accordingly $M = \{0\}.$

To prove (ii), assume $u$ is an eigenvector for $D^{*}$ (a similar reasoning applies if $v$ is an eigenvector for $D$). Since $D$ has uniform multiplicity one, there exists $n_0 \in \mathbb {N}$ such that $u\in \textrm {span}\;\{e_{n_0}\}$. Without loss of generality, we may assume $u=e_{n_0}$. Then equation (4.1) becomes

\begin{align*} [T,T^{*}] & = Dv\otimes e_{n_0} + e_{n_0}\otimes (Dv + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}e_{n_0})- \overline{\lambda_{n_0}}e_{n_0} \otimes v - v \otimes (\overline{\lambda_{n_0}}e_{n_0} + v) \\ & = (D-\lambda_{n_0} I)v\otimes e_{n_0} + e_{n_0}\otimes ( (D-\lambda_{n_0}I)v+\left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}e_{n_0}) - v\otimes v. \end{align*}

Now, $v=\sum _k \beta _k e_k$ and $\beta _k \neq 0$ for every $k \neq n_0$ and each eigenvalue $\lambda _k$ of $D$ is of multiplicity one. Hence, $e_{n_0}$ is linearly independent of $(D-\lambda _{n_0} I)v$ and $v$. Moreover, assume there exists $\beta \in \mathbb {C}\setminus \{0\}$ such that

\[ (D-\lambda_{n_0} I )v = \beta v. \]

Then $(\lambda _k - \lambda _{n_0}) = \beta$ for every $k \neq n_0$ which is a contradiction. Accordingly, $(D-\lambda _{n_0} I)v,\, e_{n_0}$ and $v$ are linearly independent vectors, again lemma 4.5 yields that $v,\, e_{n_0} \in M^{\perp }$ and $M\neq H$. Thus, proposition 4.1 yields that $M = \{0\}$, as stated in (ii).

Let us prove now statement (iii). Assume $u = \alpha v$ for some $\alpha \in \mathbb {C}\setminus \{0\}$ (observe that $\alpha \neq 0$ since $u$ is a nonzero vector).

Assume $M$ is a non-zero reducing subspace. We may assume $M\neq H$ since on the contrary, the result is just a consequence of Ionascu theorem condition (i$'$). We are required to show that $\Lambda (D)$ is contained in a line and that $T$ is a normal operator.

First, observe that if $v \in M^{\perp }$ then $u \in M^{\perp }$ and $M = \{0\}$ by proposition 4.1. Assume, therefore, that $v \notin M^{\perp }.$

Equation (4.1) turns out to be in this case

(4.2)\begin{equation} [T,T^{*}] = (\overline{\alpha}D-\alpha D^{*})v\otimes v + v\otimes (\overline{\alpha}D-\alpha D^{*})v. \end{equation}

Now, we claim that the fact that $v \notin M^{\perp }$ implies $(\overline {\alpha }D-\alpha D^{*})v = \beta v$ for some $\beta \in \mathbb {C}$. Indeed, if $x\in M$ such that $\langle {x,\,v}\rangle \neq 0$, (4.2) yields

\[ 0=\langle{x,v}\rangle (\overline{\alpha}D-\alpha D^{*})v + \langle{x,(\overline{\alpha}D-\alpha D^{*})v}\rangle v, \]

from where the claims follows having $\beta =-\langle {x,\,(\overline {\alpha }D-\alpha D^{*})v}\rangle /\langle {x,\,v}\rangle$.

So, equation (4.2) becomes

\[ [T,T^{*}] = 2\textrm{Re}\, (\beta)v\otimes v. \]

But, once again, since $T\mid _M$ is normal and $v \notin M^{\perp }$, it follows that $\textrm {Re}\; (\beta ) = 0$. Hence, $\beta = i t$ with $t \in \mathbb {R}.$

Finally, observe that $(\overline {\alpha }D-\alpha D^{*})v = \beta v= i t v$ implies

\[ \textrm{Im}\; (\alpha \overline{\lambda_n}) = t, \]

so $\Lambda (D)$ lies in the line $\{z \in \mathbb {C}: \textrm {Im}\; ( \alpha \overline {z}) = t \}$ and by Ionascu theorem condition (i$'$) it follows that $T$ is normal.

This proves statement (iii).

In order to show condition (iv), assume $Dv -\alpha v = \lambda u$ for some $\alpha,\, \lambda \in \mathbb {C}$. If $\lambda = 0$, case (ii) yields that $M=\{0\}$, which is a contradiction. So, we may assume $\lambda \neq 0$. Clearly, without loss of generality, we can also assume $\alpha = 0$ since $D-\alpha I$ is a diagonal operator of uniform multiplicity one. Thus, $Dv=\lambda u$ for $\lambda \neq 0$.

In addition, observe that if $u$ and $v$ are linearly dependent then $Dv$ and $v$ are linearly dependent, and $M=\{0\}$ by (ii), which is a contradiction. So we may also suppose that $u$ and $v$ are linearly independent vectors.

So, assume $M$ is a non-zero reducing subspace, $Dv=\lambda u$ for $\lambda \neq 0$ and $\{u,\,v\}$ are linearly independent. As in the proof of condition (iii), we may assume $M\neq H$ and the goal is to show that $\Lambda (D)$ is contained in a circle and that $T$ is normal.

Now, equation (4.1) becomes

(4.3)\begin{align} [T,T^{*}]& =u \otimes (2\textrm{Re}\, (\lambda) + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2})u - (D^{*}-\overline{\alpha}I)u \otimes v - v \otimes ( (D^{*}-\overline{\alpha}I)u + \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}v) \nonumber\\ & = u \otimes (2\textrm{Re}\, (\lambda) + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2})u - D^{*} u \otimes v - v \otimes (D^{*}u + \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}v) \end{align}

First, let us assume that $v \in M^{\perp }$. Upon applying (4.3) to any vector $x\in M$, we deduce

\[ 0= (2\textrm{Re}\, (\lambda) + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}) \langle x, u\rangle u- \langle x, D^{*}u \rangle v, \]

which along with the linear independence of $\{u,\, v\}$ yields that $(2\textrm{Re}\, (\lambda ) + \left \lvert \left \lvert {v}\right \rvert \right \rvert ^{2}) = 0$ and $D^{*}u \in M^{\perp }.$ In addition, since $M$ is reducing for $T$ if follows that

\[ Tv = Dv + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}u = \lambda u + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}u \in M^{{\perp}}. \]

Having in mind that $\textrm {Re}\; (\lambda ) = - \left \lvert \left \lvert {v}\right \rvert \right \rvert ^{2}/2$, we deduce that $Tv \neq 0$. Thus, $u \in M^{\perp }$. At this point, we may argue either using proposition 4.1 which would yield that $M = \{0\}$ since $v\in M^{\perp }$ or as follows. First, $M$ is a reducing subspace for $D$ since $M$ is reducing for $T$ and $u,\,v \in M^{\perp }$. Then, by means of Wermer theorem, there exists $n_0\in \mathbb {N}$ such that $e_{n_0} \in M$. In particular, this implies that $\langle {u,\,e_{n_0}}\rangle = \langle {v,\,e_{n_0}}\rangle = 0$, or equivalent, $\alpha _{n_0} = \beta _{n_0} = 0$, what contradicts the hypotheses of the lemma. This latter argument will be of use in a remark after finishing the proof of lemma 4.4.

Now, let us assume $v\notin M^{\perp }.$ Once again, upon applying (4.3) to any vector $x\in M$ with $\langle x,\, v\rangle \neq 0$, it follows that

(4.4)\begin{equation} D^{*}u = \mu u + \beta v, \end{equation}

where $\mu,\, \beta \in \mathbb {C}$. It can be assumed that $\beta \neq 0$. We claim that $\alpha _n \neq 0$ for every $n \in \mathbb {N}$. Indeed, from $D^{*}u = \mu u+\beta v$, it follows that for every $n\in \mathbb {N}$

\[ \overline{\lambda_n}\alpha_n = \mu \alpha_n + \beta \cdot \beta_n. \]

So, if $\alpha _{n_0} = 0$ for some $n_0 \in \mathbb {N}$, it would imply that $\beta _{n_0} = 0$ which contradicts the hypotheses.

We will show that $\mu =0$ in (4.4) since otherwise we are led to a contradiction.

Assume, on the contrary, that $\mu \neq 0$ in (4.4). Applying $D$ to (4.4) we obtain

\[ DD^{*}u = \mu Du + \beta Dv = \mu Du + \beta \lambda u. \]

Then, the eigenvalues $\lambda _n$ of $D$ satisfy

\[ |\lambda_n|^{2} = \mu \lambda_n + \beta \lambda, \]

for every $n \in \mathbb {N}$, since $\alpha _n\neq 0$ for every $n\in \mathbb {N}.$ Dividing out by $-\beta \lambda$ (which is different from 0), we deduce that there exist complex numbers $a,\,b \in \mathbb {C}\setminus \{0\}$ such that every $\lambda _n$ with $n\neq n_0$ lies in

\[ A= \{z \in \mathbb{C}: a|z|^{2} + bz ={-}1 \}. \]

The equation $a|z|^{2} +bz = -1$, $z \in \mathbb {C}$ is equivalent to the system

\[ A=\left \{ \begin{array}{@{}l} \textrm{Re}\; (a)(x^{2}+y^{2})+ \textrm{Re}\; (b)x-\textrm{Im}\; (b)y ={-}1,\\ \textrm{Im}\; (a)(x^{2}+y^{2})+ \textrm{Re}\; (b)y + \textrm{Im}\; (b)x = 0, \end{array} \right. \]

where $x = \textrm {Re}\; (z),\, y = \textrm {Im}\; (z) \in \mathbb {R}$. Observe that $A$ is therefore the intersection of two different conics, and hence $A$ is a finite set of points, what contradicts the uniform multiplicity one of $D$.

Hence, $\mu = 0$ as claimed and (4.4) becomes

\[ D^{*}u = \beta v. \]

Note that $u$ is linearly independent of $D^{*}u$ because $\{u,\, v\}$ are linearly independent. Having in mind this and applying (4.3) to any vector $x\in M$ once more time we deduce that

\[ (2 \textrm{Re}\; (\lambda) + \left\lvert\left\lvert{v}\right\rvert\right\rvert^{2}) = 0. \]

That is, $\textrm {Re}\; (\lambda ) = - {\left \lvert \left \lvert {v}\right \rvert \right \rvert ^{2}}/{2}.$ Then, equation (4.3) turns out to be

\[ [T,T^{*}]=\beta v \otimes v + v \otimes (\beta +\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2})v= (2 \textrm{Re}\; (\beta) + \left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}) v\otimes v. \]

On the other hand, having in mind that $Dv = \lambda u$, we deduce

\[ DD^{*}u = D(\beta v)= \beta \lambda u. \]

Now, the fact that $\alpha _n \neq 0$ for every $n \in \mathbb {N}$ along with the previous equality implies that

\[ DD^{*} = \beta \lambda I. \]

Note that $\beta \lambda$ is a positive number. Accordingly, the spectrum of $D$ is contained in a circle of center $0$ and radius $r=\sqrt {\beta \lambda }>0$, which is half of the statement we wished to prove. Let us see also that $T$ is normal.

Now, an easy computation involving coordinates in the expression $Dv = \lambda u$ leads to

\[ r\left\lvert\left\lvert{v}\right\rvert\right\rvert= |\lambda|\left\lvert\left\lvert{u}\right\rvert\right\rvert, \]

and therefore

(4.5)\begin{equation} |\lambda| = \frac{r\left\lvert\left\lvert{v}\right\rvert\right\rvert}{\left\lvert\left\lvert{u}\right\rvert\right\rvert}. \end{equation}

Since $Dv = \lambda u= |\lambda |{\rm e}^{i\theta }u$ for $\theta \in [0,\,2\pi )$, it follows

\[ |\lambda|u = {\rm e}^{{-}i\theta} Dv, \]

and by (4.5)

\[ \frac{ru}{\left\lvert\left\lvert{u}\right\rvert\right\rvert} = {\rm e}^{{-}i\theta} \frac{D v}{\left\lvert\left\lvert{v}\right\rvert\right\rvert}. \]

Moreover, from $\textrm {Re}\; (\lambda ) = -\left \lvert \left \lvert {v}\right \rvert \right \rvert ^{2}/2$ one deduces $\textrm {Re}\; ( {r{\rm e}^{-i\theta }}/({\left \lvert \left \lvert {u}\right \rvert \right \rvert \left \lvert \left \lvert {v}\right \rvert \right \rvert }) ) = - ({1}/{2}),\,$ which implies that $T$ is normal by Ionascu theorem condition (ii$'$). This completes the proof of the statement (iv).

Finally, let us prove the statement (v). Assume $Dv = \alpha u + \beta v + \mu D^{*}u$ for some $\alpha,\, \beta,\,\mu \in \mathbb {C}$, and $T$ is not normal. Since we can express

\[ (D-\beta I)v = (\textbf{a}-\mu\overline{\beta})u + \mu(D^{*}-\overline{\beta}I)u \]

where $\textbf {a}=\alpha + 2\mu \overline {\beta }$, we can assume with no loss of generality that $\beta = 0$.

Observe that, in this case, equation (4.1) becomes

\begin{align*} [T,T^{*}] & = u \otimes ((2 \textrm{Re}\; (\alpha)+\left\lvert\left\lvert{v}\right\rvert\right\rvert^{2})u +\mu D^{*}u)\\ & \quad - D^{*}u\otimes (\overline{\mu}u-v) - v\otimes (D^{*}u+\left\lvert\left\lvert{u}\right\rvert\right\rvert^{2}v). \end{align*}

Moreover, we can assume that $D^{*}u,\, u$ and $v$ are linearly independent and $\mu \neq 0$, otherwise cases (iii) and (iv) would yield that $T$ is normal, a contradiction.

By hypotheses, $[T,\,T^{*}]x = 0$ for every $x \in M$, so lemma 4.5 yields that $(2 \textrm {Re}\; (\alpha )+\left \lvert \left \lvert {v}\right \rvert \right \rvert ^{2})u +\mu D^{*}u$, $\overline {\mu }u-v$ and $D^{*}u+\left \lvert \left \lvert {u}\right \rvert \right \rvert ^{2}v$ belongs to $M^{\perp }.$

We will argue by contradiction. Assume, on the contrary, that the reducing subspace $M$ has dimension strictly larger than 1. Then there exists $x \in M\setminus \{0\}$ such that $\langle {x,\,u}\rangle = 0$. Since $\langle {x,\,\overline {\mu }u - v }\rangle =0,$ we have $\langle {x,\,v}\rangle = 0.$ Moreover, since $x$ is also orthogonal to $D^{*}u+\left \lvert \left \lvert {u}\right \rvert \right \rvert ^{2}v$, we have $\langle {x,\,D^{*}u}\rangle = 0.$ The fact that $M$ is invariant under $T$ implies, in particular,

\[ Tx = Dx + \langle{x,v}\rangle u = Dx \in M. \]

So, for every $x \in M \cap (\textrm {span}\;\{u\})^{\perp }$, $Dx \in M$ and $\langle {Dx,\,u}\rangle = \langle {x,\,D^{*}u}\rangle = 0$. Hence, the closed subspace $M\cap (\textrm {span}\;\{u\})^{\perp }$ is invariant under $D$.

On the other hand, $M$ is also invariant under $T^{*}$ because it is reducing and, therefore,

\[ T^{*}x = D^{*}x + \langle{x,u}\rangle v = D^{*}x \in M. \]

Moreover, since $Dv = \alpha u + \mu D^{*}u$ we deduce $\langle {D^{*}x,\,v}\rangle = \langle {x,\,Dv}\rangle = 0$. In addition, from $D^{*}x \in M$, it follows

\[ \langle{D^{*}x,\overline{\mu}u-v}\rangle = 0, \]

and therefore, $\langle {D^{*}x,\,u}\rangle = 0.$ Accordingly, $M\cap (\textrm {span}\;\{u\})^{\perp }$ is a reducing subspace for $D^{*}$.

Now, since every reducing subspace of $D$ contains an eigenvector because of Wermer theorem, it follows that there exists $n_0 \in \mathbb {N}$ such that $e_{n_0} \in M\cap (\textrm {span}\;\{u\})^{\perp }$.

Now, recalling that $\overline {\mu } u - v \in M^{\perp }$, we deduce $\langle {e_{n_0},\,v}\rangle = 0$. Thereby, $e_{n_0}$ is orthogonal to $u$ and $v$, that is, $\alpha _{n_0} = \beta _{n_0} = 0$, which contradicts the hypotheses of the lemma. Hence, the dimension of $M$ is equal or less than one, as we wished to prove. This concludes the proof of the statement (v) and hence that of lemma 4.4.

Proof of theorem 4.3. Proof of theorem 4.3

If $T$ is normal, the spectral theorem for normal operators provides plenty of reducing subspaces for $T$. In addition, if condition (b) is satisfied, a simple computation shows that $M$ is reducing for $T$, and since $M$ has dimension $1$, $T\mid _M$ is normal.

Assume now there exists a non-trivial reducing subspace $M$ for $T$ such that $T\mid _M$ is normal. By lemma 4.4 we deduce that either $T$ is normal or $\dim M = 1$. Assume $T$ is not normal. Hence, $\dim M = 1$ and therefore, there exists a non-zero vector $x \in M$ and $\alpha,\,\beta \in \mathbb {C}$ such that $Tx = \alpha x$ and $T^{*}x = \beta x.$ From $Tx = \alpha x$ we have $(D-\alpha I)x + \langle {x,\,v}\rangle u = 0,$ so

\[ (D-\alpha I)x ={-} \langle{x,v}\rangle u. \]

From $(T^{*}-\beta I)x = 0$ we have $(D^{*}-\beta I)x + \langle {x,\,u}\rangle v = 0.$ Having into account that $u \notin M^{\perp }$, we have $\langle {x,\,u}\rangle \neq 0$, and therefore

\[ v ={-} \frac{1}{\langle{x,u}\rangle}(D^{*}-\beta I)x. \]

Hence,

\[ (D-\alpha I)x = \frac{1}{\langle{u,x}\rangle}\langle{x, (D-\beta I)x}\rangle u, \]

which yields the result.

Proof. It suffices to prove that if $T$ has a non-trivial reducing subspace, then $T$ is normal since the converse is straightforward. In addition, we may assume that $D$ is not self-adjoint, since otherwise $T$ would be self-adjoint and the result holds trivially.

Thus, let $M$ be a non-trivial reducing subspace and suppose that $D$ is not self-adjoint.

First, let us assume $T\mid _M$ is normal. By theorem 4.3 we have that either $T$ is normal or $\dim M = 1$. Let us suppose that $\dim M = 1$ since otherwise we would be done.

Then there exist $\alpha,\, \beta \in \mathbb {C}$ such that $Dx + \langle {x,\,u}\rangle u = \alpha x$ and $D^{*}x + \langle {x,\,u}\rangle u = \beta x$. Hence

\[ x_n := \langle{x,e_n}\rangle \neq 0 \]

for every $n \in \mathbb {N}$. Now,

\[ (D-D^{*})x = (\alpha-\beta)x, \]

from where $\textrm {Im}\; (\lambda _n)$ is constant for every $n\in \mathbb {N}$. Thus, the spectrum of $D$ is contained in a parallel line to the real axis and by Ionascu's theorem (i$'$), it follows $T$ is normal.

If $T\mid _{M^{\perp }}$ is normal, we can argue equivalently to show that $T$ is normal.

Now, assume that neither $T\mid _M$ nor $T\mid _{M^{\perp }}$ are normal operators. If $u$ and $(D-D^{*})u$ are linearly dependent, then $u$ is an eigenvector for $(D-D^{*})$, associated with an eigenvalue $\lambda \in \mathbb {C}$. Since $\langle {u,\,e_n}\rangle \neq 0$ for every $n \in \mathbb {N}$, $\lambda _n - \overline {\lambda _n} = \lambda$ for every $n\in \mathbb {N},$ and hence $D-D^{*} = \lambda I$ and $T$ is normal. So, we may assume that $u$ and $(D-D^{*})u$ are linearly independent.

By assumption $T\mid _M$ and $T\mid _{M^{\perp }}$ are not normal, so there exist $x_0 \in M$ and $y_0 \in M^{\perp }$ such that

(4.6)\begin{equation} 0\neq [T,T^{*}]x_0 \in M \mbox{ and } 0\neq [T,T^{*}]y_0 \in M^{{\perp}}. \end{equation}

Note that in this case

(4.7)\begin{equation} [T,T^{*}] = (D-D^{*})u\otimes u + u \otimes (D-D^{*})u \end{equation}

and $u \notin M\cup M^{\perp }$ because of proposition 4.1.

If $\langle {x_0,\,(D-D^{*})u}\rangle = \langle {y_0,\,(D-D^{*})u}\rangle = 0$, it follows that $(D-D^{*})u \in M \cap M^{\perp },$ so $(D-D^{*})u = 0$ and this occurs if and only if $D = D^{*}$ since $\langle {u,\,e_n}\rangle \neq 0$ for every $n \in \mathbb {N}$. Hence, we can assume $\langle {x_0,\,(D-D^{*})u}\rangle \neq 0$.

In addition, $u \notin M$ so, by means of (4.6) and (4.7), $\langle {x_0,\,u}\rangle \neq 0$. By considering $P_M$ the orthogonal projection onto $M$ and $Q_M = I-P_M$, it follows that

\[ \langle{x_0,(D-D^{*})u}\rangle Q_M u ={-} \langle{x_0,u}\rangle Q_M (D-D^{*})u. \]

Moreover, since $Q_M$ commutes with $T$ and $T^{*}$ we have

\[ D Q_M = Q_M D+Q_M u\otimes u - u \otimes Q_M u \]

and

\[ D^{*}Q_M = Q_M D^{*}+ Q_M u\otimes u - u \otimes Q_Mu, \]

so $Q_M(D-D^{*}) = (D-D^{*})Q_M.$ Hence, $P_M$ and $Q_M$ belongs to the commutant ${\{{D-D^{*}}\}'}$. Moreover,

\[ Q_M u ={-} \frac{\langle{x_0,u}\rangle}{\langle{x_0,(D-D^{*})u}\rangle}(D-D^{*})Q_M u, \]

so $Q_M u$ is an eigenvector for $(D-D^{*})$.

Now, let $y_0 \in M^{\perp }$ considered in (4.6), that is, $0\neq [T,\,T^{*}]y_0 \in M^{\perp }$. If $\langle {y_0,\,(D-D^{*})u}\rangle = 0$ then $(D-D^{*})u \in M^{\perp },$ so

\[ (D-D^{*})P_M u = P_M (D-D^{*})u = 0, \]

and therefore $P_M u$ is an eigenvector associated with the eigenvalue $0$. If $\langle {y_0,\,(D-D^{*})u}\rangle \neq 0$, we can argue similarly to deduce that $P_M u$ is also a eigenvector for $(D-D^{*})$.

Now, recall that the eigenvalues of $(D-D^{*})$ are $(2i\textrm{Im}\, (\lambda _n))_n$ and for each $n \in \mathbb {N}$, the space of eigenvectors associated with $2i\textrm{Im}\, (\lambda _n)$ is given by

\[ M_n = \overline{\textrm{span}\;\{e_k : \textrm{Im}\; (\lambda_k) = \textrm{Im}\; (\lambda_n) \}}. \]

Note that the eigenspaces $M_n$ and $M_m$ are orthogonal if $n\neq m.$ Moreover, since $P_M u + Q_M u = u$ and $M_n \neq H$ for every $n \in \mathbb {N}$ ($D$ is not self-adjoint), it follows that both $P_M u$ and $Q_M u$ are eigenvectors associated with different eigenvalues.

Moreover, there exists a partition of the positive integers $\mathbb {N}$, that is, non-empty sets $N_1,\, N_2 \subset \mathbb {N}$ such that $N_1\cup N_2 = \mathbb {N}$ and $N_1\cap N_2 = \emptyset$, such that

\[ P_M u = \sum_{n \in N_1} \alpha_n e_n \qquad Q_M u = \sum_{n \in N_2} \alpha_n e_n. \]

Now,

\[ T P_M u = D P_M u + \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2} u \in M \]

since $M$ is invariant under $T$. Since $Q_M u \in M^{\perp }$, we have $\langle {T P_M u,\,Q_M u}\rangle = 0$. Now, since $D P_M u = \sum _{n\in N_1} \lambda _n \alpha _n e_n$ it follows that $\langle {D P_M u,\,Q_M u}\rangle = 0,$ so we deduce that $\langle {\left \lvert \left \lvert {P_M u}\right \rvert \right \rvert ^{2}u,\, Q_ M u}\rangle = 0,$ but

\[ \langle{\left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}u,Q_M u}\rangle = \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}\langle{u,Q_M u}\rangle = \left\lvert\left\lvert{P_M u}\right\rvert\right\rvert^{2}\left\lvert\left\lvert{Q_M u}\right\rvert\right\rvert^{2}, \]

so $P_M u = 0$ or $Q_M u = 0$, what contradicts lemma 2.3.