Orthogonal family and linear operator Let $E$ $n$-dimensional space with $\langle \cdot, \cdot \rangle$ inner product. We say that $\mathcal{F} \subseteq \{\mathcal{B}: \mathcal{B}$ is a orthogonal basis in $E\}$ is a orthogonal family. 
Given $T: E \to E$ isomorphism linear. Is there exist a orthogonal family such that $T(B) \in \mathcal{F}$  $\forall B \in \mathcal{F}$?
When $T$ is self-adjoint, the Spectral Theorem give us a positive answer. In general, it happens when $T$ is a normal operator.
Hints or solutions are greatly appreciated.
 A: Just use the orthogonal basis of The Singular Value Theorem, and show by induction that it's a orthogonal family. 
proof: 
 By Singular Value Theorem, there exists a orthonormal basis $\{u_{1},...,u_{n}\}$ and $\{v_{1},...,v_{n}\}$ and scalars $\{\lambda_{1},...,\lambda_{r}\}$ where $r =$ rank $T$ with $\lambda_{i} >0$  such that $A(u_{i}) = \lambda_{i} v_{i}$ for $i \in \{1,...,r\}$ and $A(u_{j}) = 0$ if $j >r$. But $T$ is linear isomorphism, so $r=n$.
Consider $\mathcal{F} = \{T^{k}(\mathcal{B})$: for $k \in \mathbb{N}$ where $\mathcal{B} =\{u_{1},...,u_{n}\}\}$ and $T^{k}(\mathcal{B}) = \{T^{k}(u_{1}),...,T^{k}(u_{n})\}$ for $k \geq 0$.
Claim: $\mathcal{F}$ is a orthogonal family.
We'll argue by induction. For $k=0,1$ is valid, because $\{u_{1},...,u_{n}\}$ and $\{v_{1},...,v_{n}\}$ is a orthonormal basis of $V$. Let's do the case $k=2$, note that $T(u_{i}) = \lambda_{i}v_{i}$ for $i \in \{1,...,n\}$. Using that $\{u_{1},...,u_{n}\}$ is a basis, for all $i \in \{1,...,n\}$ there exists $\alpha^{i}_{1},...,\alpha^{i}_{n} \in \mathbb{R}$ such that  $v_{i}= \sum_{t=1}^{n} \alpha^{i}_{t}u_{t}$.  So, $T(u_{i}) = \lambda_{i}v_{i} =  \lambda_{i}\sum_{t=1}^{n} \alpha^{i}_{t}u_{t}$, and $T^{2}(u_{i}) = \lambda_{i}\sum_{t=1}^{n} \alpha^{i}_{t}T(u_{t})$,  
$\langle T^{2}(u_{i}), T^{2}(u_{j}) \rangle = \langle \lambda_{i}\sum_{t=1}^{n} \alpha^{i}_{t}T(u_{t}), \lambda_{j}\sum_{s=1}^{n} \alpha^{j}_{s}T(u_{s}) \rangle $.
Using that $\{v_{1},...,v_{n}\}$ is a orthonormal basis, and $T$ is linear isomorphism, we have that $\{T(u_{1}),...,T(u_{n})\}$ is a orthogonal basis, so
$= \lambda_{i}\lambda_{j} (\sum_{t=1}^{n} \alpha^{i}_{t} \alpha^{j}_{t} \langle T(u_{t}),T(u_{t})\rangle)$ 
Taking $m_{1}= \min\{\langle T(u_{t}), T(u_{t})\rangle: t\in\{1,...,n\}\}$ and $M_{1}= \max\{\langle T(u_{t}), T(u_{t})\rangle: t\in\{1,...,n\}\}$, and using that $\lambda_{s} > 0$ for all $s \in \{1,...,n\}$, we have that
$ (\lambda_{i}\lambda_{j})m_{1} (\sum_{t=1}^{n}\alpha^{i}_{t} \alpha^{j}_{t}) \leq \langle T^{2}(u_{i}), T^{2}(u_{j}) \rangle \leq (\lambda_{i}\lambda_{j})M_{1} (\sum_{t=1}^{n}\alpha^{i}_{t} \alpha^{j}_{t})$.
But $0= \langle v_{i},v_{j} \rangle = \langle \sum_{t=1}^{n} \alpha^{i}_{t}u_{t}, \sum_{t=1}^{n} \alpha^{i}_{t}u_{t}  \rangle = \sum_{t=1}^{n}\alpha^{i}_{t}\alpha^{j}_{t}$. So,  $\langle T^{2}(u_{i}), T^{2}(u_{j}) \rangle = 0$. We showed that $T^{2}(\mathcal{B})$ is orthogonal basis.
Suppose that the result holds to $k-1$, i.e., $T^{k-1}(\mathcal{B}) = \{T^{k-1}(u_{1}),...,T^{k-1}(u_{n})\}$ is a orthogonal basis. We have that  $T^{k-1}(u_{i}) = \lambda_{i}\sum_{t=1}^{n} \alpha^{i}_{t}T^{k-2}(u_{t})$, so $T^{k}(u_{i}) = \lambda_{i} \sum_{t=1}^{n} \alpha^{i}_{t}T^{k-1}(u_{t})$ for all $i \in \{1,...,n\}$. 
$\langle T^{k}(u_{i}), T^{k}(u_{j}) \rangle = \langle \lambda_{i}\sum_{t=1}^{n} \alpha^{i}_{t}T^{k-1}(u_{t}), \lambda_{j}\sum_{s=1}^{n} \alpha^{j}_{s}T^{k-1}(u_{s}) \rangle $.
Using that $\{T^{k-1}(u_{1}),...,T^{k-1}(u_{n})\}$ is a orthogonal basis, 
$= \lambda_{i}\lambda_{j} (\sum_{t=1}^{n} \alpha^{i}_{t} \alpha^{j}_{t} \langle T^{k-1}(u_{t}),T^{k-1}(u_{t})\rangle)$ 
Taking $m_{k-1}= \min\{\langle T^{k-1}(u_{t}), T^{k-1}(u_{t})\rangle: t\in\{1,...,n\}\}$, $M_{k-1}= \max\{\langle T^{k-1}(u_{t}), T^{k-1}(u_{t})\rangle: t\in\{1,...,n\}\}$, and using that $\lambda_{s} > 0$ for all $s \in \{1,...,n\}$, we have that
$ (\lambda_{i}\lambda_{j})m_{k-1} (\sum_{t=1}^{n}\alpha^{i}_{t} \alpha^{j}_{t}) \leq \langle T^{k}(u_{i}), T^{k}(u_{j}) \rangle \leq (\lambda_{i}\lambda_{j})M_{k-1} (\sum_{t=1}^{n}\alpha^{i}_{t} \alpha^{j}_{t})$.
But $0= \langle v_{i},v_{j} \rangle = \langle \sum_{t=1}^{n} \alpha^{i}_{t}u_{t}, \sum_{t=1}^{n} \alpha^{i}_{t}u_{t}  \rangle = \sum_{t=1}^{n}\alpha^{i}_{t}\alpha^{j}_{t}$. So,  $\langle T^{k}(u_{i}), T^{k}(u_{j}) \rangle = 0$. We showed that $T^{k}(\mathcal{B})$ is orthogonal basis. This ends the induction step, and completes the proof.

But we don't know that this $\mathcal{F}$ is finite. 
If $T$ is a $k$-normal operator, we have that $T^{k}$ a is normal operator, so $T^{k}$ has a orthonormal basis of eigenvalues, $\mathcal{C}=\{v_{1},...,v_{n}\}$. If $\mathcal{C} \in \mathcal{F}$, note that $T^{k}(\mathcal{C}) = \{\lambda_{1}v_{1},...,\lambda_{n}v_{n}\}$, let $\lambda \in \mathbb{R}^{n}$ given by $\lambda = (\lambda_{1},...,\lambda_{n})$, we denote $T^{k}(\mathcal{C})$ by $\lambda C$. 
Now, $T^{k+1}(C) = T(T^{k}(C)) = \{\lambda_{1}T(v_{1}),...,\lambda_{n}T(v_{n})\} = \lambda T(C)$. So, for all $s \in \mathbb{N}$, we have   $T^{s}(\mathcal{C}) = \lambda^{r}T^{l}(\mathcal{C})$ for some $r \in \mathbb{N}$ and $0 \leq l \leq k$ where $\lambda^{r} =(\lambda_{1}^{r},...,\lambda_{n}^{r})$ . We can consider $\mathcal{F}^{*}=\{T^{s}(\mathcal{C}):$ for $s \in \mathbb{N} \}$ that is a orthogonal family , we´ll have a "finite" family.  
