Let $E$ be a real finite dimensional inner product space, and $\phi :E \to E$ be any map having $k < n$ distinct eigenvalues $\{\lambda_i\}$. Let
$$T = E_{\lambda_1} \oplus E_{\lambda_2} \oplus...\oplus E_{\lambda_k}, \quad where \quad E_{\lambda_i} = \{x \in E | \phi(x) = \lambda_i x\}, \quad i=1,...,k$$ .If $\phi$ is stable on $T$, then $\phi |_{T}$ is self-adjoint.
I have realised today that for any linear map $\phi: E \to E$, the eigenvectors associated with different eigenvalues are linearly independent, so if we consider subspace of $E$ that is the direct product of the eigenspaces of eigenvalues of $\phi$, since this space is generated by the eigenvectors of $\phi$, I thuoght $\phi$ might be a self-adjoint in this space provided that $\phi(x) \in T \quad \forall x \in T$, i.e it is stable on $T$.
Therefore, it the above theorem that I have created true ? If not, why ? and is there any additional condition(s) that makes it true ?
Consider $\phi:\mathbb{R}^n\rightarrow \mathbb{R}^n$ with eigenvalues $\lambda_1=1$ and $\lambda_2=2$ and corresponding eigenspaces $E_1=\langle\begin{pmatrix}0\\1\\0\\ \vdots \\0\end{pmatrix}\rangle$ and $E_2=\langle \begin{pmatrix}1\\1\\0\\ \vdots \\0\end{pmatrix}\rangle $, then $\phi|_{T}$ is representd by the matrix $$\begin{pmatrix}2&0\\1&1\end{pmatrix}$$ from which You can see $\phi|_{T}$ is not selfadjoint. If a linear mapping is selfadjoint eigenvectors to different eigenvalues have to be orthogonal as can be seen from $$\lambda_1(x_1,x_2)=(\phi x_1,x_2)=(x_1,\phi x_2)=\lambda_2(x_1,x_2)$$ where $\lambda_1\neq\lambda_2$ are eigenvalues with corresponding eigenvectors $x_1,x_2$.