Suppose $T$ is an operator on a Hilbert space $H$ such that $\sigma(T)=\sigma_{p}(T)$ (point spectrum of $T$), and for each $\alpha \in \sigma(T)$, the corresponding eigenspace ker$(\alpha I-T)$ is a reducing subspace for $T$. We are also given that $\sigma(T)\subset \mathbb{R}.$ Then it is to be shown that $H=\oplus_{\alpha \in \sigma(T)}\text{ker}(\alpha I -T)$, that is $H$ is spanned by the orthogonal eigenspaces of $T$
First, it can be shown that for $\alpha, \beta \in \sigma(T)(=\sigma_{p}(T)), \alpha \neq \beta$, the corresponding eigen spaces ker($\alpha I-T$) and ker($\beta I-T$) are orthogonal. For this it suffices to show that if $\lambda \in \sigma_{p}(T)$, then $\bar{\lambda}=\lambda$ (since $\sigma(T)\subset \mathbb{R}$) is in $\sigma_{p}(T^{*})$. i.e. $Tx=\lambda x$ implies $T^{*}x=\lambda x$.
Suppose $Tx=\lambda x$. Then each $y\in H$ can be written as $y=y_{1}+y_{2}$ where $y_{1}\in \text{ker}(\lambda I-T)$ and $y_{2} \in \text{ker}(\lambda I-T)^{\perp}$. Then $\langle T^{*}x,y\rangle=\langle T^{*}x,y_{1}+y_{2}\rangle=\langle x, Ty_{1}+Ty_{2}\rangle=\langle x, Ty_{1}\rangle$ for, since $T$ reduces ker($\lambda I-T$), $Ty_{2} \in \text{ker}(\lambda I-T)^{\perp}$ and $x \in \text{ker}(\lambda I-T)$.
Thus $\langle T^{*}x,y\rangle=\langle x, Ty_{1}\rangle=\langle x, \lambda y_{1}\rangle=\langle\lambda x, y_{1}\rangle=\langle\lambda x, y\rangle$.
This is true for all $y\in H$, hence $T^{*}x=\lambda x$.
This shows that eigenspaces corresponding to different eigenvalues are orthogonal.
Next we must show that these eigenspaces span $H$. Could someone help with this?
Let $\oplus_{\alpha \in \sigma(T)}\text{ker}(\alpha I -T)=M$, say, and suppose $M^{\perp}\neq \{0\}$. Then since each $M_{\alpha}=\text{ker}(\alpha I-T)$ is a reducing subspace for $T$, $T \restriction_{M^{\perp}}$ is a bounded operator on $M^{\perp}$. Also $\sigma(T \restriction_{M^{\perp}})\subset \sigma(T)$. Then $T \restriction_{M^{\perp}}$ has empty eigen spectrum. Does it also follow that $\sigma(T \restriction_{M^{\perp}})=\emptyset$? How?
If the above is true, then we have arrived at a contradiction, whence $M^{\perp}= \{0\}$.
If not, how else can it be shown that $H=\oplus_{\alpha \in \sigma(T)}\text{ker}(\alpha I -T)$?