invariant subspaces for operators on real Hilbert spaces First, let me ask a general question:  Is there a (good) textbook that discusses the theory of bounded linear operators acting on real Hilbert spaces?  Something that is heavy on spectral theory would be nice.
As for my other questions, consider the following.  Suppose $\mathcal{H}$ is a real Hilbert space and $T\in\mathcal{B}(\mathcal{H})$ is a bounded/continuous linear operator.  If $T$ has eigenvalues then the complexification $T_\mathbb{C}$ has corresponding 1-dimensional invariant subspaces.
Question #1.  Must $T$ have any invariant subspaces in the above case?
I vaguely remember something about the eigenvalues of $T_\mathbb{C}$ coming in conjugate pairs, and the corresponding 2-dimensional invariant subspace under $T_\mathbb{C}$ is also an invariant subspace under $T$. However, I can't imagine why this should be true and so I think maybe I'm remembering incorrectly.
Now let us consider the special case where $T$ is a normal operator (still acting on a real Hilbert space, not a complex one).  Then the eigenvectors for $T_\mathbb{C}$ corresponding to distinct eigenvalues are orthogonal.  This gives us orthogonal invariant subspaces under $T_\mathbb{C}$, but not necessarily $T$.
Suppose now that $(\lambda,\overline{\lambda})$ and $(\mu,\overline{\mu})$ are distinct pairs of conjugate eigenvalues for $T_\mathbb{C}$.  (We are still assuming that $T$ is normal.)
Question #2.  In this case, can we find orthogonal invariant subspaces for $T$?
Thanks!
EDIT:
So, Q's #1&2 are both answered.  But I'm still interested in a text discussing bounded linear operators on (infinite-dimensional) real Hilbert spaces.  There are lots of small things which it would help to know.  For example, does the equivalence
$$T\text{ is normal }\;\;\;\Leftrightarrow\;\;\;\text{ if }M\text{ is a }T\text{-invariant subspace then so is }M^\perp$$
hold even when $T$ is a bounded linear operator on a real Hilbert space?
Anyway, thanks for the help!
 A: For #1, $T_{\mathbb{C}}$ acts on the complexification $\mathcal{H}_\mathbb{C}$ by $T_{\mathbb{C}}(x+iy) = Tx + i Ty$, where $x,y \in \mathcal{H}$.  So if $\lambda = a+bi$ is an eigenvalue of $T_{\mathbb{C}}$ with eigenvector $u = x+iy$, we have
$$Tx + iTy = T_{\mathbb{C}} u = \lambda u = (a+bi)(x+iy) = (ax-by) + i (ay+bx).$$
Comparing real and imaginary parts we have $Tx = ax-by$, $Ty = ay+bx$.  Hence the subspace spanned by $x,y$ is invariant for $T$.
A: Okay, I believe I figured out Q#2 as well.  For reference, here is my answer to my own question.
It is known (I think) that eigenvectors for complex eigenvalues come in conjugate pairs for operators on real vector spaces.  It is also known that distinct eigenvectors for normal operators are orthogonal.
Thus, if $T$ has an eigenvalue $\lambda$, then $T_\mathbb{C}$ has a pair of eigenvectors $u\oplus iv$ and $u\oplus -iv$, possibly identical.  If $\mu$ is another eigenvalue distinct from both $\lambda$ and $\overline{\lambda}$, then $T_\mathbb{C}$ has two more eigenvectors $x\oplus iy$ and $x\oplus -iy$ (again, possibly identical), which are orthogonal to the previous two eigenvectors so long as $T$ is normal.  Hence we have the following
$$
0=\langle x\oplus iy,u\oplus iv\rangle =(\langle x,u\rangle +\langle y,v\rangle )+i(\langle x,v\rangle -\langle y,u\rangle )
$$
and
$$
0=\langle x\oplus -iy,u\oplus iv\rangle =(\langle x,u\rangle -\langle y,v\rangle )+i(\langle x,v\rangle +\langle y,u\rangle )
$$
Thus $\langle x,u\rangle =\langle y,v\rangle=-\langle x,u\rangle$ and similarly $\langle x,v\rangle=-\langle y,u\rangle=-\langle x,v\rangle$.  It follows that
$$\langle x,u\rangle =\langle y,v\rangle=\langle x,v\rangle=\langle y,u\rangle=0$$
Thus the subspaces $\text{span}\{u,v\}$ and $\text{span}\{x,y\}$ are orthogonal.  Note that these subspaces might be 1-dimensional or 2-dimensional, but either way they are both orthogonal, and, by the answer to Q#1 above, $T$-invariant.
