Relating eigenvectors of two self-adjoints operators

Question

Suppose I have a self-adjoint operator $\mathbf{L}$ which I seperate in two parts which are themselves self-adjoint. I write this in terms of their eigenvalues/eigenvectors:

$\mathbf{v} \Lambda \mathbf{v}^T = \mathbf{v}_1 \Lambda_1 \mathbf{v}^T_1 + \mathbf{v}_2 \Lambda_2 \mathbf{v}^T_2$

The two parts can also be written as

$\mathbf{v}_1 \Lambda_1 \mathbf{v}^T_1= DK_1D^T$

$\mathbf{v}_2 \Lambda_2 \mathbf{v}^T_2= DK_2D^T$

with $K_1$ and $K_2$ both symmetric, $D$ is skew-symmetric. Suppose $K_{1,2}$ are formed by the vector products $\mathbf{b}\mathbf{b}^T$ and $\mathbf{b_\bot}\mathbf{b}^T_\bot$ respectively.

How do I connect the eigenvectors $\mathbf{v_1}$ to $\mathbf{v_2}$? My guess is that $\mathbf{v_1}(i)^T\mathbf{v_2}(i)=0, \quad \forall\\, i$, but I don't know how to proof it.

---

K1,2 are formed by the vector products bbT and b⊥bT⊥ respectively and b and b⊥ are perpendicular to each other. 1) No, they can be written as such, no need for proof there.

So $D\textbf{b}\textbf{b}^TD^T$ has eigenvectors unrelated to the eigenvectors of $D \textbf{b}\bot \textbf{b}^T_\bot D^T$ ?

Answer 1

So your question seems to be : what is the connection between the eigenvectors of $A_1=Dbb^TD^T$ and $A_2=Db_\perp b_\perp^TD^T$ ?

Well it's easy to find these eigenvectors. First case : $Db,Db_\perp$ linearly independant.

Then the eigenspaces of $A_1$ are ${\mathbb R}Db$, and $(Db)^\perp$, and similarly for $A_2$. Since $D$ is skew-symmetric, in particular it does not preserve orthogonality and there is no connection between the eigenvectors of $A_1$ and $A_2$. The second case is obvious.