# Relating eigenvectors of two self-adjoints operators

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 $D \textbf{b}\bot \textbf{b}^T_\bot D^T$ ?

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I am sorry but I got a bit confused by your notation. This may be standard but I have not encountered it. Could you explain a bit more what is what? More precisely: what is your space? What objects are there ($\Lambda$...)? What is $v_1(i)$? Thanks. – András Bátkai Oct 26 '11 at 13:31
Hi András, thanks for reading :) . $\Lambda$ is a diagonal matrix filled with the eigenvalues, $\mathbf{v}$ is a matrix which columns are formed by the eigenvectors. $\mathbf{v}(i)$ is the $i^{th}$ eigenvector. My main question is basically what $\mathbf{b}\mathbf{b}^T$ versus $\mathbf{b_\bot}\mathbf{b}^T_\bot$ means for the difference between $\mathbf{v}_1$ and $\mathbf{v}_2$. – Bramiozo Oct 26 '11 at 13:53

## 1 Answer

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.

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Thanks Fabien. I was wondering, suppose we reverse it and state that the set of eigenvectors $\mathbf{R}$ is the summation of two distinct parts, say $\mathbf{R}=\mathbf{R}_1+\mathbf{R}_2$ where each column represents an eigenvector. Now I want that $\mathbf{R}_1(i)\cdot\mathbf{R}_2(i)=0,\, \forall i$ where $i$ indicates a specific eigenvector $\mathbf{R}(i)$ and of course $\mathbf{R}(i)=\mathbf{R}_1(i)+\mathbf{R}_2(i)$. (Also suppose that the eigenvectors are normalised.) – Bramiozo Aug 7 '12 at 15:28
What kind off requirement would be in place for $\mathbf{R}_1$ and $\mathbf{R}_2$ for this to be true? – Bramiozo Aug 8 '12 at 9:21

## protected by Community♦Aug 19 '14 at 23:58

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