## 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.

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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 2011 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 $\mathbf{v}_2$.

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 Because MO doesn't work like a forum (answers move up and down) you should add details like this as edits to the original question, rather than as an "answer" – Yemon Choi Jan 5 2012 at 4:10 Got it Choi, thanks :). – Bramiozo Jan 16 2012 at 9:19

Hi Bramiozo.

What you write is a bit confusing. Here are some questions in order to understand better :

1) When you say "the two parts can also be written", this is an hypothesis, is it ? (anyway without further hypothesis there is no connection between $v_1$ and $v_2$)

2) "Suppose $K_1$ and $K_2" etc. : this is another question then ? 3) Is$b$a vector ? Is$b_\perp$orthogonal to$b$? I guess your matrices are real. If yes, you should right symmetric instead of self-adjoint. - K1,2 are formed by the vector products bbT and b⊥bT⊥ respectively and b and b⊥ are perpendicular to each other. 1) No, they can be written as such, no need for proof there. So$D\textbf{b}\textbf{b}^TD^T$has eigenvectors unrelated to the eigenvectors of$D \textbf{b}\bot \textbf{b}^T_\bot D^T$? -  See my previous comment - could you not edit the original question instead? – Yemon Choi Jan 16 2012 at 10:07 Nope, something with a 60-days limit? – Bramiozo Jan 16 2012 at 14:26 I've merged your accounts. It would be best if you found a way to register your account. – S. Carnahan♦ Jan 16 2012 at 14:34 So your question seems to be : what is the connection between the eigenvectors of$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$. The second case is obvious. -  Thanks Fabien. I was wondering, suppose we reverse it and state that the set of eigenvectors$\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)$. (Also suppose that the eigenvectors are normalised.) – Bramiozo Aug 7 at 15:28 What kind off requirement would be in place for$\mathbf{R}_1$and$\mathbf{R}_2\$ for this to be true? – Bramiozo Aug 8 at 9:21