Relating eigenvectors of two self-adjoints operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:49:21Z http://mathoverflow.net/feeds/question/79160 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators Relating eigenvectors of two self-adjoints operators Bramiozo 2011-10-26T13:02:32Z 2012-02-13T14:22:12Z <p>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:</p> <p>$\mathbf{v} \Lambda \mathbf{v}^T = \mathbf{v}_1 \Lambda_1 \mathbf{v}^T_1 + \mathbf{v}_2 \Lambda_2 \mathbf{v}^T_2$</p> <p>The two parts can also be written as</p> <p>$\mathbf{v}_1 \Lambda_1 \mathbf{v}^T_1= DK_1D^T$ </p> <p>$\mathbf{v}_2 \Lambda_2 \mathbf{v}^T_2= DK_2D^T$ </p> <p>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.</p> <p>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. </p> http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/79162#79162 Answer by Bramiozo for Relating eigenvectors of two self-adjoints operators Bramiozo 2011-10-26T13:53:30Z 2011-10-26T13:53:30Z <p>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 $\mathbf{v}_2$. </p> http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/84893#84893 Answer by Fabien Besnard for Relating eigenvectors of two self-adjoints operators Fabien Besnard 2012-01-04T17:45:22Z 2012-01-04T17:45:22Z <p>Hi Bramiozo.</p> <p>What you write is a bit confusing. Here are some questions in order to understand better :</p> <p>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 $v_2$)</p> <p>2) "Suppose $K_1$ and $K_2" etc. : this is another question then ?</p> <p>3) Is$b$a vector ? Is$b_\perp$orthogonal to$b$?</p> <p>I guess your matrices are real. If yes, you should right symmetric instead of self-adjoint.</p> http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/85795#85795 Answer by Bramiozo for Relating eigenvectors of two self-adjoints operators Bramiozo 2012-01-16T09:47:21Z 2012-01-16T09:47:21Z <p>K1,2 <strong>are</strong> formed by the vector products bbT and b⊥bT⊥ respectively and b and b⊥ are perpendicular to each other. 1) No, they can be written as such, no need for proof there.</p> <p>So$D\textbf{b}\textbf{b}^TD^T$has eigenvectors unrelated to the eigenvectors of$D \textbf{b}\bot \textbf{b}^T_\bot D^T$?</p> http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/85810#85810 Answer by Fabien Besnard for Relating eigenvectors of two self-adjoints operators Fabien Besnard 2012-01-16T13:46:37Z 2012-01-16T13:46:37Z <p>So your question seems to be : what is the connection between the eigenvectors of$A_1=Dbb^TD^T$and$A_2=Db_\perp b_\perp^TD^T$?</p> <p>Well it's easy to find these eigenvectors. First case :$Db,Db_\perp$linearly independant.</p> <p>Then the eigenspaces of$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.</p>