User bramiozo - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T01:12:49Z http://mathoverflow.net/feeds/user/18798 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/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/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/85810#85810 Comment by Bramiozo Bramiozo 2012-08-08T09:21:09Z 2012-08-08T09:21:09Z What kind off requirement would be in place for $\mathbf{R}_1$ and $\mathbf{R}_2$ for this to be true? http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/85810#85810 Comment by Bramiozo Bramiozo 2012-08-07T15:28:33Z 2012-08-07T15:28:33Z 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.) http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/85795#85795 Comment by Bramiozo Bramiozo 2012-01-16T14:26:08Z 2012-01-16T14:26:08Z Nope, something with a 60-days limit? http://mathoverflow.net/questions/79160/relating-eigenvectors-of-two-self-adjoints-operators/79162#79162 Comment by Bramiozo Bramiozo 2012-01-16T09:19:14Z 2012-01-16T09:19:14Z Got it Choi, thanks :).