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I really need some help with showing that the following matrices are similar. $P_{\omega}=(D+\omega L)^{-T}((1-\omega)D-\omega L)(D+\omega L)^{-1}((1-\omega)D-\omega L)^T$ is similar to $S_{\omega}=(D+\omega L)^{-1}((1-\omega)D-\omega L)\left[(D+\omega L)^{-1}((1-\omega)D-\omega L)\right]^T$, where $L$ is a strictly lower triangular matrix, $D$ is a diagonal matrix with positive diagonal elements and $\omega\in(0,2)$ is a constant. So, I need to find a non-singular matrix $Q$ such that $S_{\omega}=QP_{\omega}Q^{-1}$. I have tried a lot of different $Q$'s and none have worked.

I have been trying to show it for a long time now and I can't do it. It's an important part of a proof that I need to understand (it's the proof to Theorem 2.1, chapter 15.2, concerning the convergence of the SSOR method in Iterative Solution of Large Linear Systems by David Young).

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  • $\begingroup$ Should that be $[(D+\omega L)^{-1}]^T$ at the start of $P_\omega$? $\endgroup$ May 3, 2017 at 23:18
  • $\begingroup$ Yes, it should. $\endgroup$
    – Eva
    May 4, 2017 at 6:33

1 Answer 1

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The two matrices are not always similar. There may be other conditions that you've missed. For a counterexample, consider $\omega=1$, $$ \begin{align*} &D=\pmatrix{1\\ &2},\ L=\pmatrix{0&0\\ 1&0},\\ &A=D+\omega L=\pmatrix{1&0\\ 1&2},\ B=(1-\omega)D-\omega L=\pmatrix{0&0\\ -1&0},\\ &P_1=A^{-T}BA^{-1}B^T=\pmatrix{0&-\frac12\\ 0&\frac12},\ S_1=(A^{-1}B)(A^{-1}B)^T=\pmatrix{0&0\\ 0&\frac14}. \end{align*} $$ $P_1$ and $S_1$ have different spectra. They are not similar.

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  • $\begingroup$ I didn't state it in my question, but the matrices $D$ and $L$ come from a matrix $A=D+L+U$ that is Hermitian. $\endgroup$
    – Eva
    May 4, 2017 at 6:33
  • $\begingroup$ @Eva What is $U$? If $U=L^\ast$ then the counterexample here still works. $\endgroup$
    – user1551
    May 4, 2017 at 7:38
  • $\begingroup$ $U$ is the strictly upper triangular part of $A$ so then your counterexample works. I don't think that I have missed any more conditions so now I'm really confused. $\endgroup$
    – Eva
    May 4, 2017 at 8:07
  • $\begingroup$ @Eva If there aren't any additional conditions, perhaps you've misunderstood something and the author doesn't mean that the two matrices are similar? $\endgroup$
    – user1551
    May 4, 2017 at 8:32
  • $\begingroup$ It explicitly says that $P_{\omega}$ is similar to $S_{\omega}$ so I don't think that I have misunderstood. $\endgroup$
    – Eva
    May 4, 2017 at 9:02

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