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Consider some $n\times n$ matrix $\mathbf{A}$; denote its dominant (a.k.a. leading) eigenvalue $\lambda_{A,d}$. Consider another matrix $${B} = \left[ \begin{array}[cc] \ \mathbf{A} & \vec{c} \\ \vec{r}^{\top} & k \end{array}\right]$$

where $\vec{c}$ and $\vec{r} $ are column vectors from $\mathbb{R}^N$ and $k\in\mathbb{R}$. Denote the dominant eigenvalue of B by $\lambda_{B,d}$.

Does it then follow that $\text{Re}(\lambda_{A,d})\leq \text{Re}(\lambda_{B,d})$? I ran some simulations that seem to suggest this is true, but I am having trouble proving this.

Note: I've been considering the case where $\text{Re}(\lambda_{A,d})>0$. I don't think it matters, but if it does and/or makes your life easier, then please do feel free to impose that.

Thanks so much for taking the time to at least read this!

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$$ \begin{pmatrix} 1 & -1\\ 2 & -1 \end{pmatrix} $$ The eigenvalues are $i$ and $-i$, whose real parts are zero, that is less than 1.

Notice: what you said is true for positive definite hermitian matrices

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  • $\begingroup$ Hi Exodd! Thanks for getting back to me so quickly on this, I really appreciate it. I'm not quite sure I understand the first part of your explanation (I assumed A was just an arbitrarily-chosen, real valued matrix). Also, is what I said necessarily true for matrices that are not positive definite and/or not hermitian? Thanks again! $\endgroup$ Sep 13 '20 at 0:16
  • $\begingroup$ @cluelessmathematician The matrix from the beginning of the answer is a matrix $B$, the matrix constituted by the left upper cell of $B$ is the matrix $A$. Next are presented eigenvalues of $B$ and $A$. $\endgroup$ Sep 23 '20 at 11:56
  • $\begingroup$ Hi Alex, thanks for responding to this. Unless I am missing something, I still don't see how this helps me prove what I am asking, since it's just an example. $\endgroup$ Sep 23 '20 at 20:49
  • $\begingroup$ @cluelessmathematician It is a COUNTER-example. It proves that what you are trying to prove is wrong $\endgroup$
    – Exodd
    Sep 24 '20 at 7:47
  • $\begingroup$ Yikes, you're right. Sorry about that, Exodd! $\endgroup$ Sep 24 '20 at 20:03

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