I am looking at a matrix $$\mathbf{M} = \left(\mathbf{I}+k\theta\mathbf{B}^{-1}\mathbf{A}\right)^{-1}\left(\mathbf{I}-k(1-\theta)\mathbf{B}^{-1}\mathbf{A}\right) $$ where $\mathbf{I}$ is the identity matrix, and all matrices are square. It is given that the eigenvalues of $\mathbf{B}^{-1}\mathbf{A}$ are all non-negative. In a text I'm reading the writer writes that we can relate the eigenvalues $\lambda_i$ of $\mathbf{M}$ and the eigenvalues $\eta_i$ of $\mathbf{B}^{-1}\mathbf{A}$ in the following way

$$\lambda_i = \frac{1-k(1-\theta)\eta_i }{1+k\theta\eta_i}. $$

My question is, what allows us to express the eigenvalues of $\mathbf{M}$ ($\lambda_i$) in terms of the eigenvalues of $\mathbf{B}^{-1}\mathbf{A}$ ($\eta_i$) in that way?

The writer does not state why this is possible, and there is no reference any theorem or anything. Unfortunately, it is not a publicized book, so I cannot link to it. The context of this problem is a finite element solver, that uses a finite difference time-discretization to solve the heat equation.

  • 2
    $\begingroup$ If $Cx = \lambda x,$ then $C^kx = \lambda^k x,$ for positive integer $k,$ so for any polynomial $p,$ $p(C)x = p(\lambda)x.$ Also, the eigenvalue of $p(C^{-1})$ is simply $p(\lambda^{-1}).$ $\endgroup$ – Ehsan M. Kermani Jan 26 '15 at 21:17
  • $\begingroup$ @EhsanM.Kermani I don't quite follow. Where is the polynomial? $\endgroup$ – Eff Jan 26 '15 at 21:30
  • $\begingroup$ @Eff $I+k \theta B^{-1} A$ is a polynomial in $B^{-1} A$. So is $I-k(1-\theta) B^{-1} A$. The whole expression isn't a polynomial, however, so I don't think this quite works by itself. A different approach is to notice that if $k \theta$ is small enough, $(I+k \theta B^{-1} A)^{-1}$ can be written as a geometric series. When you consider any partial sum of this series, you do have a polynomial, so you get the result when you take the limit. If $k \theta$ is instead large, you can take the series the other way, which still gives the result. $\endgroup$ – Ian Jan 26 '15 at 21:48

Hint: let $\mathbf {S := (B)^{-1}A}$, then the eigenvalues of $\mathbf{I}+k\theta \mathbf S$ are the roots of the following polynomial:

\begin{align} \lambda(\mathbf I + k\theta \mathbf S) &= \mathrm{det} (\lambda \mathbf I - (\mathbf I + k\theta \mathbf S)),\\ &= \mathrm{det} ([\lambda-1] \mathbf I - k\theta \mathbf S). \end{align}

Let $\eta := (\lambda -1)/(k\theta)$, the characteristic polynomial of the equation above becomes \begin{align} (k\theta)^n\mathrm{det} (\eta \mathbf I -\mathbf S) = 0, \end{align}

where $n$ is the order of $\mathbf S$. Hence $\eta$ are the eigenvalues of $\mathbf {S := (B)^{-1}A}$ and the eigenvalues of $\mathbf I + k\theta \mathbf S$ are given by

$\lambda_i = k\theta\eta_i +1$.

The rest of the proof follows from properties of determinant.

Can you continue the proof?

I hope this helps.



It seems to me that this is pretty much just bash-it-out linear algebra using definitions. I'll sketch the reasoning and let you fill in the deets: First, let $C=B^{-1}A$ for ease of notation. Now let's let $\mathbf v$ be an eigenvector of $C$ with eigenvalue $\eta$. First, $(I+k\theta C)\mathbf v = (1+k\theta\eta)\mathbf v$; multiply by the inverse of the matrix on the left and divide by the scalar on the right, and we see that $(I+k\theta C)^{-1}\mathbf v = \frac{1}{1+k\theta\eta}\mathbf v$. Since, by even more straightforward reasoning, $(I-k(1-\theta)C)\mathbf v = (1-k(1-\theta)\eta)\mathbf v$, we get the formula we want for the eigenvalue of $M$ (having, in fact, the same eigenvector).


Suppose that $B^{-1}Av = \eta v$. Then, first \begin{align} (I-k(1-\theta)B^{-1}A)v & = v - k(1-\theta)B^{-1}Av\\ & = v-k(1-\theta)\eta v \\ & = (1-k(1-\theta))\eta v. \end{align} On the other hand, similarly, you can prove that $$(I+k \theta B^{-1}A) v = (1+k\theta\eta)v.$$ This, in turn, implies $$(1+k\theta\eta)^{-1}v = (I+k \theta B^{-1}A)^{-1}v,$$ where we multiplied both sides by $(1+k\theta\eta)^{-1}(I+k \theta B^{-1}A)^{-1}$. You can combine this two facts and the thesis will follow.


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