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Let $\Omega_{\, m\times m}$ be a real square positive definite symmetric matrix, $u_{m\times 1}$ is a vector, $I_{m\times m}$ is the identity matrix.

Let $x$ be a solution of a matrix equation

$$ u^T(\Omega-x I)^{-1}u=1\tag{1}$$

I have hypothesised that the solution can be found as an eigenvalue problem of a matrix

$$\Lambda := \Omega - uu^T$$

i.e. $x$ satisfies the equation

$$\det\left(\Lambda-x I\right)=0.\tag{2}$$

Is this correct? Does the relation $(2)$ for $x$ indeed follow from $(1)$?

Motivation: I have managed to show the statement for $m=2$ and $m=3$ by diagonalisation of $\Omega$ and then concluded it can hold for higher $m$. It goes as follows... Let $\Omega=ODO^T$ be an orthonormal decomposition ($OO^T=O^TO=I$) with $D=\operatorname{diag}(\lambda_1,\lambda_2,\dots,\lambda_m)$ with $\lambda_1\leq\lambda_2\leq\dots\leq\lambda_m$. Let me denote $u=Ov$, then $(1)$ takes form

$$v^T(D-x I)^{-1}v=1\tag{3}$$

or

$$\sum_{k=1}^m \frac{v_k^2}{\lambda_k-x}=1\tag{4}.$$

For $m=2$ this becomes

$$v_1^2(\lambda_2-x)+v_2^2(\lambda_1-x)=\lambda_1\lambda_2 - x(\lambda_1+\lambda_2)+x^2,$$

which can be rearranged to

$$x^2 - x(\lambda_1+\lambda_2-(v_1^2+v_2^2))-\lambda_1\lambda_2\left(\frac{v_1^2}{\lambda_1}+\frac{v_1^2}{\lambda_1}\right)=0\tag{5}.$$

The relation $(2)$ becomes under decomposition

$$\det\left(D-vv^T-xI\right)=0\tag{6}$$

which is equal to

$$x^2-x\operatorname{Tr}\left(D-vv^T\right)+\det\left(D-vv^T\right)=0$$

which is, after some easy algebraic manipulations, equal to $(5)$. The similar proof holds for $m=3$. Is there, however, a general approach?

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Yes, it's correct since $$\begin{eqnarray} (\Lambda-xI)(\Omega-xI)^{-1}u&=&(\Omega-uu^T-xI)(\Omega-xI)^{-1}u\\ &=&(\Omega-xI)(\Omega-xI)^{-1}u-uu^T(\Omega-xI)^{-1}u\\ &=&u-u\cdot\left(u^T(\Omega-xI)^{-1}u\right)=u-u=0. \end{eqnarray}$$ Hence $$ 0\ne (\Omega-xI)^{-1}u\in \ker (\Lambda-xI) $$ and $\det (\Lambda-xI)=0$ follows.


Addendum

If $\det(\Lambda -xI)=0$ and $(\Omega-xI)^{-1}$ exists, then $u^T(\Omega-xI)^{-1}u=1$.

Proof: Let $v\ne 0$ be a vector such that $$ \Lambda v=xv,\quad\ \ \Omega v -(uu^T)v=xv. $$ This gives $$ (\Omega-xI)v=(u^Tv)u $$ and $$ v=(u^Tv)(\Omega-xI)^{-1}u. $$ In particular, $u^Tv\ne 0$. By left-multiplying $u^T$ on both sides, it follows $$ u^Tv = (u^Tv)\cdot u^T(\Omega-xI)^{-1}u. $$ Since $u^Tv\ne 0$, we have $$ 1=u^T(\Omega-xI)^{-1}u. $$

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  • $\begingroup$ That's great! Thank you very much for such a simple answer. However, I cannot see clerly how the other implications follows? I.e. from $\det(\Lambda-xI)=0$ to the original equation. $\endgroup$
    – Machinato
    Jan 20, 2019 at 20:09
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    $\begingroup$ @Machinato I've added some part explaining that. I hope this will help. $\endgroup$ Jan 20, 2019 at 21:16

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