# A matrix equation equivalence

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?

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.
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 $$1=u^T(\Omega-xI)^{-1}u.$$
• 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. Jan 20, 2019 at 20:09