Can we find the scalar $\lambda$ in the following quadratic vector-matrix form?

Question

Assume $\lambda$ and $\rho$ are scalers, $b \in \mathbb{R}^d$, $I$ is the identity $d \times d$ matrix and $D$ is a $d \times d$ diagonal positive semidefinite matrix. Is there a way to calculate $\lambda$ from the following quadratic equation?

$$\frac 14 b^T(\lambda I - D)^{-2}b = \rho^2$$

Answer 1

$ \def\d{\delta} \def\l{\lambda} \def\o{{\tt1}} \def\a{\approx} \def\p{d} \def\LR#1{\left(#1\right)} \def\trace#1{\operatorname{trace}\left(#1\right)} \def\qiq{\quad\implies\quad} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\c#1{\color{red}{#1}} \def\CLR#1{\c{\LR{#1}}} $For typing convenience, define the matrix $$\eqalign{ M &= \LR{\l I-D}^{-1} \qiq \c{dM=-M^2\,d\l} \\ }$$ Set up the nonlinear function and calculate its derivative $$\eqalign{ f(\l) &= 4\rho^2 - b^TM^2\,b \\ df &= -b^T\LR{2M\,\c{dM}}b \\ &= -b^T\LR{2M\CLR{-M^2\,d\l}}b \\ &= 2\,b^TM^3b\:d\l \\ f'=\grad{f}{\l} &= 2\,b^TM^3\,b \\ }$$ Now apply Newton's method to find the roots of $f(\l)=0,\,$ i.e. $$\eqalign{ \l_+ &= \l - \frac{f}{f'} \\ &= \l - \frac{4\rho^2 - b^TM^2\,b}{2\,b^TM^3\,b} \\ &= \l + \frac{b^T\LR{\l I-D}^{-2}b\:-\:4\rho^2}{2\,b^T\LR{\l I-D}^{-3}b} \\ }$$ This function isn't particularly tricky, so the iterations should converge rapidly. However, since there are lots of roots, different initial guesses will converge to different solutions.

Approximate Solutions

Replace $D$ by its isotropic approximation $$D\a\d I\qiq \d=\frac{\trace D}{\trace I}$$ This reduces the matrix equation to a scalar equation which is easily solved $$\eqalign{ 4\rho^2 &\a \|b\|^2\:\LR{\l-\d}^{-2} \\ \l &\a \d + \frac{\|b\|}{2\rho} \\ }$$ One Newton iteration yields a better closed-form approximation $$\eqalign{ \mu &= \d + \frac{\|b\|}{2\rho}, \qquad M = \LR{\mu I-D}^{-1} \\ \l &\a \frac{b^T\!\LR{M^2+2\mu M^3}b\:-\:4\rho^2}{2\,b^TM^3b} \\ }$$

Answer 2

Because $(\lambda I -D)^{-2}(\lambda I -D)^{-1} = I $

and $(\lambda I -D)(\lambda I -D)^{-1} = I $,

Then $(\lambda I -D)^{-2} = (\lambda I -D) $

The above equation becomes $b^T(\lambda I -D)b=(2 \rho)^2$

Then,

$\Sigma (\lambda - d_i)b_i^2 = (2 \rho)^2$

$\Sigma \lambda b_i^2 - \Sigma d_i b_i^2 = (2 \rho)^2$

$\Sigma \lambda b_i^2 = (2 \rho)^2 + \Sigma d_i b_i^2$

$\lambda = \frac{(2 \rho)^2 + \Sigma d_i b_i^2}{\Sigma b_i^2}$

Answer 3

You have

$ b^T (\lambda I - D)^{-2} b = 4 \rho^2 $

The left hand side expands as follows

$ b^T (\lambda I - D)^{-2} b = \displaystyle \sum_{i = 1 }^d \dfrac{b_i^2}{ (\lambda - D_{ii})^2} $

Define the $2d$-th polynomial

$ f( \lambda) = \displaystyle \prod_{i=1}^d (\lambda - D_{ii})^2 $

And the $2(d-1)$-th polynomials

$ f_j (\lambda ) = \displaystyle \prod_{i=1, i \ne j}^d (\lambda - D_{ii})^2 $

Then

$\displaystyle \sum_{j = 1}^d b_j^2 f_j(\lambda) = 4 \rho^2 f(\lambda) $

And this a polynomial equation of degree $2 d$, whose roots are the required $\lambda$'s.