Lorelei Stanton

2022-03-23

Let $A\left(\alpha \right)=(\alpha I+(1-\alpha )B)-1$ be the inverse of a convex combination of the n-by-n identity matrix I and a positive semidefinite symmetric matrix B, where the combination is given by $\alpha \in (0,1]$ . The matrix B is not invertible. For some n-dimensional vector c, I want to find $\alpha$ such that

${c}^{T}A\left(\alpha \right)(I-B)A\left(\alpha \right)c=0$

Is there a closed-form solution for$\alpha$ ?

Is there a closed-form solution for

paganizaxpo3

Beginner2022-03-24Added 8 answers

Step 1

As B is positive semidefinite symmetric matrix so we can write$B={P}^{-1}DP$ where P is a orthogonal matrix $(P\{-1\}=PT)$ and D is a diagonal matrix (${d}_{i}>0$ for all $i=1,\cdots ,n$ ). We have

$A\left(\alpha \right)={(\alpha I+(1-\alpha )B)}^{-1}$

$={(alha{P}^{-1}P+(1-\alpha ){P}^{-1}DP)}^{-1}$

$={\left({P}^{-1}(\alpha I+(1-\alpha )D)P\right)}^{-1}$

$={P}^{-1}{(\alpha I+(1-\alpha )D)}^{-1}P$

Hence

${c}^{T}A\left(\alpha \right)(I-B)A\left(\alpha \right)c={c}^{T}({P}^{-1}(\alpha I+(1-\alpha ){D}^{-1}P)\left({P}^{-1}(I-D)P\right)({P}^{-1}(\alpha I+(1-\alpha ){D}^{-1}P)c$

$={\left(Pc\right)}^{T}{(\alpha I+(1-\alpha )D)}^{-1}(I-D){(\alpha I+(1-\alpha ))}^{-1}\left(Pc\right)$

$={\left(Pc\right)}^{T}Q\left(Pc\right)$

$={e}^{T}Qe$

where Q is a diagonal matrix with the values

${q}_{i}=\frac{1-{d}_{i}}{{(\alpha +(1-\alpha ){d}_{i})}^{2}}\mathrm{\forall}i=1,\cdots ,n$

and e is a vector defined by$e=Pc$

Then

${c}^{T}A\left(\alpha \right)(I-B)A\left(\alpha \right)c=0\iff {e}^{T}Q\left(\alpha \right)e=0\iff$

$\sum _{i=1}^{n}\frac{(1-{d}_{i}){e}_{i}^{2}}{{(\alpha +(1-\alpha ){d}_{i})}^{2}}=0$

or

1)$\sum _{i=1}^{n}\frac{(1-{d}_{i}){e}_{i}^{2}}{{((1-{d}_{i})\alpha +{d}_{i})}^{2}}=0$

The equation (1) has real solution if and only if$\left\{(1-{d}_{i})\right\}}_{i=1,\cdots ,n$ don't have the same sign ($I-B$ is not positive definite or negative definite), in other words, there exists k such that

$1<k<n$

and$0<{d}_{1}\le {d}_{2}\le \cdots \le {d}_{k}<1<{d}_{k+1}\le \cdots \le {d}_{n}$

If this condition is satisfied, the equation (1) can be solved easily with a numerical method but there is no closed-form solution for$\alpha$

As B is positive semidefinite symmetric matrix so we can write

Hence

where Q is a diagonal matrix with the values

and e is a vector defined by

Then

or

1)

The equation (1) has real solution if and only if

and

If this condition is satisfied, the equation (1) can be solved easily with a numerical method but there is no closed-form solution for

membatas0v2v

Beginner2022-03-25Added 19 answers

Step 1

The left side of your equation is a rational function of α. The roots of its numerator are "closed form" according to some definitions, although if it has degree 5 or more they may not be expressible in radicals.

In the$3\times 3$ example in the comment, the equation is

$\frac{17424{\alpha}^{4}+14688{\alpha}^{3}-49128{\alpha}^{2}-24786\alpha +39366}{{\alpha}^{2}{(36{\alpha}^{2}+17\alpha -54)}^{2}}$

The numerator is an irreducible quartic, and the expression in radicals for its roots is not pretty.

The left side of your equation is a rational function of α. The roots of its numerator are "closed form" according to some definitions, although if it has degree 5 or more they may not be expressible in radicals.

In the

The numerator is an irreducible quartic, and the expression in radicals for its roots is not pretty.

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