Let \(\displaystyle{A}{\left(\alpha\right)}={\left(\alpha{I}+{\left({1}-\alpha\right)}{B}\right)}-{1}\) be the inverse of a

Lorelei Stanton 2022-03-23 Answered
Let A(α)=(αI+(1α)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 α(0,1]. The matrix B is not invertible. For some n-dimensional vector c, I want to find α such that
cTA(α)(IB)A(α)c=0
Is there a closed-form solution for α?
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Answers (2)

paganizaxpo3
Answered 2022-03-24 Author has 8 answers
Step 1
As B is positive semidefinite symmetric matrix so we can write B=P1DP where P is a orthogonal matrix (P{1}=PT) and D is a diagonal matrix (di>0 for all i=1,,n). We have
A(α)=(αI+(1α)B)1
=(alhaP1P+(1α)P1DP)1
=(P1(αI+(1α)D)P)1
=P1(αI+(1α)D)1P
Hence
cTA(α)(IB)A(α)c=cT(P1(αI+(1α)D1P)(P1(ID)P)(P1(αI+(1α)D1P)c
=(Pc)T(αI+(1α)D)1(ID)(αI+(1α))1(Pc)
=(Pc)TQ(Pc)
=eTQe
where Q is a diagonal matrix with the values
qi=1di(α+(1α)di)2i=1,,n
and e is a vector defined by e=Pc
Then
cTA(α)(IB)A(α)c=0eTQ(α)e=0
i=1n(1di)ei2(α+(1α)di)2=0
or
1) i=1n(1di)ei2((1di)α+di)2=0
The equation (1) has real solution if and only if {(1di)}i=1,,n don't have the same sign (IB is not positive definite or negative definite), in other words, there exists k such that
1<k<n
and0<d1d2dk<1<dk+1dn
If this condition is satisfied, the equation (1) can be solved easily with a numerical method but there is no closed-form solution for α
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membatas0v2v
Answered 2022-03-25 Author has 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 3×3 example in the comment, the equation is
17424α4+14688α349128α224786α+39366α2(36α2+17α54)2
The numerator is an irreducible quartic, and the expression in radicals for its roots is not pretty.
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