Let A(α)=(αI+(1−α)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 thatcTA(α)(I−B)A(α)c=0Is there a closed-form solution for α?

Question

paganizaxpo3 · Accepted Answer

Step 1As B is positive semidefinite symmetric matrix so we can write B=P−1DP where P is a orthogonal matrix (P{−1}=PT) and D is a diagonal matrix (di&amp;gt;0 for all i=1,⋯,n). We haveA(α)=(αI+(1−α)B)−1=(alhaP−1P+(1−α)P−1DP)−1=(P−1(αI+(1−α)D)P)−1=P−1(αI+(1−α)D)−1PHencecTA(α)(I−B)A(α)c=cT(P−1(αI+(1−α)D−1P)(P−1(I−D)P)(P−1(αI+(1−α)D−1P)c=(Pc)T(αI+(1−α)D)−1(I−D)(αI+(1−α))−1(Pc)=(Pc)TQ(Pc)=eTQewhere Q is a diagonal matrix with the valuesqi=1−di(α+(1−α)di)2∀i=1,⋯,nand e is a vector defined by e=PcThencTA(α)(I−B)A(α)c=0⇔eTQ(α)e=0⇔∑i=1n(1−di)ei2(α+(1−α)di)2=0or1) ∑i=1n(1−di)ei2((1−di)α+di)2=0The equation (1) has real solution if and only if {(1−di)}i=1,⋯,n don&#039;t have the same sign (I−B is not positive definite or negative definite), in other words, there exists k such that1&amp;lt;k&amp;lt;nand0&amp;lt;d1≤d2≤⋯≤dk&amp;lt;1&amp;lt;dk+1≤⋯≤dnIf this condition is satisfied, the equation (1) can be solved easily with a numerical method but there is no closed-form solution for α

membatas0v2v · Accepted Answer

Step 1The left side of your equation is a rational function of α. The roots of its numerator are &quot;closed form&quot; 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 is17424α4+14688α3−49128α2−24786α+39366α2(36α2+17α−54)2The numerator is an irreducible quartic, and the expression in radicals for its roots is not pretty.

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

Answered question

Answer & Explanation

New Questions in Algebra I