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

Question

Let

A (α) = (α I + (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 \in (0, 1]

. The matrix B is not invertible. For some n-dimensional vector c, I want to find

α

such that

c^{T} A (α) (I - B) A (α) c = 0

Is there a closed-form solution for

α

?

Answer 1

Step 1
As B is positive semidefinite symmetric matrix so we can write

B = P^{- 1} D P

where P is a orthogonal matrix

(P {- 1} = P T)

and D is a diagonal matrix (

d_{i} > 0

for all

i = 1, \dots, n

). We have

A (α) = {(α I + (1 - α) B)}^{- 1}

= {(a l h a P^{- 1} P + (1 - α) P^{- 1} D P)}^{- 1}

= {(P^{- 1} (α I + (1 - α) D) P)}^{- 1}

= P^{- 1} {(α I + (1 - α) D)}^{- 1} P

Hence

c^{T} A (α) (I - B) A (α) c = c^{T} (P^{- 1} (α I + (1 - α) D^{- 1} P) (P^{- 1} (I - D) P) (P^{- 1} (α I + (1 - α) D^{- 1} P) c

= {(P c)}^{T} {(α I + (1 - α) D)}^{- 1} (I - D) {(α I + (1 - α))}^{- 1} (P c)

= {(P c)}^{T} Q (P c)

= e^{T} Q e

where Q is a diagonal matrix with the values

q_{i} = \frac{1 - d_{i}}{{(α + (1 - α) d_{i})}^{2}} \forall i = 1, \dots, n

and e is a vector defined by

e = P c

Then

c^{T} A (α) (I - B) A (α) c = 0 \Leftrightarrow e^{T} Q (α) e = 0 \Leftrightarrow

\sum_{i = 1}^{n} \frac{(1 - d_{i}) e_{i}^{2}}{{(α + (1 - α) d_{i})}^{2}} = 0

or
1)

\sum_{i = 1}^{n} \frac{(1 - d_{i}) e_{i}^{2}}{{((1 - d_{i}) α + d_{i})}^{2}} = 0

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

{(1 - d_{i})}_{i = 1, \dots, 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} \leq d_{2} \leq \dots \leq d_{k} < 1 < d_{k + 1} \leq \dots \leq 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

α

Answer 2

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 α^{4} + 14688 α^{3} - 49128 α^{2} - 24786 α + 39366}{α^{2} {(36 α^{2} + 17 α - 54)}^{2}}

The 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

Answers (2)

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