Closed-form solution to intricate matrix equation

Question

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

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

(I'm aware that a solution might not always exist.)

(In case it's helpful, note that $\mathbf{B}$ can be written as $\mathbf{D}^T \mathbf{D}$ for some $m$-by-$n$ matrix $\mathbf{D}$, where $m < n$.)

Answer 1

As $\mathbf{B}$ is positive semidefinite symmetric matrix so we can write $\mathbf{B} = \mathbf{P}^{-1}\mathbf{D}\mathbf{P}$ where $\mathbf{P}$ is a orthogonal matrix ($\mathbf{P}^{-1} =\mathbf{P}^T$) and $\mathbf{D}$ is a diagonal matrix ($d_i > 0$ for all $i=1,...,n$). We have \begin{align} \mathbf{A}(\alpha) &= \big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{B}\big)^{-1} \\ &= \big(\alpha \mathbf{P}^{-1}\mathbf{P} + (1 - \alpha) \mathbf{P}^{-1}\mathbf{D}\mathbf{P}\big)^{-1} \\ &= \big(\mathbf{P}^{-1}\big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)\mathbf{P}\big)^{-1} \\ &= \mathbf{P}^{-1}\big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)^{-1}\mathbf{P} \\ \end{align}

Hence \begin{align} \mathbf{c}^T \mathbf{A}(\alpha) \big( \mathbf{I} - \mathbf{B} \big) \mathbf{A}(\alpha) \mathbf{c} &= \mathbf{c}^T \left(\mathbf{P}^{-1}\big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)^{-1} \mathbf{P} \right) (\mathbf{P}^{-1}(\mathbf{I}-\mathbf{D})\mathbf{P}) \left(\mathbf{P}^{-1}\big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)^{-1} \mathbf{P} \right)\mathbf{c} \\ &= (\mathbf{P}\mathbf{c})^T \big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)^{-1} (\mathbf{I}-\mathbf{D}) \big(\alpha \mathbf{I} + (1 - \alpha) \mathbf{D}\big)^{-1} (\mathbf{P}\mathbf{c}) \\ &=(\mathbf{P}\mathbf{c})^T \mathbf{Q} (\mathbf{P}\mathbf{c})\\ &=\mathbf{e}^T \mathbf{Q} \mathbf{e} \end{align}

where $ \mathbf{Q}$ is a diagonal matrix with the values

$$q_{i} = \frac{1-d_i}{\big(\alpha+(1-\alpha)d_i\big)^2} \qquad \forall i=1,...,n$$

$\qquad$ and $\mathbf{e}$ is a vector defined by $\mathbf{e} = \mathbf{P}\mathbf{c}$

Then $$\mathbf{c}^T \mathbf{A}(\alpha) \big( \mathbf{I} - \mathbf{B} \big) \mathbf{A}(\alpha) \mathbf{c} = 0 \iff \mathbf{e}^T \mathbf{Q}(\alpha) \mathbf{e} =0 \iff \sum_{i=1}^n \frac{(1-d_i)e_i^2}{\big(\alpha+(1-\alpha)d_i\big)^2} = 0$$ $\qquad$ or $$\sum_{i=1}^n \frac{(1-d_i)e_i^2}{\big((1-d_i)\alpha+d_i\big)^2} = 0 \tag{1}$$

The equation $(1)$ has real solution if and only if $\{(1-d_i)\}_{i=1,...,n}$ don't have the same sign ($\mathbf{I}-\mathbf{B}$ is not positive definite or negative definite), in other words, there exists $k$ such that $$1<k<n$$ $\qquad$ and $$0<d_1\le d_2 \le ... \le d_k < 1 < d_{k+1} \le ...\le d_n$$

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

Answer 2

The left side of your equation is a rational function of $\alpha$. 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 $$ \frac{17424 \alpha^{4}+14688 \alpha^{3}-49128 \alpha^{2}-24786 \alpha +39366}{\alpha^{2} \left(36 \alpha^{2}+17 \alpha -54\right)^{2}}$$

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