 # 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 $A\left(\alpha \right)=\left(\alpha I+\left(1-\alpha \right)B\right)-1$ 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 $\alpha \in \left(0,1\right]$. The matrix B is not invertible. For some n-dimensional vector c, I want to find $\alpha$ such that
${c}^{T}A\left(\alpha \right)\left(I-B\right)A\left(\alpha \right)c=0$
Is there a closed-form solution for $\alpha$?
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Step 1
As B is positive semidefinite symmetric matrix so we can write $B={P}^{-1}DP$ where P is a orthogonal matrix $\left(P\left\{-1\right\}=PT\right)$ and D is a diagonal matrix (${d}_{i}>0$ for all $i=1,\cdots ,n$). We have
$A\left(\alpha \right)={\left(\alpha I+\left(1-\alpha \right)B\right)}^{-1}$
$={\left(alha{P}^{-1}P+\left(1-\alpha \right){P}^{-1}DP\right)}^{-1}$
$={\left({P}^{-1}\left(\alpha I+\left(1-\alpha \right)D\right)P\right)}^{-1}$
$={P}^{-1}{\left(\alpha I+\left(1-\alpha \right)D\right)}^{-1}P$
Hence
${c}^{T}A\left(\alpha \right)\left(I-B\right)A\left(\alpha \right)c={c}^{T}\left({P}^{-1}\left(\alpha I+\left(1-\alpha \right){D}^{-1}P\right)\left({P}^{-1}\left(I-D\right)P\right)\left({P}^{-1}\left(\alpha I+\left(1-\alpha \right){D}^{-1}P\right)c$
$={\left(Pc\right)}^{T}{\left(\alpha I+\left(1-\alpha \right)D\right)}^{-1}\left(I-D\right){\left(\alpha I+\left(1-\alpha \right)\right)}^{-1}\left(Pc\right)$
$={\left(Pc\right)}^{T}Q\left(Pc\right)$
$={e}^{T}Qe$
where Q is a diagonal matrix with the values
${q}_{i}=\frac{1-{d}_{i}}{{\left(\alpha +\left(1-\alpha \right){d}_{i}\right)}^{2}}\mathrm{\forall }i=1,\cdots ,n$
and e is a vector defined by $e=Pc$
Then
${c}^{T}A\left(\alpha \right)\left(I-B\right)A\left(\alpha \right)c=0⇔{e}^{T}Q\left(\alpha \right)e=0⇔$
$\sum _{i=1}^{n}\frac{\left(1-{d}_{i}\right){e}_{i}^{2}}{{\left(\alpha +\left(1-\alpha \right){d}_{i}\right)}^{2}}=0$
or
1) $\sum _{i=1}^{n}\frac{\left(1-{d}_{i}\right){e}_{i}^{2}}{{\left(\left(1-{d}_{i}\right)\alpha +{d}_{i}\right)}^{2}}=0$
The equation (1) has real solution if and only if ${\left\{\left(1-{d}_{i}\right)\right\}}_{i=1,\cdots ,n}$ don't have the same sign ($I-B$ is not positive definite or negative definite), in other words, there exists k such that
$1
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 $\alpha$
###### Not exactly what you’re looking for? membatas0v2v
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
$\frac{17424{\alpha }^{4}+14688{\alpha }^{3}-49128{\alpha }^{2}-24786\alpha +39366}{{\alpha }^{2}{\left(36{\alpha }^{2}+17\alpha -54\right)}^{2}}$
The numerator is an irreducible quartic, and the expression in radicals for its roots is not pretty.