Can anyone solve this vector and matrix simultaneous equation?

Question

I want to solve a simltaneous equation below.

$\left(K+\lambda_1I\right)\alpha = \frac{1}{n}\left(y-Ah\right)$

$\left(A^2+\lambda_2I\right)h=A\left(y-nK\alpha\right)$

Here, $n, \lambda_1, \lambda_2>0$ are scalar. $\alpha, h$, and $y \in \mathbb{R}^n$, $A \in \mathbb{R}^{n\times n}$.

$K \in \mathbb{R}^{n\times n}$ is positive definite.

I want to solve for $\alpha$ and $h$, fixing other variables.

My answer is $\alpha=\frac{\lambda_2}{n}\left(\lambda_1\left(A^2+\lambda_2I\right)+\lambda_2K\right)^{-1}y$, and $h=\lambda_1 A \left(\lambda_1\left(A^2+\lambda_2I\right)+\lambda_2K\right)^{-1}y$. Is this correct and unique solution?

Answer 1

Write as

$$\begin{cases}\left(K+\lambda_1I\right)&\alpha+\frac{1}{n}A&h& = \frac{1}{n}y\\ nA^2K&\alpha+\left(A^2+\lambda_2I\right)&h&=Ay\end{cases}$$ which is an ordinary linear system.

If you solve it block-wise, for instance by Gaussian elimination, you will see negative terms appear, so I don't think that your solution is correct.

Answer 2

$ \def\a{\alpha} \def\L{\lambda} \def\Lo{\lambda_1} \def\Lt{\lambda_2} \def\LR#1{\left(#1\right)} $Multiply the first equation by $nA$ and subtract the second equation to eliminate $y$ $$\eqalign{ Ay &= nA\LR{K+\Lo I} \a &\;+\; \quad\;\;\quad\LR{A^2}h \\ Ay &= \qquad\quad\LR{nAK} \a &\;+\; \LR{A^2+\Lt I}h \\ }$$ $$\eqalign{ 0 &= n\LR{\Lo A} \a \;-\; \Lt h \qquad\qquad\qquad \\ }$$ Solve for $h$ and substitute into the first equation to obtain a linear system for $\a$ $$\eqalign{ h &= \frac n\Lt\LR{\Lo A} \a \\ M &\doteq n\LR{K+\Lo I} + \frac n\Lt\LR{\Lo A^2} \\ y &= M\a \quad\implies\quad \a = M^{-1}y \\ }$$ This is different than your proposed solution.