Finding a matrix $\mathbf{A}$ that projects a point to an eigenvector of $\mathbf{A}\mathbf{C}\mathbf{A}^T$ Suppose $\mathbf{b}=[b_1,b_2]'$ is $2\times 1$ and $\mathbf{C}$ is a full-rank symmetric $2\times 2$ matrix which both are real and given. Now, consider the problem of finding a $2\times 2$ matrix $\mathbf{A}$ such that $\mathbf{b'}=\mathbf{A}\mathbf{b}$ is in the same direction of one of the eigenvectors of $\mathbf{A}\mathbf{C}\mathbf{A}^T$ (lets say the eigenvector correspodning to the bigger eigenvalue). 
I have been looking a lot to find a solution for $\mathbf{A}$ through formulating the problem as an optimization problem. However, I have not been able to see what type of optimization problem this falls into. I appreciate it if you could introduce me to some references addressing a similar problem to this one, or outline the appropriate approach for solving such examples.
 A: Often, it's easiest (and insightful) to solve problems of linear algebra using algebraic or geometric methods rather than trying to cast them as optimization problems.
This meta-principle is justified by the fact that many linear algebraic problems, such as this one, make sense for arbitrary ground fields (and often for mere rings of scalars) but the optimization framework requires the ground field to be the real numbers or one of their field extensions, making optimization a (much) more limited technique.

Assume $b$ is nonzero (for otherwise, if we understand "in the same direction as" means "is a nonzero multiple of," there is no solution).
Pick any $\lambda\gt 0$ and let $A$ be a square root of $\lambda C^{-1};$ that is, choose $A$ such that $$A^\top A = \lambda C^{-1}.$$
(This strategy works very generally.  In the present case, because $C$ is real symmetric, such a matrix $A$ exists and can be found with, say, a Cholesky decomposition.)
Since $C$ is of full rank so therefore is $C^{-1},$ whence $A$ must be invertible.  Consequently
$$\left(ACA^\top\right)b^\prime= A\left(C^{-1}\right)^{-1}A^\top\,b^\prime = A\left(\lambda^{-1} A^\top A\right)^{-1}A^\top\,b^\prime = A \lambda A^{-1} \left(A^\top\right)^{-1} A^\top\,b^\prime = \lambda\, b^\prime$$
demonstrates every vector $b^\prime$ satisfies the eigenvector equation for $ACA^\top$ (with eigenvalue $\lambda$).  Because $b^\prime = A b$ is nonzero if and only if $b$ is nonzero (due to the invertibility of $A$), $b^\prime$ is an eigenvector of $ACA^\top$ with eigenvalue $\lambda,$ QED.
