Find positive definite $X$ such that $AXA^T=\alpha X$. Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$
Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $(1-\alpha)>0$ this looks like the Lyapunov equation here. However, the analytic solution to the Lypunov equation has the form
$$
X = \sum_{k=0}^\infty A^kQ(A^T)^k = (1-\alpha)\sum_{k=0}^\infty A^kX(A^T)^k
$$
So, even if such equation have a solution, I don't have a clue on how to obtain it from here.
Do you have other thoughts/suggestions?

Consider $A = \pmatrix{2 & 1 \\ 0 & 1}$. Then you seek a matrix
$$
X = \pmatrix{a & b \\ c & d}
$$
and a constant $K$ such that
$$
\pmatrix{2 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
i.e.,
$$
\pmatrix{2a+c & 2b+d \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
so
$$
\pmatrix{4a+2c+2b+d  & 2b+d \\ 2c+d & d} = K \pmatrix{a & b \\ c & d}
$$
Case 1: $d \ne 0$. Then $K = 1$, and we have
$$
\pmatrix{4a+2c+2b+d  & 2b+d \\ 2c+d & d} = \pmatrix{a & b \\ c & d}
$$
whence $2b+d = b$ so $b = -d$, and similarly $c = -d$; thus our equation becomes
$$
\pmatrix{4a+2(-d)+2(-d)+d  & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}\\
\pmatrix{4a-3d  & -d \\ -d & d} = \pmatrix{a & -d \\ -d & d}
$$
so $4a - 3d = a$, hence $a = d$, and we have
$$
X = \pmatrix{d & -d \\ -d & d}
$$
which is not positive definite, because its determinant is zero.
Case 2: $d = 0$. Then we want
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = K \pmatrix{a & b \\ c & 0}
$$
For $X$ to have full rank, we need $c \ne 0$, so $K = 2$, and we can write
$$
\pmatrix{4a+2c+2b & 2b \\ 2c & 0} = \pmatrix{2a & 2b \\ 2c & 0}
$$
which reduces to $4a + 2c + 2b = 2a$, so $a + b + c = 0$. That gives
$$
X = \pmatrix{-(b+c) & b \\ c & 0}
$$
or, replacing $b$ and $c$ with their negatives,
$$
X = \pmatrix{(b+c) & -b \\ -c & 0}
$$
The determinant here is $bc$, so $b$ and $c$ must have the same sign for $X$ to be positive definite; the trace is $b+c$, so that sign must be "+". But then the quadratic formula gives the eigenvalues as
$$
x = \frac{(b+c) \pm \sqrt{ (b+c)^2 + 4bc}}{2}
$$
and because $(b+c)^2 + 4bc > (b+c)^2$, one of the two choices in the numerator leads to a negative eigenvalue, so $X$ is not positive definite.
In short: for the matrix $A$ given above, there is no solution $X$.
