Special case of Lyapunov equation $AP+PA=-I$

Question

If we consider the Lyapunov equation

$$A^TP+PA = -Q$$

with a real symmetric system matrix $A$ (which is invertible and all eigenvalues have a strictly negative real part) and if we set $Q$ to the identity matrix $I$ then we obtain the simplified Lyapunov equation:

$$AP + PA = -I.$$

It turns out that the solution to this problem is given by $$P = -\dfrac{1}{2}A^{-1},$$ which can be verified by plugging this into the simplified Lyapunov equation. I did not like this way of obtaining this solution because it feels like guessing.

I came up with a slightly better algebraic derivation

$$AP + PA = -I \implies AP + PA = -\dfrac{1}{2}A^{}A^{-1}-\dfrac{1}{2}A^{-1}A^{}$$ $$\implies A\left[ P+\dfrac{1}{2}A^{-1}\right] +\left[P+\dfrac{1}{2}A^{-1}\right]A=0 $$

From the last equation, it is easy to see that

$$P+\dfrac{1}{2}A^{-1}=0 \implies P =-\dfrac{1}{2}A^{-1}$$

is a solution to the simplified Lyapunov equation.

I do not like both ways of showing that $P = -1/2A^{-1}$ because both exploit some knowledge about the solution. Is there a direct way to solve the simplified Lyapunov equation for $P$ without implicitly or explicitly using the knowledge of the final solution? It is also verifiable that the result implies that $A$ and $P$ commute, hence directly using commutativity is also not allowed (but showing commutativity from the equation only is permitted).

Answer 1

This isn't a solution, but it is at least the beginning of an analysis. Let $[A,B]=AB-BA$ be the commutator. Since $I$ commutes with $A^{-1}$, we can take the commutator of both sides of are equation with $A^{-1}$ (and rearranging) to get

$$APA^{-1}=A^{-1}PA.$$

Conjugating, we get $P=A^{-2}PA^2$. Thus, $P$ commutes with $A^2$. There are pathological cases where something can commute with $A^2$ but not with $A$ (e.g., if $A=\operatorname{diag}(1,-1)$), but ignoring those cases, let us assume that $P$ commutes with $A$. Then our equation simplifies to

$$(-2)AP=I$$

and we can multiply by $A^{-1}$ to solve for $P$.

And obviously, the solution we get can be plugged in and seen to work without any conditions on $A$. The remaining question is, if $A$ is such that there exists a $B$ such that $B$ commutes with $A^2$ but not with $A$, then might there be additional solutions? And right now, I do not know.

Answer 2

Putting $X=P+1/2A^{-1}$, our equation reduces to $AX+XA=0$, that is $f(X)=0$ where $f=A\otimes I+I\otimes A: M_n(\mathbb{R})\rightarrow M_n(\mathbb{R})$. cf.

https://en.wikipedia.org/wiki/Kronecker_product

If $spectrum(A)=(\lambda_i)_i$, then $spectrum(f)=(\lambda_i+\lambda_j)_{ij}$. Here $A$ is invertible and consequently, the $\lambda_i$ are not $0$. Since $A$ is diagonalizable, $f$ too and $dim(\ker(f))$ is the number $k$ of $i\not= j$ s.t. $\lambda_i=-\lambda_j$; note that $k$ is even because if $(i,j)$ is convenient, then $(j,i)$ too.

If $A>0$, then the $\lambda_i+\lambda_j$ are $>0$ and $\ker(f)$ is $0$. Otherwise, $\ker(f)$ may be $\not= 0$.

For example, if $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$, then $k=2$ and $\ker(f)=\{\begin{pmatrix}0&b\\c&0\end{pmatrix};b,c\in\mathbb{R}\}$. If you ask only for symmetric solutions $X$, then the dimension of the set of solutions is $1$:

$\{\begin{pmatrix}0&b\\b&0\end{pmatrix};b\in\mathbb{R}\}$

EDIT. Answer to MrYouMath . An equation in the form $AP+PA=B$ where $AB=BA$ admits always an "obvious" solution. Indeed, assume that $AP=PA$; (we will verify that the obtained solution satisfies this condition). Then $2PA=B$ and $P=1/2BA^{-1}$, a matrix that commutes with $A$.

When $B=-I$, it's even easier; just a little intuition. I see that "you try to create science videos for pupils and students". In particular, your role is to ask them to work their intuition; without intuition, there is no good mathematician!