Editing Lyapunov equation (section)

===Discrete time===
Using the result that <math> \operatorname{vec}(ABC)=(C^{T} \otimes A)\operatorname{vec}(B) </math>, one has
:<math> (I_{n^2}-\bar{A} \otimes A)\operatorname{vec}(X) = \operatorname{vec}(Q) </math>
where <math>I_{n^2}</math> is a [[conformable]] identity matrix and <math>\bar{A}</math> is the element-wise complex conjugate of <math>A</math>.<ref>{{cite book |first=J. |last=Hamilton |year=1994 |title=Time Series Analysis |at=Equations 10.2.13 and 10.2.18 |publisher=Princeton University Press |isbn=0-691-04289-6 }}</ref> One may then solve for <math>\operatorname{vec}(X)</math> by inverting or solving the linear equations.  To get <math>X</math>, one must just reshape <math>\operatorname{vec} (X)</math> appropriately.

Moreover, if <math>A</math> is stable (in the sense of [[Stable polynomial|Schur stability]], i.e., having eigenvalues with magnitude less than 1), the solution <math>X</math> can also be written as
:<math> X = \sum_{k=0}^{\infty} A^{k} Q (A^{H})^k </math>.

For comparison, consider the one-dimensional case, where this just says that the solution of <math> (1 - a^2) x = q </math> is 
:<math> x = \frac{q}{1-a^2} = \sum_{k=0}^{\infty} qa^{2k} </math>.