If $A\in\mathbb R^{N\times p}$, for $p<N$, is there anything simple I can say about the solutions $X\in \mathbb R^{N\times p}$ of the equation $AX^\top+XA^\top=0$ ?
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Your quesion is too inespecific: what does "simple" mean? For example, it is easy to see that the solutions form a vector space. Is that good enough? – Jose Brox Jan 23 '19 at 14:25
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That is already something. Now, what is the dimension of this space? For instance, by the rank-nullity theorem, this space has dimension at least $N(p-\frac{N+1}{2})$. Can one characterize the dimension of this space in terms of the rank of $A$ only? – TrivialPursuit Jan 23 '19 at 15:02
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When dealing with matrix equations, a powerful tool is the conversion to a equivalent linear problem (in a bigger vector space) via the Kronecker product, as explained in Chapter 4 of Topics in matrix analysis by Horn and Johnson. As inferred from Problem 4.3.3 of the same book, the solutions of the system $AX^T+XA^T=0$ are the transposes of the solutions of $$\left(I\otimes A^T + (A^T\otimes I)P(N,N)\right)\cdot\text{vec}(X)=0,$$ where $P(N,N)=(E_{ij}^T)_{1\leq i,j\leq N}$ and $E_{ij}$ are the elementary matrices.
Hence, the space of solutions is wholly determined by the matrix $$I\otimes A^T + (A^T\otimes I)P(N,N).$$
Jose Brox
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