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JunningHuang · Nov 24, 2024 · 95867ed · 95867ed
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 It's easy to see that as long as $a_{j}\neq \beta_{i}$, namely $A$ and $B$ shares no common eigen-values, there's always a solution of $t_{i,j}$.
+
+## Reference
+- [Observing the State of a Linear System](https://ieeexplore.ieee.org/document/4323124)
+- [Constrained Matrix Sylvester Equations](https://www.cs.umd.edu/users/oleary/reprints/j34.pdf)