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A given real square matrix $M$ has a decomposition

$M = T^{-1} S T$

where $S$ and $T$ are lower and resp. upper triangular with '$1$'s in the diagonal. Does this type of decomposition have a known classification/name? I could not find something similar under the wikipedia entry for matrix decomposition.

Thank you

Aryaman Maithani
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  • It's something like a special type of Jordan decomposition. I'm not aware of a name. – Joshua P. Swanson Feb 20 '21 at 07:55
  • ... according to this source, it could also be a Schur decomposition where however $S$ is upper triangular. In the case here $S$ is lower triangular but I imagine that it is easy to change a Schur decomposition so that it fits the latter? – cookie Feb 21 '21 at 10:23
  • Sure. (Hah.) As for flipping triangularity, just do a Schur decomposition of $M^\top$ and transpose afterwards. – Joshua P. Swanson Feb 22 '21 at 02:40
  • I see...so assuming that M has the decomposition above where S is upper triangular, then $M^{\top}=(T^{-1}S T)^{\top}=(ST)^{\top}(T^{-1})^{\top}=T^{\top} S^{\top} (T^{\top})^{-1}$. Hence there is a corresponding lower triangular decomposition!! Thank you – cookie Feb 22 '21 at 07:28

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