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They both seems to decompose a square matrix into a upper triangular matrix, but what's the fundamental difference between these two decompositions?

Thanks!

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The Schur decomposition uses unitary transformations. This makes it reasonable from a numerical computation standpoint. It also gives an idea of how close the matrix is to being normal.

It can be basically impossible to compute the Jordan form numerically when eigenvalues are repeated and eigenvectors are 'close to being parallel'.

The Jordan form is often easier to work with from a theoretical standpoint (fewer numbers to deal with).

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Except the computation complexity, the only difference between those two decompositions is that the Schur decomposition uses unitary transformations, which are not always used by the Jordan decomposition. Is that right? –  chaohuang May 17 '12 at 22:30
    
@chaohuang Yes. The Schur decomposition is $A = QUQ^*$ where $Q$ is unitary and $U$ is upper triangular. The Jordan decomposition is $A = PJP^{-1}$. $J$ is diagonal if the matrix is diagonalizable else it will have few super diagonals depending on the geometric multiplicity of the eigenvalues. The matrix $P$ has no structure in general except for the fact that the columns of $P$ are the eigenvectors of the matrix $A$. –  user17762 May 17 '12 at 22:37
    
Another related question: is the sum of geometric multiplicities equal to the difference between the the dimension of the vector space and the nullity of the matrix? –  chaohuang May 18 '12 at 16:17

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