# What's the difference between Jordan and Schur decomposition?

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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@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