# corollary of Schur decomposition

theorem:$A\in\mathbb{C}^{n\times n}$ is a normal matrix  if and only if there exists a unitary matrix  $Q\in\mathbb{C}^{n\times n}$ such that $Q^{H}AQ=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$ where ${}^{H}$ is the conjugate transpose  . [GVL]

proof: Firstly we show that if there exists a unitary matrix $Q\in\mathbb{C}^{n\times n}$ such that $Q^{H}AQ=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$ then $A\in\mathbb{C}^{n\times n}$ is a normal matrix. Let $D=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$ then $A$ may be written as $A=QDQ^{H}$. Verifying that A is normal follows by the following observation $AA^{H}=QDQ^{H}QD^{H}Q^{H}=QDD^{H}Q^{H}$ and $A^{H}A=QD^{H}Q^{H}QDQ^{H}=QD^{H}DQ^{H}$. Therefore $A$ is normal matrix because $DD^{H}=\operatorname{diag}(\lambda_{1}\bar{\lambda_{1}},\ldots,\lambda_{n}\bar% {\lambda_{n}})=D^{H}D$.
Secondly we show that if $A\in\mathbb{C}^{n\times n}$ is a normal matrix then there exists a unitary matrix $Q\in\mathbb{C}^{n\times n}$ such that $Q^{H}AQ=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{n})$. By Schur decompostion we know that there exists a $Q\in\mathbb{C}^{n\times n}$ such that $Q^{H}AQ=T$. Since $A$ is a normal matrix then $T$ is also a normal matrix. The result that $T$ is a diagonal matrix comes from showing that a normal upper triangular matrix is diagonal (see theorem for normal triangular matrices).
QED

## References

• GVL Golub, H. Gene, Van Loan F. Charles: Matrix Computations (Third Edition). The Johns Hopkins University Press, London, 1996.
Title corollary of Schur decomposition CorollaryOfSchurDecomposition 2013-03-22 13:43:38 2013-03-22 13:43:38 Daume (40) Daume (40) 7 Daume (40) Corollary msc 15-00