# Schur decomposition, proof of

The columns of the unitary matrix^{} $Q$ in Schur’s decomposition theorem form an orthonormal basis^{} of ${\u2102}^{n}$. The matrix $A$ takes the upper-triangular form $D+N$ on this basis. Conversely, if ${v}_{1},\mathrm{\dots},{v}_{n}$ is an orthonormal basis for which $A$ is of this form then the matrix $Q$ with ${v}_{i}$ as its $i$-th column satisfies the theorem.

To find such a basis we proceed by induction^{} on $n$. For $n=1$ we can simply take $Q=1$. If $n>1$ then let $v\in {\u2102}^{n}$ be an eigenvector^{} of $A$ of unit length and let $V={v}^{\u27c2}$ be its orthogonal complement^{}. If $\pi $ denotes the orthogonal projection onto the line spanned by $v$ then $(1-\pi )A$ maps $V$ into $V$.

By induction there is an orthonormal basis ${v}_{2},\mathrm{\dots},{v}_{n}$ of $V$ for which $(1-\pi )A$ takes the desired form on $V$. Now $A=\pi A+(1-\pi )A$ so $A{v}_{i}\equiv (1-\pi )A{v}_{i}\phantom{\rule{veryverythickmathspace}{0ex}}(modv)$ for $i\in \{2,\mathrm{\dots},n\}$. Then $v,{v}_{2},\mathrm{\dots},{v}_{n}$ can be used as a basis for the Schur decomposition^{} on ${\u2102}^{n}$.

Title | Schur decomposition, proof of |
---|---|

Canonical name | SchurDecompositionProofOf |

Date of creation | 2013-03-22 14:04:01 |

Last modified on | 2013-03-22 14:04:01 |

Owner | mps (409) |

Last modified by | mps (409) |

Numerical id | 6 |

Author | mps (409) |

Entry type | Proof |

Classification | msc 15-00 |