What is the difference between the two definitions of Schur decomposition?

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Discussion Overview

The discussion revolves around the differences between two definitions of Schur decomposition as presented by Wikipedia and Mathworld. It explores the implications of the definitions and the relationship between the matrices involved, focusing on theoretical aspects of linear algebra.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that Wikipedia defines the Schur decomposition as A = Q U Q^{-1}, while Mathworld defines it as Q^H A Q = T, highlighting the difference in the placement of the matrix A.
  • Another participant points out that since Q^H is equal to Q^{-1} for unitary matrices, they attempt to derive the equivalence of the two definitions through a series of algebraic manipulations.
  • A third participant expresses realization and gratitude for the clarification provided by the second participant's explanation.
  • A fourth participant suggests that the concept of A being a change of basis should be reviewed in the context of linear algebra.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on the definitions, as participants express differing views on their implications and relationships, indicating that multiple interpretations remain.

Contextual Notes

Participants do not fully resolve the implications of the definitions or the algebraic steps taken, leaving some assumptions and dependencies on definitions unaddressed.

junglebeast
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Wikipedia defines the Shur decomposition of matrix A as

A = Q U Q^{-1}

where Q is unitary and U is upper triangular.

http://en.wikipedia.org/wiki/Schur_decomposition

Mathworld defines the Shur decomposition of matrix A as

Q^H A Q = T,

where Q is unitary and T is upper triangular.

http://mathworld.wolfram.com/SchurDecomposition.html

Because Q is unitary, the inverse is the same as the conjugate transpose...but they still seem like completely different definitions because the matrix is either on the inside or the outside. What's the truth?
 
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Since Q^H = Q^-1, you have

Q Q^H A Q = Q T
=>
Q Q^-1 A Q = Q T
=>
I A Q = Q T
=>
A Q = Q T
=>
A Q Q^-1 = Q T Q^-1
=>
A I = Q T Q^-1
=>
A = Q T Q^-1
 
Wow, I feel stupid for not noticing that! Thanks
 
A is just a change of basis, I'd recommend reviewing change of basis in lin alg
 

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