Why P^-1AP forms a triangular matrix

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Discussion Overview

The discussion revolves around the conditions under which the expression (P^-1)AP results in a triangular matrix, particularly in the context of linear algebra and matrix theory. Participants explore specific examples and the implications of eigenvalues and eigenvectors in relation to triangular forms and Jordan normal forms.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions the formation of a triangular matrix from (P^-1)AP, suggesting that additional information is needed regarding the matrices involved.
  • Another participant provides a specific example with matrices P and A, calculating P^-1 and the resulting product (P^-1)AP, which is triangular, and seeks to understand why this occurs.
  • A later reply emphasizes that (P^-1)AP is not generally triangular for arbitrary A and P, indicating the need for specific conditions.
  • One participant discusses the relationship between eigenvalues and the structure of the resulting matrix, mentioning that the matrix is not diagonalizable and can be expressed in Jordan normal form.
  • Another participant reflects on the broader significance of triangular matrices and Jordan forms, indicating a lack of clarity on their properties and implications.

Areas of Agreement / Disagreement

Participants express differing views on the conditions necessary for (P^-1)AP to be triangular, with some asserting that it is not universally true while others provide specific examples where it holds. The discussion remains unresolved regarding the generality of the claim.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about matrices P and A, as well as the definitions of triangular and Jordan forms. The mathematical steps leading to the conclusions are not fully resolved.

d.vaughn
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why does (P^-1)AP form a triangular matrix?
 
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d.vaughn said:
why does (P^-1)AP form a triangular matrix?

If P=I, and A is not triangular, then clearly (P^-1)AP=A is not triangular.

It appears you need to supply more information.
 
for P, I have the matrix P = (4,-9; 4,-8) and the A matrix is A = (3,2; 2,1)
I found P^-1 to be (-1,2; 2,-3)
When I performed P^1AP, I got (-2,1; 0,-2) and I want to know why this formed a triangular matrix
 
Please refine your question. "P^{-1}AP" for general A and P is NOT "triangular" and it is not clear what conditions you intend on A and P.
 
d.vaughn said:
for P, I have the matrix P = (4,-9; 4,-8) and the A matrix is A = (3,2; 2,1)
I found P^-1 to be (-1,2; 2,-3)
When I performed P^1AP, I got (-2,1; 0,-2) and I want to know why this formed a triangular matrix

Do you understand eigenvalues? -2 is an eigenvalue with algebraic multiplicity 2, geometric multiplicity 1.

I.e. the charactersitic polynomial is (λ+2)2, but there is only eigenvector. In this case, the matrix is not diagonalizable. However, all matrices can be put into Jordan-Normal form, which a diagonal matrix is a special case of. If the matrix is diagonalizable, then P is a matrix with columns the eigenvectors. If it is not diagonalizable, then some of the columns of P will be eigenvectors, some of them will be what are called generalized eigenvectors.

Let J denote your Jordan matrix, J=P-1AP. Then PJ=AP. Then let P=[v w], where v and w are column vectors. Then PJ=[-2v v-2w], so Av=-2v, and Aw=v-2w. So v is an eigenvector, while w is "almost" an eigenvector.
 
d.vaughn said:
why does (P^-1)AP form a triangular matrix?

So in general, it is more than a triangular matrix, it is of the jordan form. That is one reason we did not recognize it sooner.

That is, when I hear triangular matrix, my brain goes into a freeze because I do not feel I know the full significance of those objects, with respect to where they arise and what properties they have.

Happy hunting!
 

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