Why P^-1AP forms a triangular matrix

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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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