Recent content by pibeta

Thank you for all of this help (I know what diagonalisable means, but didn't realize they was more than one way to define it). And I've decided to come back too. Anyways, a hint I was given was to use [T]C=PC<-B[T]B(PC<-B)-1 and prove the contradiction, which is the matrix is diagonalisable...

Looks like I've caused too much trouble here. Thanks for trying. Pibeta.

Homework Statement If the matrix of a linear transformation relative to a basis is non-diagonalisable, then for any other choice of basis, it too will be non-diagonalisable. Prove this is the case. Homework Equations Similar matrices. The Attempt at a Solution Let APB=[T]BPB, where A is a...