Office_Shredder said:What is your definition of diagonalizable and non-diagonalizable? It makes a pretty big difference since there are a couple ways to approach this depending on what we can assume and what we want to prove
Dick said:If you mean real diagonalizable vs complex diagonalizable, I'm not sure that's what this is about.
I would assume that a diagonalisable matrix would have a diagonal and the off-diagonal entries are all zero, and then the non-diagonalisable matrix would have off-diagonal entries not all zero - that's how I would define it (and I'm hoping that's the way my teacher wanted us to define it!)
Office_Shredder said:No, I mean, for example, the definition of non-diagonalizable that I would use, and a lot of other people, is something like "a matrix which is not diagonal under any change of basis". Making the whole question kind of moot. Someone thought this question was worth asking, so obviously there's a different definition floating around.
pibeta said:Looks like I've caused too much trouble here. Thanks for trying.
Pibeta.
"Doesn't have a closed form for its eigenvalue equation"? What does that mean?keniwas said:There might be an easier way to show this, but in physics diagonalizing matricies always leads to eigenvalue problems... I believe your statement implies the matrix doesn't have a closed form for its eigenvalue equation.
That doesn't follow. If an operator (on an n dimensional vector space) has n distinct roots then it is diagonalizable, but if not it still may be diagonalizable. As Dick said, it depends on the number of independent eigenvectors, not eigenvalues.So
[tex]|A-\omega I|=0[/tex] wouldn't have n roots and hence can't be diagonalized.
No, office-shredder did not say that. He said there is more than one way to define "diagonalizable".But as office-shredder says, there is more than one way to diagonalize a matrix...
I'm glad to know that you know what "diagonalizable" means- that helps a lot! But to help you we still need to know what your definition of "diagonalizable" is! Please tell us.pibeta said: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.
To show it is diagonalisable, I got (using the hint) [T]C=PBAPB-1 (where A=PB[T]B(PB)-1 )
but I'm not sure if this is right and where to go from here...
A non-diagonalisable matrix is a square matrix that cannot be transformed into a diagonal matrix through a similarity transformation. In other words, it cannot be written as a product of a diagonal matrix and an invertible matrix.
A matrix is non-diagonalisable if it does not have a full set of linearly independent eigenvectors. This can be determined by finding the eigenvalues of the matrix and checking if their corresponding eigenvectors span the entire vector space.
Yes, a non-diagonalisable matrix can still have eigenvalues. However, it will not have a full set of linearly independent eigenvectors, which is a requirement for a matrix to be diagonalisable.
Non-diagonalisable matrices have many important applications in various fields, such as quantum mechanics and control theory. They can also provide insights into the properties of linear transformations and their relationships with vector spaces.
The Jordan canonical form of a non-diagonalisable matrix can be found by first finding its Jordan normal form, which is a block diagonal matrix with Jordan blocks along the diagonal. The Jordan blocks are then filled in with the appropriate eigenvalues and the resulting matrix is the Jordan canonical form.