Diagonalisable matrix proof (1 Viewer)

[Librah]

Need some help understanding why if an nxn matrix M has 'n' distinct eigenvalues, then M is diagonalisable. I believe the proof involved using the fact that the eigenvectors of the columns of 'P' in D=P^-1MP were linearly independent and therefore proving P was invertible, but I'm not sure why this is?

 

[RenegadeMx]

cant remember exactly but think had something to do with the span of the eigenvectors needing to cover R^n, so if u have repeated evalues, cant have the full span

 

[Silly Sausage]

LOL I hate matrices (use back substitution when I can :rolf but I think it has something to do with the nxn matrix in that it has to have n numbers of unique eigenvalues for it to be fully diagonalisable.


It also has something to do with the geometric multiplicity of the eigenvalue (not sure if they teach that in MATH1X02.)
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Pls correct me if I'm wrong .

 

[VBN2470]

A matrix is diagonalisable if the direct sum of the eigenspaces is equal to , where is an matrix. This means that will consist of linearly independent eigenvectors (which follows from having distinct eigenvalues) which will form the columns of your invertible matrix (and form a basis for your vector space) which you can then use to find the expression for your diagonal matrix (consisting of the eigenvalues in it's diagonal entries). The true test of whether a matrix is diagonalisable, if I am not mistaken, is whether if the algebraic and geometric multiplicities each eigenvalue coincide.