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Eigenvalues/vectors diagonalization

  • Thread starter emyt
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Homework Statement



Suppose that [tex] A \in Mnxn(F) [/tex] has two distinct eigenvalues [tex]\lambda_{1}[/tex] and [tex]\lambda_{2}[/tex] and that [tex] dim(E_{\lambda_{1}}) = n - 1. [/tex] Prove that A is diagonalizable

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The Attempt at a Solution


hmm, I'm not sure.. how would I start this?

thanks
 

Answers and Replies

  • #2
HallsofIvy
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If the dimension of the eigenvector space corresponding to eigenvalue [itex]\lambda_1[/itex] is n- 1, then there are n-1 independent eigenvectors corresponding to [itex]\lambda_1[/itex]

That, together with an eigenvector corresponding to eigenvalue [itex]\lambda_2[/itex] gives you n independent vectors (eigenvectors corresponding to distinct eigenvalues are always independent) and form a basis for the vector space. The matrix for the linear transformation, written in that basis, is a diagonal matrix having [itex]\lambda_1[/itex] n-1 times on the main diagonal and [itex]\lambda_2[/itex] once.
 
  • #3
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thanks, I was thinking about that too, but for some reason my mind strayed and thought that I would need 2n independent vectors..
I was also looking at a couple of the diagonal "if and only if" theorems. I saw a couple that might've been useful, like:
T is diagonalizable if and only if the multiplicity of[tex]\lambda[/tex] is equal to [tex] dim(E_{\lambda})[/tex].
And now I have to ask: how can you deduce the multiplicity of an eigenvalue without working out the characteristic polynomial and looking at the multiplicity of the factors?
thanks :)
 

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