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Can zero be a distinct eigernvalue?

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- #1

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Can zero be a distinct eigernvalue?

- #2

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[tex]\lambda (\lambda-1)(\lambda -2)=0[/tex]

during your calculations. Perfectly valid. Now what would your eigenvalues be?

- #3

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Yes the eigenvalues are 0 1 and 2 but is 0 distinct?

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- #5

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I have to prove the following;

Let V be a finite dimensional vector space and T: V to V be a linear transformation. Let a1,...,ak be distinct eigenvalues of T and let v1,..,vk be corresponding eigenvectors. Prove that {v1,...,vk} is linearly independent.

If it was the case that 0 cannot be a distinct eigenvalue then I can complete the proof but since this isn't the case I'm a liitle stuck! Could anyone possible help me out here?

Thanks.

- #6

Landau

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- #7

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Now taking x element of V, x = sum from i=1 to k of alpha(i)v(i)

then T(x) = sum from i=1 to k of alpha(i)T(v(i)) = sum from i=1 to k of alpha(i)a(i)v(i)

Now for this sum to equal zero then either aplha(i)=0 (as required) OR a(i)v(i)=0, if it was the case that a(i) could not be equal to zero then the prrof would be complete, but since there could be one case where a(i)=0 I am stuck and don't know how to move on from here.

Sorry I haven't Latexed anything, I don't know how!

- #8

Landau

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Yes, so far so good. However:Well basically the proof (as far as I can see) is asking to show that [itex]\sum _{i=1}^{k}\alpha_iv_i =0[/itex] implies that [itex]\alpha_i=0[/itex] for all i, alpha_i being an element of the field F.

How so?! No, this is not a correct conclusion. You are basically saying if a+b=0 then a=0 and b=0.[tex]\sum _{i=1}^{k}\alpha_iTv_i =\sum _{i=1}^{k}\alpha_ia_iv_i [/tex]

Now for this sum to equal zero then either aplha(i)=0 (as required) OR a(i)v(i)=0,

A more fruitful approach: by contradiction, assume that {v1,...,vk} is dependent. Let [itex]j\leq k[/tex] be the

[tex]\alpha_1v_1+...+\alpha_j v_j=0[/tex].

Now apply T to this equation, and think some more.

- #9

HallsofIvy

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phyzmatic, 0

We need to have 0 an eigenvector in order to say "the set of all eigenvectors of T, corresponding to eigenvalue [itex]\lambda[/itex], is a subspace".

- #10

Landau

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That's just a matter of definition. A lot of people exclude 0 from being an eigenvector. They then say "the set of all eigenvectors, together with the zero vector, is a subspace". Other authors, like Axler, do the same as you. It doesn't matter, as long as you be clear on your definition.phyzmatic, 0canbe an eigenvector. A number, [itex]\lambda[/itex] is defined to be an eigenvalue of operator T if and only if there exist anon-zerovector v such that [itex]Tv= \lambda v[/itex] but once we have thatanyvector, including 0, such that [itex]Tv= \lambda v[/itex] is an eigenvector of T.

We need to have 0 an eigenvector in order to say "the set of all eigenvectors of T, corresponding to eigenvalue [itex]\lambda[/itex], is a subspace".

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