# Distinct Eigenvalues

Can zero be a distinct eigernvalue?

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Imagine you get to a characteristic equation

$$\lambda (\lambda-1)(\lambda -2)=0$$

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

Yes, it is no more special than having 1 or 2. All it means is that your matrix is singular. What we can't have though is a 0 eigenvector (for reasons that I can't recall right now).

This is what I thought, problem is that it messes with a proof I'm trying to complete.

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.

Landau
I don't understand how one number can be 'distinct', but anyway: 0 is perfectly allowed as eigenvalue, i.e. one of the a_i may or may not equal 0. Please write out explicitly what proof you had in mind and where you're stuck, then we can help you!

Well basically the proof (as far as I can see) is asking to show that the sum from i=1 to k of alpha(i)v(i) =0 implies that alpha(i)=0 for all i, alpha being an element of the field F.
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!

Landau
Hmm, this doesn't look right.
Well basically the proof (as far as I can see) is asking to show that $\sum _{i=1}^{k}\alpha_iv_i =0$ implies that $\alpha_i=0$ for all i, alpha_i being an element of the field F.
Yes, so far so good. However:
$$\sum _{i=1}^{k}\alpha_iTv_i =\sum _{i=1}^{k}\alpha_ia_iv_i$$

Now for this sum to equal zero then either aplha(i)=0 (as required) OR a(i)v(i)=0,
How so?! No, this is not a correct conclusion. You are basically saying if a+b=0 then a=0 and b=0.

A more fruitful approach: by contradiction, assume that {v1,...,vk} is dependent. Let $j\leq k[/tex] be the smallest index such that [itex]v_j\in\mbox{span}(v_1,...v_{j-1})$, i.e. j is minimal with the property of having a non-trivial linear combination
$$\alpha_1v_1+...+\alpha_j v_j=0$$.
Now apply T to this equation, and think some more.

HallsofIvy
Homework Helper
"Distinct" numbers just means different numbers. If a and b are eigen values of operator T and $a\ne b$ then they are "distinct" eigenvalues. If they happen to be 0 and 1, then, since they are different, they are "distinct". That's why Landau said "I don't understand how one number can be 'distinct'". The word "distinct" doesn't apply to a single number, it applies to sets of numbers- the numbers in the set are "distinct" if and only if they are all different.

phyzmatic, 0 can be an eigenvector. A number, $\lambda$ is defined to be an eigenvalue of operator T if and only if there exist a non-zero vector v such that $Tv= \lambda v$ but once we have that any vector, including 0, such that $Tv= \lambda v$ 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 $\lambda$, is a subspace".

Landau
phyzmatic, 0 can be an eigenvector. A number, $\lambda$ is defined to be an eigenvalue of operator T if and only if there exist a non-zero vector v such that $Tv= \lambda v$ but once we have that any vector, including 0, such that $Tv= \lambda v$ 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 $\lambda$, is a subspace".