# Distinct Eigenvalues

1. May 9, 2010

### Juggler123

Can zero be a distinct eigernvalue?

2. May 9, 2010

### phyzmatix

Imagine you get to a characteristic equation

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

3. May 9, 2010

### Juggler123

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

4. May 9, 2010

### phyzmatix

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).

5. May 9, 2010

### Juggler123

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.

6. May 9, 2010

### 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!

7. May 9, 2010

### Juggler123

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!

8. May 9, 2010

### Landau

Hmm, this doesn't look right.
Yes, so far so good. However:
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.

9. May 10, 2010

### HallsofIvy

"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".

10. May 10, 2010

### Landau

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.