2
$\begingroup$

I am learning diagonalization of matrices. We are given the following theorem:

If $A$ is an $n\times n $ matrix with $n$ distinct eigenvalues, then $a$ is diagonalizable

Now the proof is:

let $v_1,v_2,\dots,v_n$ be eigenvectors corresponding to the $n$ distinct eigenvalues of $a$. (Why could there not me more than $n$ such eigenvectors?) $v_1, v_2,\ldots,v_n$ are linearly independent, so by (a specific theorem) A is diagonalizable.

Now the question posed in the proof has gotten my thinking, why cant there be more than n eigenvectors correspdonign to n distinct eigenvalues? because an eigenvalue can correspond to only one eigenvalues? because they are distinct?

$\endgroup$
  • 1
    $\begingroup$ If there is one eigenvector belonging to someeigenvalue then there are as many (belonging to the same eigenvalue) as non-zero scalars the field has. What the posting actually says (apparently, at least, this is the intention conveyed by that word "such" there) is that eigenvectors belonging to different eigevalues are linearly independent ... $\endgroup$ – DonAntonio Jul 1 '13 at 12:16
  • 1
    $\begingroup$ If $A$ is a $n \times n$ matrix over $\mathbb R$ or $\mathbb C$, then the maximum number of eigenvalues is given by the degree of the characteristic polynomial $p(\lambda)=det(A-\lambda I)$, i.e. $n$. As the eigenvalues of $A$ are the roots $p(\lambda)=0$, then you have $n$ in general complex eigenvalues, counted with multiplicities. $\endgroup$ – Avitus Jul 1 '13 at 12:43
  • $\begingroup$ @Avitus, he's not asking about eigenvalues but about eigevectors... $\endgroup$ – DonAntonio Jul 1 '13 at 13:01
  • $\begingroup$ @DonAntonio you are right: there are already answers on eigenvectors down under; I thought that fixing ideas on eigenvalues could be useful :) $\endgroup$ – Avitus Jul 1 '13 at 13:05
  • $\begingroup$ always helps :) $\endgroup$ – WiseStrawberry Jul 1 '13 at 13:10
2
$\begingroup$

this is late, and the answers given here are great, but based on the poster's last comment, and the comment left by epoweritheta, I think there is still some misunderstanding here.

It is true that you can get more than one eigenvector for a single eigenvalue: consider the matrix $$A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 2 & 4 \end{pmatrix}$$

However, you can have at most $n$ linearly independent eigenvectors for a given matrix. There are many ways to see why this has to be so, but here's one: An $n \times n $ represents a linear operator, that is, it is a linear transformation from an $n$-dimensional vector space to itself. Now, in an $n$-dimensional vector space, any set of vectors having more than $n$ vectors is DEPENDENT. To see this, you could set up the linear independence equation, $$c_1\vec{v_1} + c_2 \vec{v_2} + \cdots + c_n \vec{v_n} + \cdots + c_{n+m} \vec{v_{n+m}} = \vec{0}$$ You will find that when you augment the matrix for that system, you will get an $n \times (n+m)$ matrix and hence you will have less pivots than there are columns upon performing Gaussian elimination. This gives rise to a solution that depends on parameters, which means you won't have a unique solution and hence you get dependence.

So having $n$ eigenvalues, each will have at least one eigenvector (by construction--if you want to know why each eigenvalue will have at least one, you can ask me--the short answer is we find eigenvalues by setting up a homogeneous system whose coefficient matrix has determinant zero, hence by the equivalence theorem, this system is guaranteed to have a non-zero vector-solution, in fact, it will have infinitely many such solutions), but there will be only $n$ such vectors that are linearly independent. And then you can apply the theorem that says having $n$ linearly independent vectors implies being diagonalizable (and vice versa).

$\endgroup$
  • $\begingroup$ You absolute hero. This was in my undergraduate year, I graduated in January and I sitll enjoy your answer very much. $\endgroup$ – WiseStrawberry Aug 9 '18 at 17:16
2
$\begingroup$

Who said there can't be more than $n$ eigenvectors? Nothing in the passage you quoted depends on there being exactly $n$ eigenvectors; all that's required is that there are (at least) $n$ eigenvectors.

Having said that, eigenvectors belonging to distinct eigenvalues are linearly independent. If you have more than $n$ eigenvectors, then any two belonging to the same eigenvalue will be linearly dependent, and so of no use for diagonalization.

$\endgroup$
  • 1
    $\begingroup$ oh i get it, the n distinct eigenvalues need at least n corresponding eigenvectors, because an eigenvalue needs an eigenvector to satisfy $Av=\lambda v$ , doesnt really matter which. as long as it satisfies the equation. $\endgroup$ – WiseStrawberry Jul 1 '13 at 13:08
2
$\begingroup$

Why could there not be more than n INDEPENDENT eigenvectors for a n×n matrix ?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.