5
$\begingroup$

I have two related questions. My question is hypothetical, i.e. not from an actual physical problem. If I give you a matrix and tell you that it has a repeated eigenvalue, can you say anything about whether the eigenvectors of that repeated eigenvalue are linearly independent or not? Is there a general procedure to check? I don't know the elements of the matrix itself so I can't work out the eigenvectors in the usual way.

On that note, can someone provide me an example of a matrix with a repeated eigenvalue but where the eigenvectors are not linearly independent?

Assume the field is complex numbers.

$\endgroup$
  • $\begingroup$ Only distinct eigenvalues produce distinct eigenvectors. So two of the same eigenvalues will produce the same eigenvectors (or linear combinations of each other). $\endgroup$ – dreamer Nov 26 '13 at 10:40
  • 2
    $\begingroup$ The thing to note here, perhaps, is that if an eigenvalue has algebraic multiplicity greater than $1$, then there's no 'the' eigenvector belonging to each 'instance' of the eigenvalue. There's simply the eigenspace (and generalized eigenspace). There aren't heuristics to finding any particular element of these spaces rather than any other: you simply find the spaces themselves along with (provided we're speaking of matrices) a basis for them. $\endgroup$ – Jonathan Y. Nov 26 '13 at 11:30
  • 2
    $\begingroup$ The question is confused, because "whether the eigenvectors of that repeated eigenvalue are linearly independent or not" does not mean anything unless you say which eigenvectors you are talking about. You can always find linearly dependent eigenvectors, just take one eigenvector, and then twice that vector. The question how many linearly independent eigenvectors one can find at least makes sense; however the answer is $\dim\ker(A-\lambda I)$ which just means you have to describe the eigenspace for$~\lambda$, but apparently you don't want to do that. $\endgroup$ – Marc van Leeuwen Nov 26 '13 at 13:11
4
$\begingroup$

Once we have one eigenvector $v$ corresponding to the eigenvalue $\lambda$, any non-zero scalar multiple of $v$ is also an eigenvector corresponding to $\lambda$. So rather than talking about the number of eigenvectors, we talk about the number of linearly independent eigenvectors ( for example corresponding to a particular eigenvalue), equivalently the dimensions of the eigenspaces (corresponding to a particular eigenvalue).

One useful thing is that for an eigenvalue, its geometric multiplicity cannot exceed its algebraic multiplicity. In general, one is often forced to solve for the eigenspace. Depending on the information you have, you may be able to make some deductions without solving for the eigenspace/s.

Again in regard to your last question, you need to be clear on what you are asking.

For example $A = \left(\begin{matrix} 1 & 1\\0 & 1\end{matrix}\right)$ has one repeated eigenvalue (multiplicity two) $\lambda =1$. Solving for the corresponding eigenspace, we get $v = t \left(\begin{matrix} 1\\0\end{matrix}\right)$, $t\in\mathbb{R}$. So we can write down only one eigenvector, if we are asked to give the set of linearly independent eigenvectors (any nonzero scaler multiple of $\left(\begin{matrix} 1\\0\end{matrix}\right)$ will do).

On the other hand, the $2\times 2$ identity matrix again has only one eigenvalue (multiplicity two) but its corresponding eigenspace has dimension $2$ (note any nonzero vector is an eigenvector). So we have two linearly independent eigenvectors.

$\endgroup$
3
$\begingroup$

Let $A$ be an $n \times n$ matrix with distinct eigenvalues $L_1, ..., L_k$ (note that some of these may have multiplicity > $1$, i.e, be repeated). Let $B_i$ be a basis for the eigenspace of $L_i$ and let $B$ be the union of $B_1, ..., B_k$. Then elements of $B$ are linearly independent.

$\endgroup$

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.