4
$\begingroup$

This question already has an answer here:

I have a question about the following matrix:

$$ \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3 \\ \end{bmatrix} $$

Find the eigenvalues without calculations and define your answer. Now, I was thinking about this problem. And I thought, yeah ok if you try the vector (1,1,1), you can find 6 as one eigenvalue (and I know you have a double multiplicity 0 too). But than you are doing sort of guessing/calculation work.

I see that the columns are linearly dependant. So I know the dimension of the column space and of the null space.

Thank you in advance.

EDIT: follow up question:

Ok, so you find that the dimension of the null space is 2, so there are 2 eigenvectors when the eigenvalue is 0. Now my question is, can the dimension of the eigenspace be bigger than the amount of eigenvalues? I guess not. I know it can be smaller

|cite|improve this question
$\endgroup$

marked as duplicate by Marc van Leeuwen linear-algebra Jan 16 '15 at 12:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ You didn't under stood me, I said you need to edit your question in the comment:" can the dimension of the eigenspace be bigger than the amount of eigenvalues"? It dosent make any sense. $\endgroup$ – Ofir Schnabel Jan 16 '15 at 12:12
  • $\begingroup$ If you have an eigenspace for $\lambda$ of dimension$~d$, then clearly $(X-\lambda)^d$ divides the characteristic polynomial. So the "amount of eigenvalues" (I suppose you mean the multiplicity of the eigenvalue as root of the characteristic polynomial) cannot be less than the dimension of the eigenspace. $\endgroup$ – Marc van Leeuwen Jan 16 '15 at 12:42
  • $\begingroup$ That is what I meant. So to make it a hundred percent clear: You cannot have: 3 eigenvalues for example: lambda = 3,4,5 for a 3x3 matrix and have 4 linearly independent eigenvectors. $\endgroup$ – Thomas Jan 16 '15 at 13:44
6
$\begingroup$

Notice that rank=1 and hence $0$ is an eigenvalue of multiplicity $2$. Then trace=sum of eigenvalue and hence the last eigenvalue is $6$.

It is also rather easy to find all eigenvectors without a lot of work. For $6$ the vector is $(1,1,1)$. For $0$ you can take basis $(2,-1,0),(3,0,-1)$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Ok, can you also say. The determinant of this matrix is 0, so zero must be an eigenvalue (ok that is the same you said). But the multiplicity, is it always amount of rows minus the rank ? $\endgroup$ – Thomas Jan 16 '15 at 11:52
  • $\begingroup$ That is right. This is the Null dimension of $A$ or the dimension of kernel $A$ when thinking of $A$ as operator. $\endgroup$ – Ofir Schnabel Jan 16 '15 at 11:53
  • $\begingroup$ Ok, so you find that the dimension of the null space is 2, so there are 2 eigenvectors when the eigenvalue is 0. Now my question is, can the dimension of the eigenspace be bigger than the amount of eigenvalues? I guess not. I know it can be smaller. $\endgroup$ – Thomas Jan 16 '15 at 11:56
  • 1
    $\begingroup$ I think you need to edit your question. $\endgroup$ – Ofir Schnabel Jan 16 '15 at 11:57
  • $\begingroup$ You cannot have more than 1 linearly independant eigenvector corresponding to 1 eigenvalue which has the multiplicity 1. $\endgroup$ – Thomas Jan 16 '15 at 12:20

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