1
$\begingroup$

If a matrix $A$ has repeated eigenvalues, its eigenspace matrix would be dependent because there isn't enough eigenvector to tag along with the eigenvalue.

But if a matrix $A$ has repeated eigenvalues of $0$, that means the matrix $A$ is singular, then the dimension of the nullspace of matrix $A$ is the number of $0$ eigenvalues that matrix $A$ has. Although I read that the eigenvectors of the matrix A are dependent if there are repeated eigenvalues, I'm thinking why couldn't the eigenvectors be independent if the repeated eigenvalues were $0$?

Say in a matrix that has 2 eigenvalues of $0$, this implies that $dim(N(A))=2$, then wouldn't the eigenvector to each of this eigenvalue of $0$ be the 2 independent nullspace vectors? Then when putting all the eigenvectors together to form an eigenspace matrix, they will all be independent, wouldn't it?

What do I don't understand that lead me to this thought? Thanks for any help.

Update:

With the example suggested by Didier Piau, $$A=\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}$$ The $rank(A)=1$, $N(A)=\begin{bmatrix} 1\\ 0 \end{bmatrix}$, $eigenvals(A)=0, 0$ and true enough, the eigenvector can only be $\begin{bmatrix} 1\\ 0 \end{bmatrix}$ and therefore the eigenspace is dependent. There isn't enough eigenvectors to tag to the other eigenvalue of zero.

But say with another example of matrix $B$, $$B=\begin{bmatrix} 1 & 2 & 3\\ 1 & 2 & 3\\ 2 & 4 & 6 \end{bmatrix}$$ The $rank(B)=1$, the $N(A)=c_{1}\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} -3\\ 0\\ 1 \end{bmatrix}, c_{1}, c_{2} \in \mathbb{R}$, $eigenvals(B)=9, 0,0$ then the eigenvectors are... $$B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=9\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}$$ Then for the next two eigenvalues of $0$, which eigenvector do I choose to use as their eigenvector? Do I use one for each or only just one of the two or both of the two? $$B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=0\begin{bmatrix} -2\\ 1\\ 0 \end{bmatrix}$$ and it could also be... $$B\begin{bmatrix} 0.5\\ 0.5\\ 1 \end{bmatrix}=0\begin{bmatrix} -3\\ 0\\ 1 \end{bmatrix}$$ Then for the eigenspace, should it be: $\begin{bmatrix} 0.5 & -2 & 3\\ 0.5 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}$ or $\begin{bmatrix} 0.5 & -2 & -2\\ 0.5 & 1 & 1\\ 1 & 0 & 0 \end{bmatrix}$ or $\begin{bmatrix} 0.5 & 3 & 3\\ 0.5 & 0 & 0\\ 1 & 1 & 1 \end{bmatrix}$?

|cite|improve this question
$\endgroup$
  • $\begingroup$ You could test the successive phases of your understanding of your question on the matrix A=[0,1|0,0], since this example already exhibits many of the relevant features of the problem. For eample, I doubt you can find two linearly independent vectors x and y such that Ax=Ay=0. $\endgroup$ – Did Aug 16 '11 at 9:20
  • $\begingroup$ Say when a matrix A has an eigenvalue of $0$, the eigenvector to this zero eigenvalue will directly be the nullspace vector of matrix A, right? Then if the nullspace has more than one dimension, wouldn't there naturally have more than one independent nullspace vector to attach to the zero eigenvalues too? $\endgroup$ – xenon Aug 16 '11 at 9:24
  • $\begingroup$ I'm still a little confused with that eigenspace has dimension grater than 1. I have the example suggested by Didier Piau, and indeed, there is only one eigenvector to both the zero eigenvalues. But funnily, in another example that I tried, it isn't the case. I've updated my question to show what I'm confused with. Thanks. $\endgroup$ – xenon Aug 16 '11 at 10:11
  • $\begingroup$ So, since there are 2 independent eigenvectors for the 2 eigenvalues of zero, I can pick any one, just only one, of the two eigenvectors as the eigenvector tagged to the eigenvalue of $0$. Then, the eigenvalue of $0$ would be a combination of this chosen eigenvector. Therefore, the eigenspace would be dependent because the last two column vectors are the same eigenvectors. Am I getting this right? $\endgroup$ – xenon Aug 16 '11 at 10:29
  • $\begingroup$ You found a perfectly good basis for the nullspace right at the beginning, and you correctly wrote every element of the nullspace as a linear combination of that basis. If you are looking for a matrix $M$ so that $M^{-1}BM$ is diagonal, then the columns of $M$ need to be a basis of eigenvectors for $B$. So in your case, you can use the eigenvector for 9 as the first column and the two independent eigenvectors for $0$ which you correctly found as the other two columns. Hope this helps. $\endgroup$ – Geoff Robinson Aug 16 '11 at 10:44
2
$\begingroup$

One can test one's understanding of this question on the matrix A=[0,1|0,0], since this example already exhibits many of the relevant features of the problem. For eample, one cannot find two linearly independent vectors v and w such that Av=Aw=0 (try to solve for v the equation Av=0) although 0 is a double root of the characteritic polynomial det(XI-A)=X^2.

This is not specific to the eigenvalue 0. Recall that for any eigenvalue one distinguishes its algebraic multiplicity n (in this case, n=2) from its geometric multiplicity m (in this case, m=1). See here for more details and a worked out example.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I'm still a little confused this. I have updated my question with another example which I'm not sure which eigenvector to choose for the eigenvalues. $\endgroup$ – xenon Aug 16 '11 at 10:12
  • $\begingroup$ For B the eigenvalues are 9 (AM=1) and 0 (AM=2) (with AM=algebraic multiplicity). For the eigenvalue 9, GM=1 (GM=geometric multiplicity) because GM is always between 1 and AM hence when AM=1, AM=GM. For the eigenvalue 0, one could a priori have GM=1 or GM=2 but Bv=0 means that the coordinates of v are such that x+2y+3z=0 hence the eigenspace is a plane, in other words GM=2. Since AM=GM for every eigenvalue of B, B is diagonalizable (this is equivalent). (You should read carefully the WP page I linked to, it explains all this and much more.) $\endgroup$ – Did Aug 16 '11 at 10:28
-1
$\begingroup$

you have 2 eigenvectors that represent the eigenspace for eigenvalue = 1 are linear independent and they should both be included in your eigenspace..they span the original space... note that if you have 2 repeated eigenvalues they may or may not span the original space, so your eigenspace could be rank 1 or 2 in this case.

|cite|improve this answer
$\endgroup$

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.