0
$\begingroup$

I have the following matrix $A$

$$ A = \begin{bmatrix} 4 & 3 & -3 \\ -3 & -2 & 3 \\ 3 & 3 & -2 \end{bmatrix} $$

and it has eigenvalues $\lambda_1 = -2$ with multiplicity $k = 1$, and $\lambda_2 = 1$ with $k = 2$. I found the corresponding eigenvector $\mathbf{k_1}$ of $\lambda_1$ as $\mathbf{k_1} = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}. $ For the eigenvectors $\mathbf{k_{2,1}, k_{2,2}}$ of $\lambda_2$, I substituted $\lambda_2$ to $(A - \lambda I)\mathbf{k} = \mathbf{0}$ and get

$$ \begin{bmatrix} 3 & 3 & -3 \\ -3 & -3 & 3 \\ 3 & 3 & -3 \end{bmatrix} \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $$

which tells us that $\mathbf{k} = \begin{bmatrix} k_1 \\ k_2 \\ k_3 \end{bmatrix} = \begin{bmatrix} k_1 \\ k_2 \\ k_1 + k_2 \end{bmatrix} = k_1 \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + k_2 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} $.

My question is, would it be reasonable to say that the two corresponding eigenvectors of $\lambda_2$ is $\mathbf{k_{2,1}} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $ and $\mathbf{k_{2,2}} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} $ since $ \text{det}\left(\begin{bmatrix} \mathbf{k_1} & \mathbf{k_{2,1}} & \mathbf{k_{2,2}} \end{bmatrix}\right) = 1 \neq 0$? Or are there more conditions needed to find the eigenvectors of a repeated eigenvalues? Lastly, may I ask what are the general procedures in finding the eigenvectors of a repeated eigenvalues?

$\endgroup$
3
  • 2
    $\begingroup$ Your work seems fine (I didn't check the calculations though). I have an issue with you saying that (101) and (011) are the eigenvectors corresponding to $\lambda_2$. These two specific eigenvectors form a basis of the eigenspace corresponding to $\lambda_2$. That's the best you can say. If you choose any other basis, it would have been equally correct and useful. Other than that, everything you say is correct and the general procedure is precisely what you are doing. $\endgroup$ Apr 11, 2021 at 8:54
  • $\begingroup$ @Mathematician42 But isn't it enough to prove that if I found two eigenvectors that are linearly independent of each other and each solves the equation $(A-\lambda I)\mathbf{k} = \mathbf{0}$, then I have found the eigenvectors of $\lambda$ given that it is a repeated eigenvalue of $A$? $\endgroup$
    – sinclair
    Apr 11, 2021 at 8:59
  • $\begingroup$ In general, the dimension of the eigenspace $E_\lambda=\{X\mid (A-\lambda I)X=0\}$ is bounded above by the multiplicity of the eigenvalue $\lambda$ as a root of the characteristic equation. In this example, the multiplicity of $\lambda=1$ is two, so $\dim(E_\lambda)\leq 2$. Hence $\dim(E_\lambda)=1$ or $\dim(E_\lambda)=2$. As you explicitly find two independent eigenvectors belonging to $\lambda=1$, you indeed found that $\dim(E_\lambda)=2$. $\endgroup$ Apr 11, 2021 at 9:11

1 Answer 1

1
$\begingroup$

Your procedure is correct. You asked about a general procedure. It turns out that not all matrices have full sets of eigenvectors. If a matrix has all distinct eigenvalues with algebraic multiplicity $1$, then it does have a full set. But sometimes the procedure you used comes up short. You can always find one eigenvector corresponding to a given eigenvalue (otherwise it wouldn't be an eigenvalue) but the geometric multiplicity (number of linearly independent eigenvectors, or dimension of the eigenspace) can sometimes be less than the algebraic multiplicity.

In that situation, you search for generalized eigenvectors, by solving powers of the equation you used: $(A-\lambda I)^n = 0$. There is more detail to it, but that is the topic, and it leads to the Jordan Canonical Form, which is "as close to diagonal as you can get" for a matrix that cannot be diagonalized (because it doesn't have enough eigenvectors.)

$\endgroup$

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.