-2
$\begingroup$

If I have an eigenvalue $\lambda$ with algebraic multiplicity $m_i$ and I find more than one eigenvector when computing $Ker(A - \lambda \, \mathbb{1})$, is it true that they are always linearly independent? I don't mean any scalar multiples of an eigenvector, but the two different ones I may find by solving the above equation.

Example:

$$A = \left(\begin{matrix} -5 & -6 & 3\\3 & 4 & -3\\0 & 0 & -2\end{matrix}\right)$$

$$\lambda_1 = -2 \text{ with } m_1 = 2 \\ \lambda_2 = 1 \text{ with } m_2 = 1$$

$$v_{11} = \left(\begin{matrix} -2 \\1 \\0 \end{matrix}\right) \text{ and }v_{12} = \left(\begin{matrix} 1 \\0 \\1 \end{matrix}\right)$$

$$v_2 = \left(\begin{matrix} -1 \\1 \\0 \end{matrix}\right)$$

Here $v_{11}$ and $v_{12}$ are linearly independent. But is this true in general?

$\endgroup$

closed as unclear what you're asking by JMoravitz, Daniel W. Farlow, ahulpke, Shailesh, Namaste Aug 25 '17 at 0:26

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ For just two vectors, either they are scalar multiples of each other or they are linearly independent... $\endgroup$ – Ian Aug 23 '17 at 16:45
  • 1
    $\begingroup$ If you have two nonzero vectors and the first one is not a scalar multiple of the other one, then they are linearly independent. This conclusion is no longer valid if you have 3 or more vectors. All this has nothing to do with eigenvectors and eigenvalues. $\endgroup$ – Thomas Aug 23 '17 at 16:45
  • $\begingroup$ "And I find two different eigenvectors when computing...are they always linearly independent?" That depends entirely on you and the method you learned. It is certainly possible that you pick up the problem, find one eigenvector, write it down and forget what it was, pick up the problem later and calculate another eigenvector without remembering what the first was, and it be exactly the same eigenvector and then obviously not independent. So, the question isn't so much if the eigenvectors are independent so much as whether your technique for the kernel produces independent vectors. $\endgroup$ – JMoravitz Aug 23 '17 at 16:46
  • $\begingroup$ The question of "If I pick $k$ different vectors from a kernel of dimension $k$, will they all be linearly independent?" (or equivalently if I pick $k$ different eigenvectors for the eigenvalue $v$ of geometric multiplicity $k$) the answer depends on how you picked them, but if you picked them randomly the answer is that they might be dependent. $\endgroup$ – JMoravitz Aug 23 '17 at 16:49
-1
$\begingroup$

I found the answer and the example is so simple I should have thought about it before asking. Anyways... The above statement is not true. Why?

Consider the eigenvectors of the $2 \times 2$ identity matrix:

$$A = \left(\begin{matrix} 1 & 0\\0 & 1\end{matrix}\right)$$ $$char_A(\lambda) = \det(\left(\begin{matrix} 1-\lambda & 0\\0 & 1-\lambda\end{matrix}\right)) = (1-\lambda)^2$$ $$\lambda_1 = 1 \text{ with } m_1 = 2$$

To find the eigenvectors corresponding to $\lambda_1$ we need to solve $Av = \lambda_1v = 1v = v$. But since $A$ is the identity matrix every vector $v \in \mathbb{R^2}$ satisfies this equation so every $v \in \mathbb{R^2}$ is an eigenvector to $\lambda_1$.

So $v_1 = \left(\begin{matrix} 1 \\0\end{matrix}\right)$ is one eigenvector as well as $v_2 = \left(\begin{matrix} 1 \\1\end{matrix}\right)$ as well as $v_3 = \left(\begin{matrix} 0 \\1\end{matrix}\right)$. Every pair of them is linearly independent but all three together are not. This answers the question.

EDIT: The original question could have as well been if the eigenspace of an eigenvalue contains only scalar multiples of the eigenvectors or also linear combinations. Since the eigenspace is a vector space, which naturally needs to contain also linear combinations, there are also eigenvectors which are themselfs linear combinations.

$\endgroup$

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