0
$\begingroup$

For an $n\times n$ matrix $A$ of rank $r$, for $\lambda = 0$, I think the dimension of the eigenspace (equal to the null space of $A$) is always $n-r$. Is it possible to show whether the multiplicity of $\lambda = 0$ can exceed $n-r$?

If the multiplicity of $\lambda = 0$ always equal the dimension of its eigenspace ($n-r$), why is this true yet for a matrix like $\begin{bmatrix}3&1\\0&3\end{bmatrix}$, we have $\lambda = 3$ with a multiplicity of 2 but only one eigenvector?

$\endgroup$
  • $\begingroup$ That matrix has characteristic polynomial $(3-\lambda)^2 - 1$, whose roots are $2$ and $4$. $\endgroup$ – Kaj Hansen Jul 26 '19 at 20:29
  • $\begingroup$ @KajHansen I think the characteristic polynomial is just $(3-\lambda)^2$? At least that's what I get from this $\endgroup$ – Yandle Jul 27 '19 at 23:02
1
$\begingroup$

The multiplicity of an eigenvalue known as algebraic multiplicity is $\ge $ than the geometric multiplicity (geometric multiplicity is $n-r$ for your exemple of $\lambda=0$). A classic fact.

$\endgroup$
  • $\begingroup$ My question arises from this video where the professor had a rank 1 3x3 matrix and immediately wrote down eigenvalue = 0 twice after stating that the nullity = 2. I'm guessing for this example he knows the third eigenvalue one is not zero because the trace is 1? $\endgroup$ – Yandle Jul 28 '19 at 21:40
  • $\begingroup$ If rank is $1$ at least two eigenvalues are zeros (algebraic multiplicity but they could be three zeros), yes it means $\ker(matrix)$ has dimension two (geometric multiplicity) the last eigenvalue is concluded as you said. $\endgroup$ – Toni Mhax Jul 28 '19 at 22:06
1
$\begingroup$

You may be under the impression that R^n is spanned by the eigenvectors of a matrix this is actually not true however see https://en.wikipedia.org/wiki/Jordan_normal_form
It is typically true however one can think of large blocks in the JOrdan normal form as a "degeneracy" so to speak. For a simple counter example to your claim consider the right shift operator where the right most element is deleted one can easily see any eigenvalue must be 0 but the dimension of the nullspace is merely one you may also want to look up idempotent and nilpotent operators

$\endgroup$

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.