3
$\begingroup$

Consider the matrix

$$ A = \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ k & 3 & 0 \end{array} \right) $$

where k is an arbitrary constant. For which values of k does A have three distinct real eigenvalues? For which k does A have two distinct eigenvalues?

Hint: Graph the function $$ g(\lambda) = \lambda^3 - 3\lambda $$

Find its local maxima and minima.

So far I've tried this:

Let $$ B = A - \lambda * I = \left( \begin{array}{ccc} -\lambda & 1 & 0 \\ 0 & -\lambda & 1 \\ k & 3 & -\lambda \end{array} \right) $$ where I is the identity matrix.

Then:

$$ p(\lambda) = det(B) = -\lambda^3 + 3\lambda + k $$

Then I found the critical points by setting the derivative to 0.

$$ p'(\lambda) = -3\lambda^2 + 3 = 0 \\ \lambda = 1 \\ \lambda = -1 $$

Plugging this back into the original equation I got $$ p(1) = -1^3 + 3(1) + k > 0 \\ p(-1) = 1^3 + 3(-1) + k < 0 \\ $$

This gave me the values of k $$ k > -2 \\ k < 2 $$

This is only one set of values for k. Is the value of k such that there are three distinct real eigenvalues or two? Also, if this is correct, how do I find the other set of values for k? And finally, did I missing something since the hint gave me an equation that I didn't find. I'm not really sure where that equation came from.

Edit: I realize where that equation that's given in the hint came from. I didn't realize it was the characteristic polynomial that I calculated but just in a different form.

$\endgroup$
  • $\begingroup$ The characteristic equation is $-\lambda^3 + 3\lambda + k = 0$. $\endgroup$ – hjpotter92 Apr 26 '15 at 2:57
  • $\begingroup$ @hjpotter92 I'm not really sure what to do with that. I was under the impression that you had to solve for lambda first, but you can't with just that equation $\endgroup$ – user3370201 Apr 26 '15 at 3:06
  • $\begingroup$ Two distinct eigenvalues is for $p'(\lambda)=0$ then $p(\lambda)=0$ has only 2 solutions for values of $k$, as graph is tangent to the $p=0$ line at these points of intersection (they are "doubled") $\endgroup$ – Alexey Burdin Apr 26 '15 at 3:09
  • $\begingroup$ @AlexeyBurdin So what I've done so far is finding k for two distinct eigenvalues right? If so, then how do I find k for three distinct eigenvalues? $\endgroup$ – user3370201 Apr 26 '15 at 3:17
1
$\begingroup$

the characteristic equation is $$3\lambda - \lambda^3 = k.$$ the graph has a local max $(1,2)$ and local min at $(-1-2).$

so if $$ -2 < k < 2, \text{the char equation has three distinct roots}\\ k = \pm 2, \text{the char equation has a repeated real root}\\ |k| >2 \text{char equation has one real root and two complex conjugate roots.}$$

$\endgroup$
  • $\begingroup$ This makes a lot of sense. Just to clarify, this means that by increasing the range of k, there are more distinct roots right? $\endgroup$ – user3370201 Apr 26 '15 at 3:32
  • $\begingroup$ @user3370201, you have three distinct roots for $k$ in the range $-2$ to $2.$ $\endgroup$ – abel Apr 26 '15 at 3:34
2
$\begingroup$

By plotting $p(\lambda)$ for the case where $k = 0$, we obtain the following curve:

enter image description here

By varying $k$, we are free to move the curve up or down. Now recall that having three distinct real eigenvalues corresponds to making $p(\lambda)$ cross the horizontal axis exactly three times. By computing the local extrema, notice that this can happen by vertically translating up by no more than $2$ units or vertically translating down by no more than $2$ units. So $k \in (-2, 2)$ corresponds to three distinct real eigenvalues and $k \in \{-2, 2\}$ corresponds to two distinct real eigenvalues and $k \in (-\infty, -2) \cup (2, \infty)$ corresponds to one real eigenvalue.

$\endgroup$

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.