# Find value of k for distinct eigenvalues

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.

• The characteristic equation is $-\lambda^3 + 3\lambda + k = 0$. – hjpotter92 Apr 26 '15 at 2:57
• @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 – user3370201 Apr 26 '15 at 3:06
• 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") – Alexey Burdin Apr 26 '15 at 3:09
• @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? – user3370201 Apr 26 '15 at 3:17

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 $$-2 < k < 2, \text{the char equation has three distinct roots}\\ k = \pm 2, \text{the char equation has a repeated real root}\\ |k| >2 \text{char equation has one real root and two complex conjugate roots.}$$
• @user3370201, you have three distinct roots for $k$ in the range $-2$ to $2.$ – abel Apr 26 '15 at 3:34
By plotting $p(\lambda)$ for the case where $k = 0$, we obtain the following curve:
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.