How do I restrict k to ensure my matrix has exactly 3 distinct eigenvalues? $$A=\begin{bmatrix}-1&-1&0\\-12&3&-1\\k&0&0\end{bmatrix}$$
How do I restrict $k$ to ensure that my matrix has 3 distinct real eigenvalues? I tried going about it the long way by using the characteristic polynomial and factorizing, but I'm sure there is a faster way.
 A: Another approach: 
The characteristic polynomial is $c(x) = x^3 - 2x^2 - 15x - k$. 
Using basic calculus, $p(x) = x^3 - 2x^2 - 15x$ has three distinct real roots with local  max and min of $400/27$ and $-36$ (respectively at $x = -5/3$ and $x = 3$). 

Hence $c(x) = p(x) - k$ will likewise have three distinct real roots iff $k \in (-36,400/27)$.
A: Hint: 


*

*Find the characteristic polynomial.

*A polynomial has a multiple root if and only if its discriminant equals to $0$. Or, in other words, a polynomial has a multiple root if and only if it has a common root with its derivative.
A: Since you don't have access to the cubic discriminant during the exam, it looks to me like consulting the characteristic polynomial and factorizing is your best bet. Ultimately this is an exercise in calculus. Luckily the matrix is given to you so that this polynomial isn't too bad. If my calculation is correct I find that $$\det(A-I\lambda) = -\lambda^3+2\lambda^2+15\lambda+k$$ The only way you'll get duplicate eigenvalues is if your cubic bounces off the $x$-axis. We know at that point the derivative will be equal to zero, so we can set $$3\lambda^2-4\lambda-15 = 0 $$ and get roots $\lambda = 3,-\frac{5}{3}$ where the characteristic polynomial has slope zero. Using the first derivative test we see that $\lambda = 3$ is our local maximum and $-\frac{5}{3}$ is a local minimum. Now we want to choose a $k$ so that we force that minimum underneath the $x$ axis, but keeps the local maximum above the $x$-axis. Then we'll know the cubic will pass through zero on the way to the minimum, pass through again on the way to the maximum and pass through a third time as the cubic tends toward $-\lambda^3$, yielding three distinct roots. Letting $f(\lambda) = -\lambda^3+2\lambda^2+15\lambda+k$ this simply means we want $$f(3)>0 \quad \text{and} \quad f(-5/3)<0$$ You'll get two intervals dependent on $k$ from solving each inequality. The intersection of these two intervals will tell you where you can pick $k$ from. I probably messed up my math somewhere along the way, but the idea should still hold.
