Finding real, distinct eigenvalues for arbitrary constants

Question

Let $A= \begin{bmatrix} 1 & 1 & 0 \\ -4 & -3 & 1 \\ k & 0 & 0 \end{bmatrix}$. Find all values of $k$ such that $A$ has three real distinct eigenvalues. I have obtained the characteristic polynomial of this matrix, it is $\lambda^3 + 2\lambda^2 + \lambda - k$. I can't seem to figure out how to solve this polynomial such that all three eigenvalues are distinct and real.

Answer 1

The polynomial $p(l) = l^3+2l^2+l-k$ will have at-least one root.

Now lets us examine the shape of the polynomial. We have $p(l) \to \infty$ as $l \to \infty$ and $p(l) \to -\infty$ as $l \to -\infty$.

For it to have three real roots, we need the local maximum of the polynomial to be positive and the local minimum of the polynomial to be negative. The critical points are obtained by setting the derivative to zero, which gives us $$3 l^2 + 4l+1 = 0 \implies l = -1, -1/3$$ Hence, we need $k$ such that the polynomial is negative for $l=-1/3$ and positive for $l=-1$, i.e., $$-1/27+2/9-1/3 - k <0 \implies k > -\dfrac4{27}$$ $$-1+2-1 - k >0 \implies k <0$$

Answer 2

The cubic's discriminant must be positive:

$$\Delta:=2^2\cdot 1^2-4\cdot 1\cdot1^3-4\cdot2^3\cdot(-k)-27\cdot1^2\cdot k^2+18\cdot1\cdot2\cdot1\cdot(-k)>0\iff$$

$$4-4+32k-27k^2-36k>0\iff 27k^2+4k<0\iff\;\ldots$$