The following 3x3 matrix has three real unique eigenvalues if and only if k is between what range?

Question

I need to find a range for possible values of k in matrix A so that it has three REAL and UNIQUE eigenvalues

\begin{equation*} A = \begin{pmatrix} -1 & 1 & 0 \\ -9 & 5 & 1 \\ k & 0 & 0 \end{pmatrix} \end{equation*}

I know that the characteristic equation of this matrix is:

\begin{equation*} -\lambda^3+4\lambda^2-4\lambda+k=0 \end{equation*}

Where lambda represents the eigenvalues

I then used the following equation:

\begin{equation*} -(\lambda-a)(\lambda-b)(\lambda-c)=0 \end{equation*}

Where a, b, and c are the three unique eigenvalues

To get that: \begin{equation*} a+b+c=4 \end{equation*} \begin{equation*} abc=k \end{equation*} \begin{equation*} -ab-ac-bc=-4 \end{equation*}

I have no idea what to do now.

Answer 1

Given a cubic in the form $$f(x) = A x^3 + B x^2 + C x + D$$ the discriminant is defined as $$D(f) = -27 A^2 D^2 + 18 ABCD - 4 A C^3 - 4 B^3 D + B^2 C^2.$$ $D(f)>0\iff \text{f has 3 distinct real roots}.$

We compute $D(f)$ for your cubic, with $A= -1, B= 4, C= -4, D =k$. This gives us
$$D(f) = -27k^2+32k >0\iff 0<k<32/27.$$

Answer 2

The discriminant of that polynomial is $32k-27k^2$. So, your matrix will have three real distinct roots if and only if $32k-27k^2>0$, wich means that $k\in\left(0,\frac{32}{27}\right)$.

Answer 3

Consider the function $f(x)=x^3-4x^2+4x$. Then $$ f'(x)=3x^2-8x+4 $$ that vanishes at $2/3$ and $2$. Since $$ f(2/3)=\frac{8}{27}-\frac{16}{9}+\frac{8}{3}=\frac{32}{27} $$ and $$ f(2)=0 $$ the equation $f(x)=k$ has three distinct solution if and only if $$ 0<k<\frac{32}{27} $$