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.
Best Answer
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
$$D(f) = -27k^2+32k >0\iff 0<k<32/27.$$