I have the following matrix
$$A = \begin{pmatrix}0 & 1 & -k & 0 \\\ k & 2 & 0 & -k \\\ -1 & 0 & -2 & 1 \\\ 0 & 1 & k & 0 \end{pmatrix}$$
I want to find all complex values for $k$ such that $A$ is diagonalizable
What I've done so far: computed the characteristic polynomial to be $\chi_A = x(x^3 - (4+2k)x + 4k)$
If $k=0$ then $0$ has an algebraic multiplicity of $2$. Replacing k with 0 in the matrix shows three columns that are multiples of each other. Then, $ker(A) $ has dimension $2$ so $A$ is diagonalizable
For $k \ne 0$, what I did was compute $A^2$ and $A^3$. Since the characteristic polynomial has 0 as a root, so does the minimal polynomial $m_A$
This implies that, were the minimal polynomial to be of degree 2, the coefficient $\beta$ of the identity in the linear combination $A^2 = \alpha A + \beta I$ is $0$. And computing $A^2$ clearly shows that $A^2$ is not a multiple of A
A similar thing happens when asking if the minimal polynomial could have a degree of 3. When $k \ne 0$, $A^3$ is not a linear combination of previous powers of $A$
Then, the minimal polynomial is the characteristic polynomial
We know, then, that A is diagonalizable if and only if $m_A$ has all simple roots. I tried this approach by trying to find $gcd(m_A, m_A')$ but I end up getting a cubic polynomial on k in the denominator (which complicates things a lot) so I would like a simpler approach
I also looked at the trace of $A$ which is $0$. If I wanted to find real values for k this would be really helpful: $m_A$ has three unknown roots (0 is always a root) and $m_A$ would then be a polynomial in $\mathbb R[X]$. Then, a poltnomial of third degree always has a real root, which would mean that either there's a real root and two complex conjugate roots and then $A$ is diagonalizable, or that all three roots are real
But, if all three roots are real, I can find the values for k which are NOT diagonalizable by setting two roots to be equal.
Alas, I'm supposed to find all complex values of k, so this is only good for finding strictly real ones
I've exhausted all the approaches I can think of so some tip or guidance is much appreciated
We can do calculus on the characteristic polynomial. I will assume that you found it correctly.
$$\chi_A'(x) = 4x^3-(8+4k)x+4k$$
It's zeros are where $$x^3-(2+k)x+k = 0$$
For geometric multiplicity to be larger than $1$, multiple roots must exist at any $x$ and for polynomials, derivative must be zero at the point of multiple roots.
So we seek simultaneous solution $$\cases{x(x^3-(4+2k)x+4k=0\\x^3-(2+k)x+k=0}$$
We can now eliminate $x^3$ by substituting second into first:
$$x(-(2+k)x + 3k) = 0$$ Now this should be easier to solve.