How to solve for x in matrix? I solved a couple of "regular" matrices but now I ran into this:
$$\begin{vmatrix}
x & 0  & 1  & 0 \\ 
0 & 1 & 0 & x\\ 
1 & 0 & x & 0 \\ 
0 & x & 0 & 1
\end{vmatrix} = 0$$
And the question is to solve the equation. However I do not recognize this format. Are there any special techniques to solve a system as this one above? What is it called? Im not asking for you to solve this particular one, but to point me in a direction where I can read/watch on how to solve these type of equations.
Cheers
 A: Evaluating the determinant
$$
\begin{vmatrix}
x & 0  & 1  & 0 \\ 
0 & 1 & 0 & x\\ 
1 & 0 & x & 0 \\ 
0 & x & 0 & 1
\end{vmatrix} = -x^4+2 x^2-1=-((x-1) (x+1))^2=-(x^2-1)^2=0
$$
I think you can solve it.
P.S. Your matrix is a Hankel matrix (or catalecticant matrix).
A: There are two ways of solving this equation.


*

*Simply calculate what $$\begin{vmatrix}
x & 0  & 1  & 0 \\ 
0 & 1 & 0 & x\\ 
1 & 0 & x & 0 \\ 
0 & x & 0 & 1
\end{vmatrix}$$ is equal to. You should find it's some fourth degree polynomial of $x$.

*Think about what it means for a matrix to have a $0$ determinant. It means that the columns must be linearly dependent, i.e. there must exist some linear combination of the four columns that equals $0$. But the columns have a very special structure, so, if $v_i$ is the $i$-th column, you can (with a simple argument I leave to you) show that $$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3+\alpha_4v_4$$
implies that both $\alpha_1v_1+\alpha_3v_3$ and $\alpha_2v_2+\alpha_4v_4$ must be equal to $0$, which simplifies your problem a lot.
