Calculate minimal polynomial of a matrix \begin{bmatrix}0&1&0&1\\1&0&1&0\\0&1&0&1\\1&0&1&0\end{bmatrix}
I have calculated characteristic polynomial as $x^2(x^2-4)$ but I don't know what is minimal polynomial please solve
 A: As the given matrix is symmetric it is diagonalizable $\Rightarrow$ its minimal polynomial has distinct roots  $\Rightarrow$ minimal polynomial $= x(x-2)(x+2)$.
A: All the distinct roots of the characteristic polynomial are also the roots of the minimal polynomial, hence the minimal polynomial has roots $0,2,-2$
Hence $x(x^2-4)$ divides the minimal polynomial, 
Also all roots of the minimal polynomial is also a root of the characteristic polynomial, so the minimal polynomial must divide the characteristic polynomial.
Hence all these implies that the minimal polynomial is either $x(x^2-4)$ or $x^2(x^2-4)$.
Now by putting the matrix in the equation $x(x^2-4)$ if it comes $0$ then $x(x^2-4)$ is the minimal polynomial else $x^2(x^2-4)$ is the minimal polynomial.
Another way to decide on the last part:
The dimension of the null space of the above matrix is 2, hence it has a basis consisting of the eigenvectors of the matrix, hence it is diagonalizable, hence it's minimal polynomial spilts into distinct linear factors, hence it cannot be  $x^2(x^2-4)$, hence the answer is  $x(x^2-4)$. 
A: These examples they give are always way too simple. Here you can spot by inspection the kernel (which is the eigenspace for $\lambda=0$), which is a huge give-away. But I'll apply a general method instead.
Take some nonzero vector, and apply the matrix repeatedly to it, until the images become linearly dependent. I'll just take the first standard basis vector $e_1$ and call the matrix $A$, which gives
$$\pmatrix{1\\0\\0\\0}\overset A\mapsto
  \pmatrix{0\\1\\0\\1}\overset A\mapsto
  \pmatrix{2\\0\\2\\0}\overset A\mapsto
  \pmatrix{0\\4\\0\\4}
$$
with obvious linear dependency $-4Ae_1+A^3e_1=0$. This (and the fact that this is the first linear dependency) tells you the polynomial $P=X^3-4X$ is the smallest degree monic polynomial to satisfy $P[A](e_1)=0$. Thus $P$ divides the minimal polynomial, and the (unknown at this point) quotient of that division is the minimal polynomial of the restriction of (the linear map defined by) $A$ to the image of $P[A]$. But it turns out the $P[A]=0$ already (you were lucky), so (its image is the zero space, the mentioned quotient is $1$, and) $P$ is itself the minimal polynomial.
As you see, one can do entirely without the characteristic polynomial.
