Classification PDE, 2 order Classificate the following Partial differential equations of order 2:
$$a)\sum_{i=1}^n(-1)^{i(i+1)}\frac{\partial^2 u}{\partial x_i^2}=0$$$$b) \sum_{i,j=1}^n (-1)^{i+j} \frac{\partial^2 u}{\partial x_i \partial x_j}=0$$
For the classification i look at the eigenvalues of the coefficient-matrix A with $A_{ij}=a_{ij}$ of $\sum_{i,j=1}^n a_{ij} \frac{\partial^2 u}{\partial x_i^2}$. 
In a) i become $A=E_n$, so $(\lambda_1,..,\lambda_n)=(1,..,1) \rightarrow $elliptic
In b) i have $A= \begin{pmatrix}
1 & -1 &1 &\dots \\
-1 & 1 & -1 & \dots \\
1 & -1 & 1 & \dots\\
\vdots & \vdots & \ddots & \ddots
\end{pmatrix}$
I try to calculate the determinant and think $det(A)=0$, so it exists a eigenvalue $0$ and the equation isn´t elliptic. The Hurwitz criterium doesn´t help me to decide if the equation is hyperbolic or parabolic and i don´t know how i can calculate the eigenvalues of A.
For n=1 the equation is elliptic.
 A: To classify the PDE $\sum a_{ij} \frac{\partial^2u}{\partial x_i \partial x_j} + \sum b_i \frac{\partial u}{\partial x_i} = f(x_1,\ldots,x_n)$ we need to find the eigenvalues of the matrix $A$ with coefficients $a_{ij}$. If $A$ has $0$ as en eigenvalue then the PDE is parabolic (caveat; some books require all non-zero eigenvalues to have the same sign for it to be parabolic). If none of the eigenvalues are zero and all or none are positive then the PDE is elliptic. Otherwise it's hyperbolic or ultrahyperbolic. See e.g. Wiki::PDE Classifications (and these notes for a slightly different definition than on Wikipedia).
Let's apply this to your problem (b) where $a_{ij} = (-1)^{ij}$ so 
$$A = \pmatrix{1 & -1 & 1 & -1 & \cdots \\ -1 & 1 & -1 & 1 & \cdots \\ 1 & -1 & 1 & -1 & \cdots \\ \cdots & \cdots & \cdots & \cdots & \cdots}$$
If $n > 2$ then (atleast) two of the rows in $A$ are seen to be equal which implies $\det A = 0$ and $0$ is an eigenvalue making the PDE parabolic. The same applies for $n=2$ as a direct computation shows. When $n=1$ then the PDE ($u_{xx} = 0$) is elliptic.
Depending on the classification definition you use you might not need to explicitly compute all the eigenvalues of $A$ to solve the problem at hand: either it's parabolic of it does not fall within the classification. For completeness I'll add some notes on how one might proceed to compute all the eigenvalues of $A$ as it's a nice little problem. 

Consider the matrix $A - \lambda I$. Start by adding row $k+1$ to row $k$ for $k=1,2,\ldots,n-1$ to arrive at the matrix
$$\pmatrix{-\lambda & -\lambda & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 &-\lambda & -\lambda & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & -\lambda & -\lambda & 0 & \cdots & 0 & 0\\\cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots\\ 0 & 0 & 0 & 0 & 0 & \cdots & -\lambda & -\lambda\\(-1)^{n+1} & (-1)^{n} & (-1)^{n+1} & (-1)^{n} & (-1)^{n+1} & \cdots & -1 & 1-\lambda}$$
Now assume $\lambda \not= 0$ and add $\pm \frac{1}{\lambda}$ of row 1 to the last row (sign depending on the parity of $n$), then $\pm\frac{2}{\lambda}$ of row 2 to the last row, then $\pm\frac{3}{\lambda}$ or row $3$ and so on until we have transformed it to the form
$$\pmatrix{-\lambda & -\lambda & 0 & 0 & 0 & \cdots & 0 & 0 \\ 0 &-\lambda & -\lambda & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & -\lambda & -\lambda & 0 & \cdots & 0 & 0\\\cdots & \cdots & \cdots & \cdots & \cdots & \cdots &\cdots & \cdots\\ 0 & 0 & 0 & 0 & 0 & \cdots & -\lambda & -\lambda\\0 & 0 & 0 & 0 & 0 & \cdots & 0 & n-\lambda}$$
for which we can read off that the characteristic equation $0 = \det(A-\lambda I)$  is proportional to $\lambda^{n-1}(\lambda -n)$ so $\lambda = n$ is the only possible non-zero eigenvalue. The eigenvalue $0$ must therefore have multiplicity $n-1$.
