Multiplying 3D matrix I was wondering if it is possible to multiply a 3D matrix (say a cube $n\times n\times n$) to a matrix of dimension $n\times 1$? If yes, then how. Maybe you can suggest some resources which I can read to do this. Thanks!
 A: Multiplication of regular matrices arises from their interpretation as linear transformations. For a square matrix you get a map $T:V\to V$ (after having chosen a basis for $V$.) Since the domain and range of $T$ are the same, you can compose linear transformations, and this gives you matrix multiplication.
A cubical matrix can correspond to a linear map $V\to V\otimes V$ or $V\otimes V\to V$ (or a number of other possibilities such as $V\otimes V\otimes V\to \mathbb R$). (As Milind points out, this type of thing is called a tensor, and the different possibilities correspond to different placement of the indices up or down on the tensor, $t^i_{jk}, t^{ij}_k, t_{ijk}$, etc. 
An $n\times 1$ matrix can represent a map from $V$ to $\mathbb R$. So if you think of the 3D array as a map from $V\otimes V\to V$, then you can compose it with the map $V\to\mathbb R$. The resulting map is a map $V\otimes V\to \mathbb R$, which can be thought of as an $n\times n$ matrix.
A: Tensors are very relevant to your question, as they can be represented as multi-dimensional arrays.
A tensor product of a order 3 tensor (the $n \times n\times n$ cube) and a 1st order tensor (the $n\times 1$ vector) will give you a tensor of order 4 (i.e. a 4-dimensional array).
A: If you view the $n\times n\times n$ object as $n$ different $n\times n$ matrices, then you can apply each one to the $n\times 1$ vector, yielding $n$ different $n\times1$ vectors, which could be viewed as a $n\times n$ matrix. 
This won't have interesting multiplication properties though beyond those that come from the regular matrix-vector multiplication.
