# Tensors = matrices + covariance/contravariance?

I have read several topics on tensors but it is still not clear to me. Tensors are different from matrices because they contain additional information about how do they transform.

To fully specify a $n\times n$ matrix, one need to specify its $n^2$ components.

My question is: what do someone need to specify to fully define a tensor ? Is it true to say that, to fully define a 2D tensor, one need to provide its $n^2$ components plus whether the tensor is covariant or contravariant along each dimension ? (or is there something else) ?

• Actually, a matrix is a tensor; a $(0,n)$ tensor. And even a vector is a 0-tensor; vectors are tensors, but not viceversa. AFAIK, an object is a tensor if it satisfies certain nice properties under a change of basis. Formally, in multilinear algebrathough, a tensor is a multilinear map . Maybe you're considering the physics or enginerring definition of tensor. – user99680 Nov 6 '13 at 2:34

Usually, a matrix is thought of a representation of a linear operator: a map that takes a vector and spits out another vector. Say $A$ is some linear operator and $v$ is some vector, then $A(v)$ is the output vector.
An equivalent way of looking at it, however, is to say that there is a map $B$ that takes two vectors $v, w$ and spits out a scalar, given by $B(v,w) = A(v) \cdot w$, say. Such a map is what is usually described in the literature when talking about tensors.
Where do contravariance and covariance come in? Well, the above idea of a tensor is actually a bit of a cheat. There might not be an inner product; we might not be able to freely convert between vectors and covectors using it. So instead of saying that $B$ takes two vectors as arguments, let $B$ be a map taking one vector $v$ and a covector $\alpha$ instead, so that $B(v, \alpha) = \alpha(A(v))$.
(You'll note that, if there is a way to convert from vectors to covectors, then any tensor acting on $p$ vectors and $q$ covectors could be converted to one that acts on $p+q$ vectors, for instance.)
In physics, it's common to look at the components of a tensor with respect to some basis--rather than supply whatever vectors or covectors that might be relevant to a problem, we supply a set of basis vectors and covectors instead, so we need only remember the coefficients. If $e_i$ is the $i$th basis vector and $e^j$ is the $j$th basis covector, then $B(e_i, e^j) = {B_i}^j$ takes us from the more math-inclined definition of a tensor to the more familiar form to a physicist.