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I am reading "Analysis on Manifolds" by James R. Munkres.

Definition:
Let $V$ be a vector space. Let $V^k = V \times \cdots \times V$ denote the set of all $k$-tuples $(v_1, \cdots, v_k)$ of vectors of $V$. A function $f : V^k \to \mathbb{R}$ is said to be linear in the $i$th variable if, given fixed vectors $v_j$ for $j \ne i$, the function $T : V \to \mathbb{R}$ defined by $$T(v) = f(v_1, \cdots, v_{i-1}, v, v_{i+1}, \cdots, v_k)$$ is linear. The function $f$ is said to be multilinear if it is linear in the $i$th variable for each $i$. Such a function $f$ is also called a $k$-tensor, or a tensor of order $k$, on $V$.

This is the definition of tensors.

I heard that tensors are a generalization of scalars, vectors, and matrices. But tensors don't look like scalars, vectors, and matrices at all.

For example, please show me a tensor which corresponds to a matrix.

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  • $\begingroup$ A tensor over/in a finite-dimensional vector space $V$ is an element of a tensor product of several copies of $V$ and of several copies of its dual. If you take the tensor product of a finite-dimensional vector space with its dual, you obtain the space of operators in that space ("matrices"). If you take the tensor product of no vector spaces, you obtain the space of scalars. $\endgroup$ – Alexey Apr 20 '20 at 10:30
  • $\begingroup$ You might be interested in the definition of tensor product in the general case. I suggest expository papers by Keith Conrad, like this one. $\endgroup$ – Alexey Apr 20 '20 at 10:31
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One way to think about what a tensor does is ask how many arguments it has to accept before it returns a scalar. What you have written above shows a function which is linear in $k$ arguments, and when you feed it $k$ vectors, you get back a number. Often times, it is convenient to distinguish between rows and columns as vectors, for example, if you are not in the presence of a natural inner product. In this case, instead of simply speaking about $k$-tensors, we can speak of $(i,j)$ tensors, where the tensor accepts $i$ rows and $j$ columns as argument.

A matrix can be regarded as a $2$-tensor, or more specifically a $(1,1)$ tensor. If $M$ is the matrix, and $v,w$ are vectors, then the matrix accepts two arguments. You write $v^TMw$, and this evaluates to a number.

Similarly, if $v$ is a vector, then this is a $(1,0)$ tensor, since multiplying on the left by a row vector gets you a number.

If you are in the presence of an inner product, the process of applying the inner product to turn vectors into covectors or conversely is called lowering/raising of indices.

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  • $\begingroup$ Thank you very much, Alfred Yerger. $f(v, w) = a v_1 w_1 + b v_1 w_2 + c v_2 w_1 + d v_2 w_2$ is a $2$-tensor on $\mathbb{R}^2$. Then $\begin{bmatrix}a & b\\c&d\end{bmatrix}$ corresponds to $f$. $\endgroup$ – tchappy ha Mar 10 '20 at 7:51
  • $\begingroup$ +1, but shouldn’t a vector be a (1,0) tensor? $\endgroup$ – PrudiiArca Mar 10 '20 at 9:24
  • $\begingroup$ @PrudiiArca Depends. A row vector is a (1,0)-tensor, and a column vector is a (0,1)-tensor. If you have a particular convention relating row/column status to covariance/contravariance, then one of them is a covariant vector and the other is a contravariant vector. $\endgroup$ – probably_someone Mar 10 '20 at 16:22
  • $\begingroup$ @PrudiiArca As stated, it is just a convention, but you're right that the standard convention is to go the other way, and I should switch it around to reflect that. I'll edit. Thanks! $\endgroup$ – Alfred Yerger Mar 10 '20 at 16:36

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