# Could a Rank Two Tensor be a Scalar?

The universal property of the tensor product (of vectors or vector spaces) states that a bilinear map out of a cartesian product is a linear map out of the tensor product. I've also heard that tensors are multidimensional arrays of numbers.

Let the bilinear operation be matrix multiplication and the linear operation be multiplication by two. Say these are acting on a column and a row vector that can be multiplied:

$\pmatrix{a&b}\pmatrix{x\\y}=2\pmatrix{a&b}\otimes\pmatrix{x\\y}$

When you solve for the tensor product, which should be a matrix according to what I've heard:

$\frac{1}{2}\pmatrix{a&b}\pmatrix{x\\y}=\pmatrix{a&b}\otimes\pmatrix{x\\y}=\frac{1}{2}(ax+by)=$ an element of the field

I know that a scalar is like a 1 by 1 matrix, but still, what it gives you is part of the underlying field of the vector space. Am I thinking about this the wrong way? This is all confusing and new for me!

Let $k$ denote the ground field. Let $V$ and $W$ be two vector spaces (in your example $V=k^2$ viewed as row vector, $W=k^2$ viewed as column vector).
It seems that you believe the following: For each bilinear map $b\colon V\times W\to X$ (in your example $X=k$ and $b$ is given by matrix multiplication) and for each linear map $f\colon V\otimes W\to Y$ (in your example $Y=V\otimes W$ and $f$ is given by multiplication by $2$), we have for each $v\in V$ and $w\in W$ the equation $b(v,w)=f(v\otimes w)$.
Now compare with the actual universal property of the tensor product: For each bilinear map $b\colon V\times W\to X$ there exists a unique linear map $g\colon V\otimes W\to X$ (note that the same $X$ appears again) such that we have for each $v\in V$ and $w\in W$ the equation $b(v,w)=g(v\otimes w)$.
• In the wrong statement, $b(v,w)$ is an element of $X$ and $f(v\otimes w)$ is an element of $Y$. Since $X$ and $Y$ are different vector spaces, it does not make sense to compare the two elements to each other. In the correct statement $b(v,w)$ and $g(v\otimes w)$ are both elements of $X$ so it makes sense to compare them.
• The quantifiers are different: compare "For each $b$ and for each $f$ we have..." (wrong statement) to "For each $b$ there exists a $g$ such that..." (correct statement) .
• Thanks, that's very helpful! In this example, what might $g$ look like? – Benjamin Thoburn Jul 15 '18 at 22:44