Why is a linear transformation a $(1,1)$ tensor?

Question

Wikipedia says that a linear transformation is a $(1,1)$ tensor. Is this restricting it to transformations from $V$ to $V$ or is a transformation from $V$ to $W$ also a $(1,1)$ tensor? (where $V$ and $W$ are both vector spaces). I think it must be the first case since it also states that a linear functional is a $(0,1)$ tensor and this is a transformation from $V$ to $R$. If it is the second case, could you please explain why linear transformations are $(1,1)$ tensors.

Answer 1

It's very common in tensor analysis to associate endomorphisms on a vector space with (1,1) tensors. Namely because there exists an isomorphism between the two sets.

Define $E(V)$ to be the set of endomorphisms on $V$.

Let $A\in E(V)$ and define the map $\Theta:E(V)\rightarrow T^1_1(V)$ by \begin{align*} (\Theta A)(\omega,X)&=\omega(AX). \end{align*} We show that $\Theta$ is an isomorphism of vector spaces. Let $\{e_i\}$ be a basis for $V$ and let $\{\varepsilon^i\}$ be the corresponding dual basis. First, we note $\Theta$ is linear by the linearity of $\omega$. To show injectivity, suppose $\Theta A = \Theta B$ for some $A,B\in E(V)$ and let $X\in V$, $\omega \in V^*$ be arbitrary. Then \begin{align*} (\Theta A)(\omega,X)&=(\Theta B)(\omega,X)\\ \\ \iff \omega(AX-BX)&=0. \end{align*} Since $X$ and $\omega$ were arbitrary, it follows that \begin{align*} AX&=BX\\ \iff A&=B. \end{align*} To show surjectivity, suppose $f\in T^1_1$ has coordinate representation $f^j_i \varepsilon^i \otimes e_j$. We wish to find $A\in E(V)$ such that $\Theta A = f$. We simply choose $A\in E (V)$ such that $A$ has the matrix representation $(f^j_i)$. If we write the representation of our vector $X$ and covector $\omega$ as \begin{align*} X&=X^i e_i\\ \omega&=\omega_i \varepsilon^i, \end{align*} we have \begin{align*} (\Theta A)(\omega, X)&=\omega(AX)\\ \\ &=\omega_k \varepsilon^k(f^j_i X^i e_j)\\ \\ &=f^j_i X^i \omega_k \varepsilon^k (e_j)\\ \\ &=f^j_i X^i \omega_k \delta^k_j\\ \\ &=f^k_i X^i \omega_k. \end{align*} However we see \begin{align*} f(\omega,X)&=f(\omega_k\varepsilon^k,X^ie_i)\\ \\ &=\omega_k X^i f(\varepsilon^k,e_i)\\ \\ &=f^k_i X^i \omega_k. \end{align*} Since $X$ and $\omega$ were arbitrary, it follows that $\Theta A = f$. Thus, $\Theta$ is linear and bijective, hence an isomorphism.

Answer 2

Let $T : V \mapsto W$. Then define $\tau: V \times W^* \mapsto K$ such that, for $a \in V$ and $\alpha \in W^*$, we have

$$\tau(a, \alpha) = (\alpha \circ T)(a)$$

Note that it's typical to define tensor to mean a multilinear map that is a function of vectors only in the same vector space, or of covectors in the associated dual space, or some combination of the two. So, we could identify a linear operator $T: V \mapsto V$ with a $(1,1)$ tensor $\tau: V \times V^* \mapsto K$, but in the case that $V$ and $W$ are distinct vector spaces, these would just be some construction of multilinear maps, not tensors.

Answer 3

To summarize as an answer what I wrote in various comments above: first beware that autors differ in their definition of tensor, even when using the same approach, i.e. using the tensor product in this case.

For some authors a tensor is defined only as ...

$$ T\in \underbrace{V \otimes\dots\otimes V}_{n \text{ copies}} \otimes \underbrace{V^* \otimes\dots\otimes V^*}_{m \text{ copies}}$$

From which it makes sense to speak of a type-$(n,m)$ tensor.

For others, a tensor is any...

$$T\in V_1 \otimes\dots\otimes V_d$$

where $V_1, \dots, V_d$ can be different vector spaces, however all must be over the same scalar field. And with this latter definition one can speak of an order-$d$ tensor. A type-$(n,m)$ tensor [in the former sense] is a tensor of order $d=n+m$ in the latter sense, but second definition is broader for it does not restrict us to a single vector space. In particular, a second-order tensor is an element of $V \otimes W$ where $V$ and $W$ may be two different vector spaces. Type-(1,1) tensors are tensors of second order, but the converse of this statement doesn't make sense. (N.B.: I've updated Wikipedia to reflect these different definitions.)

As for your 2nd question, endomorphisms (linear maps) from a vector space to itself are (isomorphic with) type-(1,1) tensors (detailed proof given here by beedge89), but if you consider homomorphisms (linear maps) between different vector spaces $V$ and $W$, i.e. $\mathrm{Hom}(V,W)$, these are isomorphic with only a certain class of order-2 tensors, namely with $V^* \otimes W$. If we let $(\phi, w)\in V^* \times W$, then the correspondence is given by $\phi \otimes w \leftrightarrow F_{\phi, w}$, where the latter is a (linear) map defined as $F_{\phi, w} (v) = \phi(v)w$. (Remember that covectors are themselves maps from vectors to scalars, so the formula for $F$ makes sense as it's a product of the scalar $\phi(v)$ with the vector $w$). A detailed proof of the fact that this is an isomorphism is given in Yokonuma (pp. 18-19). Apologies for not including it here.

As you may expect, the result for type-(1,1) tensors also follows as a corollary of this, i.e. $\mathrm{Hom}(V,V)$ is isomorphic with $V^* \otimes V$ (and with $V \otimes V^*$ by commutativity of the tensor product, which is also understood in the sense of an isomorphism between $V \otimes W$ and $W \otimes V$ for any vector spaces $V$ and $W$).

And one important caveat here: this is an isomorphism only for finite-dimensional vector spaces. (The introduction of Yokonuma's book actually says to assume all vector spaces in the book are finite-dimensional unless stated otherwise.) If both $V$ and $W$ are infinite-dimensional, then it turns out $V^*\otimes W$ is only a proper subspace of $\mathrm{Hom}(V,W)$, namely it is the subspace of linear transformation of finite rank.

And to tie this in with bilinear (and in general with multi-linear) maps: there's also a one-one correspondence between bilinear maps $f : V\times W \to U$ and $linear$ maps $g : V\otimes W \to U$. (For a proof see for instance http://www.landsburg.com/algebra.pdf) That's why second-order tensors are basically said to be just bilinear maps, and in general why d-order tensors are said to be just multi-linear maps.