# Conflicting definitions of tensors?

I am currently taking a course in continuum mechanics using P. Chadwick's Continuum Mechanics: Concise Theory and Problems and I am having trouble reconciling the following definitions of a tensor:

In the book, Chadwick defines a tensor in the following way:

A tensor is a linear transformation or the Euclidean vector space $$E$$ into itself.

Since I have some experience using tensors, the definition I am familiar with is:

Let $$V$$ be a vector space and $$V^*$$ be its duel. Then a $$(p,q)$$ tensor is defined as $$T:\underbrace{V^{*}\times...\times V^{*}}_\text{p-copies}\times \underbrace{V\times...\times V}_\text{q-copies} \to \mathbb{R}$$.

In other words, a tensor is a multilinear map. Now that both of these definitions are on the table, here is my question:

How does Chadwick's definition relate (if at all) to the definition that I am familiar with? To me his definition isn't very precise because according to what he wrote a tensor should be an object of the form $$T : E \to E$$ (linearly of course). Is there something that I am not seeing or is this another way of defining tensors?

note: this class is centered towards engineers (I am a mathematician by training) if that helps put things in context a little better.

• his definition is not the general one. It only works for very specific cases (which perhaps is all that is used in the book) In his definition, a linear map $E\to E$ is "isomorphically equivalent" to a $(1,1)$ tensor on $E$ (i.e $\text{Hom}(E,E)\cong T^1_1(E)$ is a natural isomorphism assuming finite dimensions). Sep 6 at 21:02
• yes, seems like he is only considering (1,1)-tensors, as these can be considered linear transformations of euclidean space into itself Sep 6 at 21:07
• It is notable that a linear transformation can be naturally considered to be a $(1,1)$ tensor. Given a linear map $T:E \to E$, the associated map $\tau:V^* \times V \to \Bbb R$ is given by $\tau(f,v) = f(T(v))$. Sep 6 at 21:07
• In the linear algebra book I read, If $V_1, \dots, V_l$ are finite-dimensional vector spaces, then the tensor product is $V_1 \otimes \dots \otimes V_l := M(V_1', \dots, V_l'; \mathbb{F})$, the space of multilinear maps from $V_1' \times \dots \times V_l'$ to $\mathbb{F}$. $'$ denotes the dual space. Then every element of $V_1 \otimes \dots \otimes V_l$ is called a tensor. This is essentially your definition since $V \approx V''$. Sep 6 at 21:10
• Remember that $e_iA^i_{\,j}e^j\in\mathfrak{L}(V;V)\cong(V\otimes V^*)$ Sep 6 at 21:30

$$\bullet$$ (peek-a-boo's comment): his definition is not the general one. It only works for very specific cases (which perhaps is all that is used in the book) In his definition, a linear map $$E \to E$$ is "isomorphically equivalent" to a $$(1,1)$$ tensor on $$E$$ (i.e., $$\text{Hom}(E,E)\cong T^1_1(E)$$ is a natural isomorphism assuming finite dimensions).
$$\bullet$$ (Ben Grossmann's comment): It is notable that a linear transformation can be naturally considered to be a $$(1,1)$$ tensor. Given a linear map $$T:E \to E$$, the associated map $$\tau:V^* \times V \to \Bbb R$$ is given by $$\tau(f,v) = f(T(v))$$.