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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.

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    $\begingroup$ 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). $\endgroup$
    – peek-a-boo
    Sep 6 at 21:02
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    $\begingroup$ yes, seems like he is only considering (1,1)-tensors, as these can be considered linear transformations of euclidean space into itself $\endgroup$
    – AlexD
    Sep 6 at 21:07
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    $\begingroup$ 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 $\tau(f,v) = f(T(v))$. $\endgroup$ Sep 6 at 21:07
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    $\begingroup$ 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 $V \approx V''$. $\endgroup$
    – Mason
    Sep 6 at 21:10
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    $\begingroup$ Remember that $e_iA^i_{\,j}e^j\in\mathfrak{L}(V;V)\cong(V\otimes V^*)$ $\endgroup$ Sep 6 at 21:30
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My question has been answered by the following comments above (thank you to everyone though who commented):

$\bullet$ (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 $\tau(f,v) = f(T(v))$.

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