A linear map $A$ can be seen as a (1,1) <tensor> because:
$$
T(w, v*) = v* A w
$$
is a number, $v*$. is a <dual vector>, and <W> is a <vector>. Furthermoe, $T$ is linear in both $v*$ and $w$. All of this makes $T$ fullfill the definition of a (1,1) tensor.


A linear map is a (1,1) tensor

{title2=$V^*$}
{wiki}

The dual space of a <vector space> $V$, sometimes denoted $V^*$, is the vector space of all <linear forms> over $V$ with the obvious addition and scalar multiplication operations defined.

Since a linear form is completely determined by how it acts on a <basis>, and since for each basis element it is specified by a scalar, at least in finite dimension, the dimension of the dual space is the same as the $V$, and so they are isomorphic because <all vector spaces of the same dimension on a given field are isomorphic>, and so the dual is quite a boring concept in the context of finite dimension.

Infinite dimension seems more interesting however, see: https://en.wikipedia.org/w/index.php?title=Dual_space&oldid=1046421278#Infinite-dimensional_case

One place where duals are different from the non-duals however is when dealing with <tensors>, because they transform differently than vectors from the base space $V$.


Dual space

{title2=$e^i$}

Dual vectors are the members of a <dual space>.

In the context of <tensors> , we use raised indices to refer to members of the dual basis vs the underlying basis:
$$
\begin{aligned}
e_1 & \in V \\
e_2 & \in V \\
e_3 & \in V \\
e^1 & \in V^* \\
e^2 & \in V^* \\
e^3 & \in V^* \\
\end{aligned}
$$
The dual basis vectors $e^i$ are defined to "pick the corresponding coordinate" out of elements of V. E.g.:
$$
\begin{aligned}
e^1 (4, -3, 6) & =  4 \\
e^2 (4, -3, 6) & = -3 \\
e^3 (4, -3, 6) & =  6 \\
\end{aligned}
$$
By expanding into the basis, we can put this more succinctly with the <Kronecker delta> as:
$$
e^i(e_j) = \delta_{ij}
$$

Note that in <Einstein notation>, the components of a dual vector have lower indices. This works well with the upper case indices of the dual vectors, allowing us to write a dual vector $f$ as:
$$
f = f_i e^i
$$

In the context of <quantum mechanics>, the <bra-ket>[bra] notation is also used for dual vectors.


Dual vector

{tag=Tensor}

<An Introduction to Tensors and Group Theory for Physicists by Nadir Jeevanjee (2011)> shows that this is a <tensor> that represents the <volume of a parallelepiped>.

It takes as input three vectors, and outputs one real number, the volume. And it is linear on each vector. This perfectly satisfied the definition of a tensor of <order of a tensor>[order] (3,0).

Given a basis $(e_i, e_j, e_k)$ and a function that return the volume of a parallelepiped given by three vectors $V(v_1, v_2, v_3)$, $\varepsilon_{ikj} = V(e_i, e_j, e_k)$.


Ciro Santilli @cirosantilli 36

 Incoming links: Tensor