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\begin{document}
\title{Tensor homework.}
\author{Renzo's math 469}
\date{}

\maketitle
%\section{Tensor products}


 
 \begin{description}
 \item[Decomposable tensors] Let $V$ and $W$ be two dimensional vector spaces with bases $e_1, e_2$, $f_1, f_2$, respectively. Prove that the tensor
 $$
 T= a\ e_1 \otimes f_1 + b\ e_1 \otimes f_2 + c\ e_2 \otimes f_1 + d\ e_2 \otimes f_2
 $$
is decomposable if and only if $ad-bc= 0$. Now replace $V$ with $V^\ast$, so that you can think of $T$ as a linear transformation from $V$ to $W$. What does $T$ being decomposable correspond to in the language of linear functions. Why? 
 \item[Changing basis].\\
 \begin{enumerate}
 \item Let $V$ have two bases $e_1, \ldots, e_n$ and $f_1, \ldots, f_n$, and for each $j$ let $f_j = \sum_i a^i_j e_i$. The tensor $A= a^i_j$ is called the matrix of base change for $V$.
 Given a vector 
 $$
 v = \sum v^k f_k,
 $$
 how do you describe the vector $v$ in the basis $e_1, \ldots, e_n$?
 \item Let $f^k = \sum_i \tilde{a}^k_l e^l$ give the transformation for the dual vectors. Prove that the matrix $\tilde{a}^k_l$ is $A^{-1}$.
 \item Given a general operator $T\in \mathcal{L}(V)$, described as a tensor in some basis $e_1, \ldots, e_n$, describe how the tensor changes when changing to a different basis $f_1, \ldots, f_n$.
  \end{enumerate}
 \item[The identity function].\\
 \begin{enumerate}
 	\item Show that the identity operator in $Id. \in\mathcal{L}(V)$ is expressed, in any basis, by the tensor $\sum e^i\otimes e_i$. 
	\item  Show that any linear operator $T\in \mathcal{L}(V,W)$ is seen as a tensor as follows:
	\begin{itemize}
		\item Consider the map $Id_V{^\ast} \otimes T: V^\ast \otimes V \to V^\ast \otimes W$.
		\item Consider the element $(Id_{V^\ast} \otimes T)(\sum e^i\otimes e_i).$
	\end{itemize}  
 \end{enumerate}
 \item[Trace] Let $T\in \mathcal{L}(V)$ be an operator, and consider it as a tensor in $V\ast \otimes V$. We can now obtain a number by contracting the upper and lower index of such tensor. Show that this produces the trace of $V$. Notice that since the contraction of tensors along factors does not depend on the choice of basis, it follows that the trace of a matrix doesn't change when you conjugate the matrix by any invertible matrix.
 \end{description}
 




%\textbf{I am imagining they will need a good amount of help on this one too...}
\end{document}


We introduce tensor products of vector spaces in two ways: first we describe the universal property that defines them, which is a high power but less concrete approach. Second, we give an explicit basis.

Given two vector spaces $V$ and $W$, the tensor product $V \otimes W$ is the unique vector space that, together with  a map $f: V\times W \to V\otimes W$
 satisfy the following universal property:

{\it For any bilinear  map  $\Phi: V\times W \to k$ there exist a unique linear function $\tilde{\Phi}: V\otimes W \to k$ such that}
$$
\Phi = \tilde{\Phi} \circ f.
$$
 
In other words, a bilinear function on $V\times W$ is equivalent information to a linear function on $V\otimes W$. Now we understand what $V\otimes W$ {\bf does} for us... can we also understand what it {\bf is}?

First off, we can make the map $f$ completely trivial by declaring that the vector space $V\otimes W$ contain elements denoted $v\otimes w$ for any $v\in V$ and $w\in W$, and then 
$$
f((v,w)):= v\otimes w.
$$ 

Now we need to give this set a vector space structure... it turns out that just the set of symbols $v\otimes w$ is both "too small" and "too large" for what we want. First, we enlarge it by introducing all formal finite linear combinations of elements of the form $v\otimes w$. Now we created a vector space, but an infinitely dimensional one. We will cut it down by introducing an equivalence relation that makes the function $f$ bilinear. Such an equivalence relation is generated by:

\begin{itemize}
	\item for any $v_1,v_2\in V, w\in W$, $$ (v_1+v_2)\otimes w \sim v_1\otimes w + v_2\otimes w .$$
	\item for any $v\in V, w_1,w_2\in W$, $$ v\otimes (w_1+w_2) \sim v\otimes w_1 + v\otimes w_2. $$
	\item for any $v\in V,  w\in W, \lambda \in k$, $$ (\lambda v)\otimes w \sim \lambda(v\otimes w) \sim v\otimes (\lambda w). $$
\end{itemize}

Notice that all of these equivalences are {\bf necessary} in order for $f$ to be bilinear. Now we can define: 
$$\tilde\Phi \left(\sum \lambda_i (v_i\otimes w_i)\right) := \sum \lambda_i \Phi( v_i, w_i).$$

The function $\tilde{\Phi}$ is definitely linear, provided that it is a well defined function. This means, that it has to be consistent with the three kind of relations that we imposed to define $V\otimes W$. But this follows from the fact that $\Phi$ is bilinear.

OK, now we have constructed $V\otimes W$ as a vector space. Do we know its dimension? Can we produce some natural bases for it? The answer is yes.
If $V$ has dimension $n$ and a basis given by $e_1, \ldots, e_n$ and $W$ has dimension $m$ and a basis given by $g_1, \ldots, g_m$, then $V\otimes W$ has dimension $nm$ and a basis given by all vectors of the form $e_i\otimes g_j$.
\begin{itemize}
	\item To show that the vectors  $e_i\otimes g_j$ generate $V\otimes W$, it suffices to show that if $v= \sum v^i e_i$ and $w= \sum w^j g_j$, then using the relations above you can show
	$$
	v \otimes w = \sum_{i,j} v^iw^j e_i\otimes g_j.
	$$
	\item To show that the vectors  $e_i\otimes g_j$ are linearly independent, define a bilinear function $\Phi: V\times W \to M(n,m; k)$ by $\Phi(e_i,g_j)=E_ij$,
	where $M(n,m; k)$ is the vector space of $n$ by $m$ matrices with coefficients in $k$ and $E_ij$ is the matrix that has one in position $ij$ and zero everywhere else.
	By the universal property, this must define a linear function $\tilde{\Phi}: V\otimes W \to M(n,m; k)$. If we restrict $\tilde{\Phi}$ to the linear span of the vectors  $e_i\otimes g_j$, we still have a surjective function, which means that the dimension of such span is at least $mn$. Then it has to be exactly $mn$ and the vectors are linearly independent.
\end{itemize}

\section{Functions are vectors, vectors are functions...and more}

We already encountered this philosophy when we introduced dual vector spaces. A linear  function $\varphi:V \to k$ is a vector in the dual space $V^\ast$ and similarly a vector can be thought of as a  linear function $v: k \to V$, namely the linear function that sends $1$ to the vector v.


Given an element $T = \sum \varphi_i \otimes w_i \in V^\ast \otimes W$, it is natural to produce from it a linear functiion from $V$ to $W$. You just define:
$$
T(v):= \sum \varphi_i(v) w_i .
$$

Similarly, if we start from a linear function $T:V \to W$, we can produce a vector in the space $V^\ast \otimes W$. 
The simplest way to think about it is by picking a basis $e_1, \ldots, e_n$ for $V$. Then let us denote by $e^1, \ldots, e^n$ the dual basis. Since we know that $T$ is uniquely determined by its values on the basis vectors, we associate to $T$ the element
$$
\sum e^i \otimes T(e_i) \in V^\ast \otimes W.
$$
Of course as a next step
