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

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.
