Studying the Peter-Weyl theorem, I've come across the following linear maps:
$\theta_E: E'\otimes E \rightarrow$ Hom$(E,E)$
Where $ E'\otimes E$ denotes the tensor product of a finite-dimensional vector space $E$ and its dual $E'$. Hom$(E,E)$ denotes the set of operators of $E$.
$\theta_E (v \otimes x) \ (y) = v(y)x = \langle v, y\rangle x$, that is, the value of $v$ at $y$ times $x$.
Chapter 3 of Hofmann, K. - The Structure of compact groups says that this linear map is an isomorphism, that is, bijective.
But can this be injective? If you choose $v = 0 \in E'$ we get that for ANY $x \in E$ the value of $\theta$.
$\theta_E(v\otimes x) (y) = 0(y)x = \langle 0, y \rangle x = 0$. Note that this is the linear map $0:E \rightarrow \mathbb{C}$ at every $y \in E$ So the kernel
$Ker \ \theta_E$ contains $0 \otimes E$ as a subspace and so, $\theta_E$ cannot be injective.
What am I seeing wrong here? I mean, even wikipedia says this.
If you have $v=0$, then for any $x\in E$
$v\otimes x=0$.
That is, $0\otimes E$ is trivial.
In more detail, let $\cdot$ denote the scalar product. If $v=0$ then we have
$0\otimes x= (0\cdot 0)\otimes x=0\cdot(0\otimes x)=0\otimes(0\cdot x)=0\otimes 0$.
The two equalities in the middle follow from the definition of the tensor product.