How to show that the linear mappings $\phi_\mu^\nu$ generates the set $L(E;V)$ In the book of Linear Algebra By Werner Greub at p 76, to show that $$\dim L(E;F) = \dim E \times \dim F$$, it constructs a map $\phi_\mu^\nu : E \to F$ by
$$\phi_\sigma^\lambda (x_\nu) = \delta_\nu^\lambda y_\sigma$$, and claims that these maps forms a basis for $L(E;F)$, but it doesn't give and proof to show that these are linearly independent maps, I'm trying to prove this.

How to show that the linear mappings $\phi_\mu^\nu$ generates the set
  $L(E;V)$

To show that $$\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu = 0_{L(E;V)}$$, we need to show that $$(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu)(x) = 0\quad \forall x\in E$$
I actually made a proof about it, and I was writing here, so that you can check for any flow, but I have found a mistake during the writing, so how can we prove this ?
 A: To pick it up from where you left off: let's prove that the mappings $\phi_\mu^\nu$ are linearly independent.
We want to show that if $\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu = 0_{L(E;V)}$, then all coefficients must be zero. That is, we need to show that 
$$
\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x) = 0\quad \forall x\in E \implies  \alpha_{\mu}^\nu = 0 \quad \forall \mu, \nu
$$
With that, we can plug in $x = x_\mu$ and note that
$$
\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x_\sigma) = 
\sum_{\mu}\alpha^\sigma_\mu x_\mu = 0
$$
However, the vectors $x_\mu$ form a basis, so we conclude that $\alpha^\sigma_\mu = 0$ for all $\mu$.  Noting that $\sigma$ was also arbitrary, we conclude that every $\alpha_\mu^\sigma = 0$.

Now, there's the separate questions of showing that these maps generate the set $L(E;F)$.   That is: given an arbitrary map $\phi:E \to F$, we want to show that there exist coefficients such that
$$
\phi = \sum_{\mu,\nu} \alpha_\mu^\nu \phi_\mu^\nu
$$
The proof is as follows: again, write out $\left(\sum_{\nu, \mu}\alpha_\mu^\nu\phi_\mu^\nu\right)(x_\sigma)$.  By the definition of a basis, there is a unique set of coefficients $\alpha_\mu^\nu$ that guarantee these equations hold.
With that, we want to show that under these constraints, we have $\phi = \sum \alpha_\mu^\nu \phi_\mu^\nu$.  To that end, it suffices to note that these are two linear maps which act the same way on a basis.
