Is it correct this reasoning? Let $E,F$ be reals vector space. Since
(1) $\dim (E\times F)=\dim E + \dim F$
(2) $\dim\ \text{Hom}(E,F)=\dim E\cdot \dim F$
Given $r>0$ integer, is it true that:
$$\text{Hom}(E\times \stackrel{(r)}{\ldots} \times E,E)=r(\dim E)^2\quad \text{?}$$
Many thanks!
 A: The dimension of $\operatorname{Hom}(F,E)$ is $\dim F\cdot\dim E$. Since $E^r=\underbrace{E\oplus E\oplus\dots\oplus E}_{\text{$r$ times}}$ has dimension $r\dim E$, the result follows.
(The notation $V\oplus W$ is more common than $V\times W$, that's used more frequently in the context of bilinear maps.)
Why is $\dim\operatorname{Hom}(F,E)=\dim F\cdot\dim E$? Represent linear maps with matrices and you'll see.
Why is $\dim\underbrace{E\oplus E\oplus\dots\oplus E}_{\text{$r$ times}}=r\dim E$? By simple induction from $\dim(E\oplus F)=\dim E+\dim F$ that follows from Grassmann's formula.
A: Well, you're pretty much there. What keeps you from thinking this result is right?
Note that, if $E$ is a vector space, then $E \times ... \times E$ is a vector space. Then, for $r > 0$,
$$\dim \left(\hom (E \times \overset{(r)}{\cdots} \times E, E)\right) \underset{(2)}{=} \dim(E \times \overset{(r)}{\cdots} \times E) \times \dim(E) \\ \underset{(1)}{=} \left( \sum_{k=1}^r \dim (E) \right) \times \dim(E) = r(\dim E)^2.$$ 
