Compute $Hom(V,W)$ and also determine its dimension over $F$ where $V$ and $W$ are vector spaces over the Field $F$ given that $V=\mathbb R^2, W=\mathbb R^2, F=\mathbb R$
I have done this: $V=\{v1=(1,0) , v2 = (0,1)\}, W=\{w1=(1,0) , w2 = (0,1)\}$
$V= a(v1)+b(v2)$ where $a,b$ belongs to $F$ then $a(v1)+b(v2) = (x,y)$ implies $a=x , b =y$. Now, we need to calculate $T11 ,T12 , T21 ,T22$. So, $T11(V) = a(w1)=x(1,0)$ similarly, $T12 =a(w2), T21=b(w1), T22=b(w2)$.
Is this correct? If not, please help.
Hint (building on the comment of Arpit Kansal):
given vector spaces $\mathcal V,\mathcal W$ over a field $F$ with $\dim(\mathcal V)=n,\dim(\mathcal W)=m$ there exists an isomorphism $\operatorname{Hom}(\mathcal V,\mathcal W)\rightarrow F^{m\times n}$. The proof for this relies basically on the fact, that every linear map $\varphi:\mathcal V\rightarrow\mathcal W$ can be identified with a matrix (you can also check out wikipedia for more information on that).