Finding the dimension of $B(U):=\{T:V \to V: T$ is linear and $T(U)\leq U\}$ where $U\leq V$ 
Possible Duplicate:
Set of linear transformations 

Let $V$ be an $n$ dimensional vector space over a field $F$ and $U$ be an $m$ dimensional subspace of $V$. Define  
$B(U):=\{T:V \to V: T$ is linear and $T(U)\leq U\}$
We know $B(U) \leq Hom(V,V):=\{T:V \to V: T$ is linear $\}$
How would we find the dimension of $B(U)$? Any help would be greatly appreciated!
 A: A hint: As $U$ is given in advance you may choose a basis for $V$ that is particularly suited for this problem. How would the matrix of $T$ look like in this basis? How many degrees of freedom do you have in chosing such a matrix?
A: Take a basis for $U$ and complete it to a basis for $V$.
The basis elements for $U$ must map to linear combinations of themselves. However, the remainder of the basis can map to anything.
Fix some notation: Let $u_1,\dots,u_m$ be a basis for $U$ and $u_{m+1},\dots,u_n$ be
vectors which complete this to a basis for $V$.
We can build up a basis for $B(U)$ as follows:
(1) Fix $1 \leq i \leq m$. Let $T(u_i)=u_j$ ($1 \leq j \leq m$). Send all other basis 
vectors $u_k$ to $0$: $T(u_k)=0$ for all $k \not=i$.
(2) Fix $m+1 \leq i \leq n$. Let $T(u_i)=u_j$ ($1 \leq j \leq n$). Send all other basis 
vectors $u_k$ to $0$: $T(u_k)=0$ for all $k \not=i$.
All of the linear maps of type (1) and (2) are members of $B(U)$. It's easy to see they're linearly independent and they span. So the dimension is $m \cdot m + (n-m) \cdot n = n^2 -mn+m^2$. 
[$U=0$ and $U=V$ give you $m=0$ or $m=n$ and $B(U)=\mathrm{Hom}(V,V)$. In these cases we find that $\dim(\mathrm{Hom}(V,V))=n^2$]
