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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!

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marked as duplicate by Srivatsan, lhf, tomasz, t.b., Chris Eagle Aug 17 '12 at 10:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
"$\leq$"?! Don't you mean "$\subseteq$"? –  kahen Sep 23 '11 at 13:20
    
@kahen: which one are you referring to? –  Freeman Sep 23 '11 at 13:21
    
@kahen, $\le$ is also used to mean subspace. –  lhf Sep 23 '11 at 13:25
    
Are you saying that $U$ is a $T-invariant$ subspace? –  user38268 Sep 23 '11 at 13:30
    
@lhf I have voted to close this question as a duplicate. –  Srivatsan Sep 23 '11 at 13:37

2 Answers 2

up vote 1 down vote accepted

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 $\dim(\mathrm{Hom}(V,V))=n^2$]

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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?

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