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March 17th, 2018, 12:12 PM  #1 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 7 Thanks: 0  Dimension of the subspace T of HomV
L<V is a subspace of vector space V. T is a set of all linear maps from HomV (HomV={f : V>V, f is linear map}) for which the restriction fL is a zero map. What is the dimension of T? I tried to find the range of f, but I'm obviously doing it wrong. Thank you and sorry for my English. Last edited by Birgitta; March 17th, 2018 at 12:43 PM. Reason: *forgot to write what is HomV 
March 20th, 2018, 08:22 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,328 Thanks: 94 
Let S be the set of all linear transformations T from V to V st TL=0 where L is a subset of V. S is a vector space: T1+T2 and aT belong to S (T1)L=0 so rows of T1 are orthogonal to columns of L. S consists of all T whose rows are LC's of rows of T1, ie, dim S = dim V  dim L. OK, it's a little skechy. I would have to look at relevant sections of https://ocw.mit.edu/courses/mathemat...aspring2010/ to clean it up. Ck out orthogonal complement. 
March 20th, 2018, 11:21 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,328 Thanks: 94 
OP describes orthogonal complement of L in V which has dimension dimV dimL. That's cleaner.

March 20th, 2018, 02:24 PM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 350 Thanks: 192 Math Focus: Dynamical systems, analytic function theory, numerics  This is wrong. Not only are you assuming that $V,L$ are finite dimensional, but you are also assuming $V$ admits an inner product. Neither of these need be the case. What is an orthogonal complement in a space which has no inner product?

March 21st, 2018, 12:24 PM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,328 Thanks: 94 
Let $\displaystyle e_{1}, e_{2}$ be a basis of L. Extend it to a basis of V: $\displaystyle e_{1},e_{2},e_{3},e_{4},e_{5}$ $\displaystyle FL=0$ and $\displaystyle Fe_{j} = \sum_{i=1}^{i=5}F_{ij}e_{i} \rightarrow F_{1j}$ and $\displaystyle F_{2j} = 0$ Then all linear transformations will be given by all linear combinations of columns 3, 4, and 5, and so dim of T is 3, or in general, dim T=dimVdimL An infinite dimensional vector space doesn't doesn't have a dimension. 
March 21st, 2018, 07:14 PM  #6 
Member Joined: Jan 2016 From: Athens, OH Posts: 86 Thanks: 44 
Unfortunately, Zylo's answer is wrong. If V has finite dimension n and L has dimension k, then dim(T)=$n^2nk$. 
March 22nd, 2018, 12:50 AM  #7 
Newbie Joined: Mar 2018 From: Split, Croatia Posts: 7 Thanks: 0 
Thank you all! I've already found the answer: https://math.stackexchange.com/quest...fvectorspace 
March 22nd, 2018, 07:26 AM  #8  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,328 Thanks: 94  Quote:
If A,B,C are the first three columns of [F}, then all linear transformations are represented by [A,0,0,0,0], [0,B,0,0,0], [0,0,C,0,0], where each column is a vector of dim 3 and there are 3 such columns. In general then, if dimV=n and dimL=m, there are (nm)cols which each have n components for a total dimension of n(nm). To see this, any OP linear transformation can be written as all linear combinations of unit "vectors" which is an nxn matrix with a 1 in one of the first three (or nm) columns. Last edited by zylo; March 22nd, 2018 at 08:15 AM.  

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algebra, dimension, homv, linear algebra, subspace, vector spaces 
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