My Math Forum Dimension of the subspace T of HomV

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March 17th, 2018, 12:12 PM   #1
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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 f|L 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.
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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: 1,431 Thanks: 105 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...a-spring-2010/ to clean it up. Ck out orthogonal complement.
 March 20th, 2018, 11:21 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,431 Thanks: 105 OP describes orthogonal complement of L in V which has dimension dimV-- dimL. That's cleaner. Thanks from Birgitta
March 20th, 2018, 02:24 PM   #4
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Quote:
 Originally Posted by zylo (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.
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: 1,431 Thanks: 105 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=dimV-dimL 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: 89 Thanks: 47 Unfortunately, Zylo's answer is wrong. If V has finite dimension n and L has dimension k, then dim(T)=$n^2-nk$. Thanks from topsquark
 March 22nd, 2018, 12:50 AM #7 Newbie   Joined: Mar 2018 From: Split, Croatia Posts: 13 Thanks: 0 Thank you all! I've already found the answer: https://math.stackexchange.com/quest...f-vector-space
March 22nd, 2018, 07:26 AM   #8
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Quote:
 Originally Posted by zylo 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=dimV-dimL An infinite dimensional vector space doesn't doesn't have a dimension.
Couldn't understand link and first part of what johng40 did was what i did. After that it became unintelligible.

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 (n-m)cols which each have n components for a total dimension of n(n-m).

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 n-m) columns.

Last edited by zylo; March 22nd, 2018 at 08:15 AM.

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