Questions about the concept of subspace of linear transformation

  • Thread starter bigheadsam
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  • #1

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Hi all,
I have some questions about the concept of subspace of linear transformation and its dimension, when I try to prove following problems:
Prove T is a finite dimensional subspace of L(V) and U is a finite dimensional subspace of V, then
T(U) = {F(u) | F is in T, u is in U} is a subspace of V,
Dim(T(U))<=(dim(T))(dim(U))

What does “T is a finite dimensional subspace of L(V)” mean?
L1(v)+L2(v) = (L1+L2)(v)
aL1(v) = (aL1)(v) ?

and what is dim(T) and dim(T(U))?

Every Linear transformation has its Matrix format, dim(T) is dim(M(T))?
I am a fish in Linear algebra. Hope I explain my questions clearly.
 

Answers and Replies

  • #2
radou
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I am a fish in Linear algebra.
You are a fish in definitions, obviously. First of all, I assume that by L(V) you mean the vector space of all linear operators from some finite dimensional space V to V?
 
  • #3
You are a fish in definitions, obviously. First of all, I assume that by L(V) you mean the vector space of all linear operators from some finite dimensional space V to V?
Yeah, L(V) means the vector space of all linear operators from finite dimensional space V to V.
 
  • #4
radou
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OK, so, first of all, I assume this is the problem statement:

Prove that, if T is a finite dimensional subspace of L(V) and U is a finite dimensional subspace of V, then
(i) T(U) = {F(u) | F is in T, u is in U} is a subspace of V,
(ii) Dim(T(U))<=(dim(T))(dim(U)) .

In general, if A : V --> W is a linear operator, and L is a subspace of V, then A(L) is a subspace of W. Do you know how to prove that?
 
  • #5
HallsofIvy
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T(U) = {F(u) | F is in T, u is in U} is a subspace of V
If T is a subspace of V, then what do you mean by T(U)? If F is in T, then F is a vector in V. What do you mean by F(u)?
 
  • #6
radou
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If T is a subspace of V, then what do you mean by T(U)? If F is in T, then F is a vector in V. What do you mean by F(u)?
He said T was a subspace of L(V), not V.
 
  • #7
In general, if A : V --> W is a linear operator, and L is a subspace of V, then A(L) is a subspace of W. Do you know how to prove that?
yeah, I know how to prove it.
I just has an idea of what does "T is a finite dimensional subspace of L(V)" mean.
Does It mean that T is a subset of one kind of operator and this subset must be closed under addition and scalar multiplication.
for example, the non-invertible operator is not a subspace of L(V), cause it is not closed under addition. [tex]\left(\begin{array}{cc}1&0\\0&0\end{array}\right)[/tex] and [tex]\left(\begin{array}{cc}0&0\\0&1\end{array}\right)[/tex] they are both non-invertible operator but their addition [tex]\left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/tex] is invertible.
 
  • #8
radou
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yeah, I know how to prove it.
I just has an idea of what does "T is a finite dimensional subspace of L(V)" mean.
Does It mean that T is a subset of one kind of operator and this subset must be closed under addition and scalar multiplication.
Yes, exactly.

For example, take some subspace M of V. We say that M is invariant under the operator A, if A(M) is in M. Now, for some fixed subspace M, take the set of all operators such that M is invariant for these operators. Then this set is a subspace of L(V).

Edit: perhaps a more instructive example of a subspace of L(V) would be the space of all continuous linear operators from V to V.
 
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  • #9
Thank you radou.
I think I've already figured out how to prove the problem.
 

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