# Questions about the concept of subspace of linear transformation

1. Oct 22, 2007

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.

2. Oct 22, 2007

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. Oct 22, 2007

Yeah, L(V) means the vector space of all linear operators from finite dimensional space V to V.

4. Oct 23, 2007

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. Oct 23, 2007

### HallsofIvy

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. Oct 23, 2007

He said T was a subspace of L(V), not V.

7. Oct 23, 2007

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. $$\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$$ and $$\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$$ they are both non-invertible operator but their addition $$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$ is invertible.

8. Oct 23, 2007

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.

Last edited: Oct 23, 2007
9. Oct 23, 2007