## Another Linear Algebra proof about linear transformations

1. The problem statement, all variables and given/known data
Given:
T is a linear transformation from V -> W and the dim(V) = n and dim(W) = m

Prove:
If β = {v1, ..., vm} is a basis of V, then { T(v1), ..., T(vm) } spans the image of T.
NOTE: because of bad hand writing I can't tell if the bold is suppose to be an 'm' or an 'n'. I think 'm' because that makes more sense to me.

3. The attempt at a solution

Let A = { T(v1), ..., T(vm) } .

We must show that the vectors in A are L.I., and that the dim(A) is m.

If we show the vectors of A are L.I. then since there are m vectors we know the dimension is m.

Then there must be only the trivial solution to c1T(v1) + ... + cmT(vm) = 0 .
Or, by linearity, T(c1v1 + ... + cmvm) = 0 .

Note: I don't think I can jump straight to the next step. (in a similar thread of mine I did so by utilizing the fact that the null space is 0).

Since the vectors v1...vm form a basis, they are L.I. and only the trivial solution exists.

Therefore span(A) = Im(T).

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 Quote by brushman 1. The problem statement, all variables and given/known data Given: T is a linear transformation from V -> W and the dim(V) = n and dim(W) = m Prove: If β = {v1, ..., vm} is a basis of V, then { T(v1), ..., T(vm) } spans the image of T. NOTE: because of bad hand writing I can't tell if the bold is suppose to be an 'm' or an 'n'. I think 'm' because that makes more sense to me.
But it doesn't make sense to me. It's given that dim(V) = n, so any basis for V must have n vectors. So β = {v1, ..., vn}.

From this, the set of image vectors must be { T(v1), ..., T(vn) }
 Quote by brushman 3. The attempt at a solution Let A = { T(v1), ..., T(vm) } . We must show that the vectors in A are L.I., and that the dim(A) is m. If we show the vectors of A are L.I. then since there are m vectors we know the dimension is m. Then there must be only the trivial solution to c1T(v1) + ... + cmT(vm) = 0 . Or, by linearity, T(c1v1 + ... + cmvm) = 0 . Note: I don't think I can jump straight to the next step. (in a similar thread of mine I did so by utilizing the fact that the null space is 0). Since the vectors v1...vm form a basis, they are L.I. and only the trivial solution exists. Therefore span(A) = Im(T).

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 Quote by brushman 1. The problem statement, all variables and given/known data Given: T is a linear transformation from V -> W and the dim(V) = n and dim(W) = m Prove: If β = {v1, ..., vm} is a basis of V, then { T(v1), ..., T(vm) } spans the image of T. NOTE: because of bad hand writing I can't tell if the bold is suppose to be an 'm' or an 'n'. I think 'm' because that makes more sense to me. 3. The attempt at a solution Let A = { T(v1), ..., T(vm) } . We must show that the vectors in A are L.I., and that the dim(A) is m.
The problem isn't asking you to show that A is a basis for W. It's asking you to show A spans Im(T). The vectors in A may not, in fact, be independent.

## Another Linear Algebra proof about linear transformations

Thanks Mark and vela. How does this look:

Let A = { T(v1), ..., T(vm) } and β = {v1, ..., vn}.

Since β is a basis of V, all vectors in V can be written as v = c1v1 + ... + cnvn.

Then T(v) = T(c1v1 + ... + cnvn) = c1 T(v1) + ... + cn T(vn).

Because the Im(T) = T(v), and T(v) can be written as a linear combination of the elements of A, A spans Im(T).

 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus That's the basic idea. I'd probably slightly reorder it: start the proof with "Let $w\in \mathrm{Im}(T)$" and show that this implies that w can be written as a linear combination of the vectors in A.
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