A problem on the Image Space

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1. Nov 15, 2015

Abtinnn

1. The problem statement, all variables and given/known data

If A is an mxn matrix, show that for each invertible nxn matrix V, im(A) = im(AV)

2. Relevant equations
none

3. The attempt at a solution
I know that im(A) can also be written as the span of columns of A.
I also know that AV = [Av1 Av2 ... Avn]
so im(AV) is the span of the columns of that matrix. However, I don't understand how the two can be equal.

2. Nov 15, 2015

Staff: Mentor

Forget the spanning vectors for a moment. What does it mean for a vector x to be in $im A$. $im (A V)$ resp.?

3. Nov 15, 2015

Abtinnn

If A is mxn and y ∈ im(A), then y can be written as Ax, where x ∈ Rn.
If y ∈ im(AV) then y can be written as (AV)x, where x ∈ Rn.

4. Nov 15, 2015

Staff: Mentor

Right. Now all you need is the associative law for linear functions for one inclusion and to put $V \cdot V^{-1} = 1$ somewhere in between for the other inclusion. $im (A \cdot V) ⊆ im A$ and $im (A \cdot V) ⊇ im A$

Actually you've already proved one inclusion by explaining to me.

5. Nov 15, 2015

Abtinnn

I believe I understand it! Could you please check if I've got it right?

Assume y ∈ im A
then y = Ax = (AVV-1)x
y = AV(V-1x)
since V-1x ∈ Rn, then y ∈ im(AV) and im(A) ⊆ im(AV)

Assume y ∈ im AV
then y = AVx = A(Vx)
since Vx ∈ Rn, then y ∈ im(A) and im(AV) ⊆ im(A)

Therefore im(A) = im(AV).

6. Nov 15, 2015

Staff: Mentor

yep

7. Nov 15, 2015

Abtinnn

Thanks a lot!! I really appreciate it :)

8. Nov 15, 2015

Staff: Mentor

You're welcome.