Is Im(A) Equal to Im(AV) for an Invertible Matrix V?

[Abtinnn]

Homework Statement

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

Homework Equations

none

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 ... Av_n]
so im(AV) is the span of the columns of that matrix. However, I don't understand how the two can be equal.

 

[fresh_42]

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

 

[Abtinnn]

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

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.

 

[fresh_42]

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. $im (A \cdot V) ⊆ I am A$ and $im (A \cdot V) ⊇ I am A$

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

 

[Abtinnn]

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. $im (A \cdot V) ⊆ I am A$ and $im (A \cdot V) ⊇ I am A$

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

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

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

Therefore im(A) = im(AV).

 

[fresh_42]

yep

 

[Abtinnn]

Thanks a lot! I really appreciate it :)

 

[fresh_42]

You're welcome.