Show that the null space of $A$ is equal to the null space of $UA$ for some invertible $m\times m$ matrix $U$ and some $m\times n$ matrix $A$? And equivalently, given some invertible $n\times n$ matrix V, how can I show that the column space of $A$ is equal to the column space of $AV$.?
I've looked through my textbook and can't seem to find anything relating to this in the section dedicated to the row and column space. The best I could find online and from other sources is that the row space of a matrix is not affected by elementary row operations. So extrapolating from this, I can see that $UA$ would have the same row space as $A$, as $U$ can be reduced to a series of elementary row operations. But I read that elementary row operations can change the column space. So I still have no understanding of this property.
 A: Another hint. By definition, the column space of $A$ consists of vectors of the form $Ax$. Similarly, the column space of $AV$ contain vectors of the form $AVu$. Now, to say that the column spaces of $A$ and $AV$ are identical, we mean for every vector of the form $Ax$ is equal to a vector of the form $AVu$ and vice versa.
A: Thanks to the advice, I think I got these done...
Proof that Null(A)=Null(UA)

$\vec{x}$ is in the null space of $A$ if and only if $\vec{x}$ is a solution to the homogeneous system $A\vec{x}=\vec{0}$. Equivalently, $\vec{x}$ is in the null space of $U\!A$ if and only if $\vec{x}$ is a solution to the homogeneous system $U\!A\vec{x}=\vec{0}$. Since $U$ is an invertable matrix, we can multiply both sides of the equation $U\!A\vec{x}=\vec{0}$ on the left by $U^{-1}$ and get $U^{-1}UA\vec{x}=U^{-1}\vec{0}$, or $A\vec{x}=\vec{0}$. Therefore, every $\vec{x}$ that is in the null space of $A$ is also in the null space of $U\!A$, and every $\vec{x}$ not in the null space of $A$ is not in the null space of $U\!A$. The null space of $A$ is equivalent to the null space of $U\!A$.

Proof that Col(A)=Col(AV)


*

*If $\vec{y}$ lies inside the column space of $AV$, then $\vec{y}=AV\vec{u}$ for some $\vec{u}\in\mathbb{R}^{n}$. Let $\vec{x}=V\vec{u}$. Then $\vec{y}=A\vec{x}$ for some vector $\vec{x}\in\mathbb{R}^{n}$. Therefore $\vec{y}\in Col(A)$, and $Col(AV)\subset Col(A)$.


*If $\vec{y}$ lies inside the column space of $A$, then $\vec{y}=A\vec{x}$ for some $\vec{x}\in\mathbb{R}^{n}$. $A\vec{x}=AI\vec{x}=AVV^{-1}\vec{x}$. Let $\vec{x}=V\vec{u}$ for some $\vec{u}\in\mathbb{R}^{n}$. $\vec{u}=V^{-1}\vec{x}$. Then $AVV^{-1}\vec{x}=AV\vec{u}$. Therefore $\vec{y}\in Col(AV)$, and $Col(A)\subset Col(AV)$


*By (1) and (2), $Col(AV)\subset Col(A)\wedge Col(A)\subset Col(AV)\iff Col(A)=Col(AV)$.

Now that I think about it though, the first proof seems insufficient. I'm not sure if I'm missing anything or if that's good enough.
A: Proof. We will show that null(A) = null(UA) by showing that null(A) ⊆ null(UA) and that
null (U A) ⊆ null (A). 
Let ⃗x ∈ null (A). Then
A⃗x =⃗0 =⇒ U(A⃗x) = U⃗0 =⇒ (UA)⃗x =⃗0
so ⃗x ∈ null (UA) and null (A) ⊆ null (UA).
Now let ⃗x ∈ null (UA). 
Then (UA)⃗x=⃗0 =⇒ U−1(UA)⃗x=U−1⃗0 =⇒ (U−1U)A⃗x=⃗0 =⇒ A⃗x=⃗0
so ⃗x ∈ null (A) and null (UA) ⊆ null (A).
Thus null (A) = null (UA) .
