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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.

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If you just write down in terms of vectors what it means to be in those null/column spaces, this should become easy. For instance $v\in\ker A$ iff $A\cdot v=0$; can you see the link with $\ker(UA)$? –  Marc van Leeuwen Sep 16 '13 at 10:00
    
Thought I mentioned that in the subject, but I just edited it to show. Thanks for pointing it out. –  agent154 Sep 19 '13 at 3:21

2 Answers 2

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.

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I've thought about your hint and I just can't seem to come up with anything. The best I could think of (which may be on the right track but I can't figure out where to go with it) is to show that if $A\vec{x}=AV\vec{u}$ then $\vec{x}=V\vec{u}$. I know that since $V$ is an invertable $n\times n$ matrix, its column space is all of $R^n$... but then I have no idea where to go. –  agent154 Sep 16 '13 at 23:22
    
@agent154 If $y$ lies inside the column space of $AV$, then $y=AVu$ for some $u$. Let $x=Vu$. Then $y=Ax$ for some vector $x$. Therefore $y$ lies inside the column space of $A$. In other words, the column space of $AV$ is a subset of the column space of $A$. The prove the set inclusion in reverse direction, you need to use the invertibility of $V$. –  user1551 Sep 17 '13 at 8:11
    
I don't see how the invertibility of $V$ comes into play. Couldn't you just do the exact same thing in reverse? I'm having an immensely hard time with this for some stupid reason. I even spoke with one of my previous profs in Linear Algebra and he gave basically the same argument but I still can't understand why the proof is actually proving what it claims. –  agent154 Sep 19 '13 at 1:02
    
If $\vec{y}$ lies inside the column space of $AV$, then $y=AV\vec{u}$ for some $\vec{u}\in\mathbb{R}^{n}$. Let $\vec{x}=V\vec{u}$, $\vec{u}=V^{-1}\vec{x}$. Then $\vec{y}=A\vec{x}$ for some vector $\vec{x}\in\mathbb{R}^{n}$. Therefore $\vec{y}$ lies inside the column space of $A$, and $Col(AV)\subset Col(A)$. If $\vec{y}$ lies inside the column space of $A$, then $\vec{y}=A\vec{x}$ for some $\vec{x}\in\mathbb{R}^{n}$. Let $\vec{x}=V\vec{u}$ for some $\vec{u}\in\mathbb{R}^{n}$. Then $\vec{y}=AV\vec{u}$, and $y\in Col(AV)$. –  agent154 Sep 19 '13 at 1:06
up vote 0 down vote accepted

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)

  1. 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 $Col(AV)\subset Col(A)$.

  2. 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 $Col(A)\subset Col(AV)$

  3. 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.

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