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Question: A Linear transformation $T: \mathbb R^4 \to \mathbb R^4$ is represented by the matrix $$\mathbf A=\begin{pmatrix} \\1&-1&2&3 \\ 2 & -3 & 4 & 5\\ 5 & -6 & 10 & 14\\ 4 & -5 & 8 & 11 \end{pmatrix}$$ (i) Show that the $\dim\big(R(\mathbf A)\big) =2$.
(ii) Let M be a given $4\times 4$ (non-singular) matrix and let $\mathcal S$ be the vector space consisting of vectors in the form of MAx where $x \in \mathbb R^4$. Show that if M is non-singular then $\dim\big(\mathcal S\big) =2$.

The first part is pretty straight-forward. All one needs to do is compute $\operatorname{rref}(\mathbf A)$.

But it is the second part that has stomped me (and for quite some time now). I did come up with a solution but it seemed very cheap to me and it didn't really satisfy me completely.
What I tried was to write A in the form, $$\mathbf A=\Big [\begin{matrix} \vec p_1& \vec p_2&\vec f_1&\vec f_2 \\ \end{matrix}\Big]$$ Where $\vec p_1$ and $\vec p_2$ are the columns corresponding to the pivot variables (there are $2$ pivot variables, since $\dim\big(R(\mathbf A)\big) =2$) and $ \vec f_1$ and $\vec f_2$ are the columns corresponding to the free variables.
Now if M be any $4\times 4$ matrix in the form, $$\mathbf M=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big]$$ Where $\vec c_1, \vec c_2, \vec c_3$ and $\vec c_4$ are the four columns of M, then $$\mathbf {MAx}=\Big[ \begin{matrix} \vec c_1& \vec c_2& \vec c_3& \vec c_4 \\ \end{matrix}\Big] \cdot x_1\vec p_1+x_2\vec p_2+ x_3\vec f_1+x_4\vec f_2 $$ I don't know what to do after this.

Is this apprach correct? If yes, how do I proceed from here? If not, where is it that I am going wrong and how do I solve this problem?

Any Help is much appreciated!
Thanks in Advance!

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  • $\begingroup$ What are pivot and free variables again? I don't think we need them to answer this question. Also, it looks like your $MA$ is just a row vector. I would approach it like: 1)Show that the rows of $MA$ are linear combinations of rows of $A$, so $r(MA)\le r(A)$. 2) Show that the the rows of $A$ that are linearly independent remain so after being multiplied by $M$, so $r(MA)\ge2$. $\endgroup$ May 17, 2012 at 10:10
  • $\begingroup$ $MA$ can't be a row vector. I don't think multiplication of two $4\times 4$ matrices can yield a row vector? $\endgroup$
    – user11470
    May 17, 2012 at 10:13
  • $\begingroup$ I agree, it cannot. But before your edit $MA$ looked like a list of four numbers. That's why I asked :-) But it has been fixed now. Sorry about being a bit trigger-happy. $\endgroup$ May 17, 2012 at 10:15
  • $\begingroup$ Oh! My bad then :) $\endgroup$
    – user11470
    May 17, 2012 at 10:16

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I didn't read your proof thoroughly so I can't say whether it will work. What I can tell is that it is way to complicated (and confusing). Maybe you should think about the question differently. The statement is a special case of the proposition:

If $M$ is non-singular then $rank(MA)=rank(A)$

If you have a basis of $Im(A)$, can you show that $M$ takes this basis to a basis of $S$? Hint: You need non-singularity only to show that the images of the base vectors are linearly independant.

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  • $\begingroup$ Yes I do agree with you on that. The proof is very messy. I don't have much experience with proof writing. The thing is I am still in high school and I am slowly getting the hang of it:) $\endgroup$
    – user11470
    May 17, 2012 at 10:21
  • $\begingroup$ Then you are doing quite advanced stuff! No worries you will get a hang of it pretty quickly if you keep working on it. Is my answer helpful? $\endgroup$ May 17, 2012 at 10:24
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That approach doesn’t look very promising; at best it’s going to be very messy. Try showing that $\mathbf{MA}$ and $\mathbf A$ have the same null space when $\mathbf M$ is non-singular; that’s not hard, and it implies that they have the same rank.

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Isn't it enough to say that non-singular matrix preserves linear independence?

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  • $\begingroup$ Could you elaborate a little bit? $\endgroup$
    – user11470
    May 17, 2012 at 10:17
  • $\begingroup$ Since you had $Ax$ of dimention 2, then it had a two vector basis. Images of those base vectors by $M$ are linearly independent in $MAx$ and you can show that they form a maximal set of independent vectors by contradiction, using the fact that $M$ is invertible. I can elaborate more if you wish, just tell me on which part. :) $\endgroup$ May 17, 2012 at 15:08

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