# The linear transformation $T: R^4 \rightarrow R^4$ is represented by matrix $A$. Dim ( Range of T ) = 2. Prove the following.

The linear transformation $T : \Bbb R^4 \rightarrow \Bbb R^4$ has $2$ dimensional range, and is represented by matrix $A$.

Let $M$ be a given $4 × 4$ matrix and let $S$ be the vector space consisting of vectors of the form $MAx$, where $x \in R^4$.

Show that if $M$ is non-singular then the dimension of $S$ is $2$.

My attempt at a solution :

$Ax = b = a_1v_1 + a_2v_2$, where $(v_1,v_2)$ are a basis of $T$. Applying $M$, we get

$MAx = Mb = M(a_1v_1) +M(a_2v_2)$

$MAx = a_1(Mv_1) + a_2 (Mv_2)$

How do I use the fact that $M$ is non-singular to prove that $Mv_1$ and $Mv_2$ are linearly independent?

• Do you want the dimension of the image of $A$ to be $2$? It's vaguely stated in the title but you should clarify – leibnewtz May 28 '17 at 21:28
• What do you mean by the basis of a matrix? – amd May 28 '17 at 23:18

So, $v_1,v_2$ is a basis of the range of $T$ (= column space of $A$).
Thus, what you proved so far, is that range of $MA$ is included in the span of $Mv_1,Mv_2$, so $\dim({\rm range}\,MA)\le 2$, and indeed their independency is left to prove only.
For that, we can simply use e.g. that $M$ is invertible.
In fact, whenever the kernel (nullspace) of $M$ is trivial, $M$ "reflects" independency of vectors: $v_1,\dots,v_n$ independent implies $Mv_1,\dots,Mv_n$ independent.