# Proving a vector space property

Let $M$ be a $4\times4$ matrix and $S$ be the vector space consisting of vectors of the form $MAx$ where $x\in ℝ^4$ and $A$ is a $4\times4$ matrix having a rank of $2$. Show that if $M$ is non singular then the dimension of $S$ is 2.

$$S=MAx$$ $$M^{-1}S=Ax$$ $M^{-1}$ exist since it is non-singular.

Can somebody complete this proof?

Moreover what's meant by "dimension" of $S$? The dimension of range space of $S$ or the dimension of null space of $S$?

• Hint: What does it mean that $A$ is of rank 2? – NickD May 24 '17 at 14:16
• Since "A is a 4 by 4 matrix having a rank of 2", A maps $R^4$ to a two dimensional subspace of $R^4$. Since "M is non-singular", M maps that two dimensional subspace of $R^4$ to a two dimensional subspace of $R^4$. – user247327 May 24 '17 at 14:16
• @Nick it means $Ax=pe_1+qe_2$ where $p$ and $q$ are scalar. The problem I am having is, it's evident that $M^{-1}S$ has a dimension of 2, but how to make a statement for $S$? – mathnoob123 May 24 '17 at 14:18
• @user247327 How do you know that M maps 2D subspace to 2D subspace? Can it not it to 1D subspace? – mathnoob123 May 24 '17 at 14:19
• As you yourself pointed out, M is non-singular, so no, it cannot go down a dimension. – NickD May 24 '17 at 15:08

Null space of $S$ is of dimension $2$, as a vector of the form $MAx$ is zero iff $Ax=0$ as $M$ is nonsingular, and the dimension of the vector space consisting of all vectors such that $Ax=0$ is $2$, as rank of $A$ is $2$, hence $dim(S)=4-2=2$.