So I have this question that I just don't know how to set up for my linear algebra class. It involves the concept of subspaces which is something I've struggled on in this class for sure, it would be very much appreciated if someone could explain it! Thanks!
State whether the set is a subspace of $\mathbb{R^4}$. If it is not a subspace, explain why not and if it is a subspace, give a basis for it.
a) Let $S_1$ be the set of vectors $\begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix}$ such that $$\begin{bmatrix} 3 & 2\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 1 & 4\\ 0 & -3 \end{bmatrix} \begin{bmatrix} z \\ w \\ \end{bmatrix}$$
I have no idea how to start on this! Please help!
From the system $$\begin{bmatrix} 3 & 2\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix} = \begin{bmatrix} 1 & 4\\ 0 & -3 \end{bmatrix} \begin{bmatrix} z \\ w \\ \end{bmatrix}$$ we deduce that $3x+2y=z+4w$ and $x-y=-3w$. Then, a little algebra shows that $y= 3w+x$ and $z=2w+5x$. In other words, whenever $(x,y,z,w)\in S_1$, then $y= 3w+x$ and $z=2w+5x$. On the other hand, it is not difficult to see that if $x,w\in \mathbb{R}$, then $(x,3w+x,2w+5x,w)\in S_1$, i.e., $$\begin{bmatrix} 3 & 2\\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ 3w+x \\ \end{bmatrix} = \begin{bmatrix} 1 & 4\\ 0 & -3 \end{bmatrix} \begin{bmatrix} 2w+5x \\ w \\ \end{bmatrix}.$$ This shows that, for every $(x,y,z,w)\in \mathbb{R}^{4}$, $(x,y,z,w)\in S_1$ if and only if $y= 3w+x$ and $z=2w+5x$. Furthermore, the last two equations kind of tell us that every element of $S_1$ "depends" only on two variables.
With this in mind, it is natural to think that $S_1$ is a 2- dimensional subspace of $\mathbb{R}^{4}$. Proving that $S_1$ is a subspace of $\mathbb{R}^{4}$ at this point should be very easy. You just have to argue now that it has dimension 2. This can be done in a number of ways, but I think a very elegant way to do it is to note that $f: \mathbb{R}^{2} \to S_1$ given by $f(x,w) = (x,3w+x,2w+5x,w)$ is a linear isomorphism (this is also simple). In particular, $\{f(1,0), f(0,1)\}$ is a basis for $S_1$.