Dimensionality of a subspace described by equations T-F questions 20-21 from http://www.math.washington.edu/~smith/Teaching/308/2013-Spring-Midterm-Answers.pdf
The set $\{(x_1, x_2, x_3, x_4) \mid x_1 + x_3 = x_2 - x_4 = 0\}$ is a two-dimensional subspace of $R^4$.  -- T
$\{x \in R^4 \mid x_1 + x_3 = x_2 - x_1\}$ is a 3-dimensional subspace of $R^4$. -- T
How do I determine the number of independent vectors being described, which will tell me the dimensionality?
 A: Note in the first question that we have $2$ separate equalities expressed compactly.  That is, we're given the information that
$$
x_1 + x_3 = 0\\
x_2 - x_4 = 0
$$
In the language of matrix multiplication, we may state that this set is the set of all $(x_1,x_2,x_3,x_4)$ such that
$$
\pmatrix{
1&0&1&0\\
0&1&0&-1}
\pmatrix{x_1\\x_2\\x_3\\x_4}=
\pmatrix{0\\0}
$$
Or in other words, we're looking for the nullspace of the matrix 
$$
A = \pmatrix{
1&0&1&0\\
0&1&0&-1}
$$
Since the null-space of any vector space is a subspace, we are already halfway there. In order to show that this null-space is two-dimensional, we note that $A$ is already row-reduced, with two pivots and two free variables.  Since there are two free variables, we can conclude that the null-space of $A$ is two dimensional.

For the next question, we can rearrange the equation given to find that the subset is determined by the equation
$$
2x_1 - x_2 + x_3 = 0
$$
In matrix notation, that is
$$
\pmatrix{2&-1&1&0}\pmatrix{x_1\\x_2\\x_3\\x_4} = 0
$$
Which is to say that we are considering the null-space of the matrix
$$
A = \pmatrix{2&-1&1&0}
$$

Note that I have used language that I have guessed you might be familiar with in the hopes of relating this questions to things you may have already learned.  If there's something that it would be helpful for me to clarify, let me know.
A: As an intuition you can count variables and subtract the number of conditions. But that doesn't give you a proof. One way of showing dimension is giving isomorphisms:
First note that both of your sets, $$V_1 = \{(x_1, x_2, x_3, x_4) | x_1 + x_3 = x_2 - x_4 = 0\} \\V_2 = \{x \in R^4 | x_1 + x_3 = x_2 - x_1\}$$ are indeed vector spaces over $\Bbb R$. In order to find the dimension of $V_1$, consider the map:
$$f_1: \Bbb R^2 \rightarrow V_1 \\ (t_1,t_2) \mapsto (t_1,t_2,-t_1,t_2).$$
This is an $\Bbb R$-linear map. It is easy to see that it is one-to-one and onto (we can give a pre-image of any element in $V_1$) so $f_1$ is indeed an isomorphism. As isomorphisms are dimension-preserving, $\dim V_1 = \dim \Bbb R^2= 2$.
The same argument works for $V_2$ with a map like
$$f_2: \Bbb R^3 \rightarrow V_2 \\ (t_1,t_2,t_3) \rightarrow
 (t_1,t_2+t_1,t_2-t_1,t_3)$$ where $t_1$ is as interpreted $x_1$, the entry $t_2$
 works as $x_1-x_3 = x_2 -x_1$ and $t_3 $ gets the unconstrained $x_4$.
 Therefore $\dim V_2 = \dim \Bbb R^3 = 3$.
