Does $W \oplus U$, for a particular $W$ and $U$, equal $\mathbb{R}^4$? Trying to teach myself some linear algebra before I even start the class, but I am a bit stuck. I have a question and honestly I don't know where to start. I'll be glad to get an answer but more important a good explanation.
By the way it's the first time I am asking a question here, so I am very sorry if I'm not following the rules.
Here is the question:
$$U = \{(x,y,z,t) \in \mathbb{R}^4 \mid x+5y+4z+t = 0 \land y+2z+t=0\}$$
$$W = \{(x,y,z,t) \in \mathbb{R}^4 \mid x+z+3t = 0 \land 2x-3y-4z+3t=0\}$$
Does $W \oplus U$ equal $\mathbb{R}^4$ (where the "+" sign is inside a circle, but it doesn't say what it means)?
Thank you very much; if the question is not written properly, please let me know.
 A: The symbol $\oplus$ stands for direct sum (of vector spaces, in this case). Basically, the question is: can you write every vector in $\Bbb{R}^4$ as a sum of a (unique) vector in $W$ and a (unique) vector in $U$?
Note that the answer is positive if and only if you can find a basis of $\Bbb{R}^4$ formed by vectors in $U$ and vectors in $W$. Since $U$ and $W$ have dimension $2$, this amounts to finding vectors $u_1,u_2 \in U$ and $w_1,w_2 \in W$ such that $u_1,u_2,w_1,w_2$ are linearly independent.

A vector $(x,y,z,t)$ is in $U$ if and only if
$$
\begin{cases}
x+5y+4z+t = 0 \\
y+2z+t = 0
\end{cases}
\leftrightarrow
\begin{cases}
x = -4y -2z \\
t = -y -2z
\end{cases}
$$
in particular, every vector in $U$ is of the form $a u_1 + b u_2$ with
$$
u_1 =
\begin{pmatrix}
-4 \\
1 \\
0 \\
-1
\end{pmatrix}
\qquad
u_2 =
\begin{pmatrix}
-2 \\
0 \\
1 \\
-2
\end{pmatrix}
$$
Similarly, a vector $(x,y,z,t)$ is in $W$ if and only if
$$
\begin{cases}
x+z+3t = 0 \\
2x−3y−4z+3t = 0
\end{cases}
\leftrightarrow
\begin{cases}
x = 3y +5z \\
t = -y -2z
\end{cases}
$$
in particular, every vector in $W$ is of the form $a w_1 + b w_2$ with
$$
w_1 =
\begin{pmatrix}
3 \\
1 \\
0 \\
-1
\end{pmatrix}
\qquad
w_2 =
\begin{pmatrix}
5 \\
0 \\
1 \\
-2
\end{pmatrix}
$$
Now, the quickest way to check if some vectors are linearly independent is to check if the determinant of the matrix which has those vectors as columns is non-zero. In our case, $u_1,u_2,w_1,w_2$ are linearly independent if and only if
$$
0 \neq \det
\begin{pmatrix}
-4 & -2 & 3 & 5\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
-1 & -2 & -1 & -2
\end{pmatrix}
$$
which isn't possible, because the second, third, and fourth line are linearly dependent. In other words, this means that $U \oplus W \neq \Bbb{R}^4$.
On the other hand, you can prove that $u_1,u_2,w_1$ are linearly independent, which means that $U \oplus W$ is isomorphic to $\Bbb{R}^3$.
A: Ok, first of all, $U\bigoplus W$, represents the direct sum as they previously mentioned. The vector space $U+W = \{u+w:u\in U, w\in W\}$. When the sum is direct, (i.e, the $\bigoplus$ symbol is used, it means that $U\cap W = \emptyset$. 
Now, going to the exercise, to see that $U\bigoplus W = \mathbb{R}^4$ you have to see that $\mathbb{R}^4 \subseteq U\bigoplus W$ and $U\bigoplus W \subseteq \mathbb{R}^4$. One of the inclusions is obvious. However the other is not. Try to prove it. (Watch out, this may be not true, as the exercise states, in this case you will have to find a vector $v \in \mathbb{R}^4$ such that $v$ is not in $U\bigoplus W$).
