Find dimension from subspace equations In order to calculate its dimension, I need the basis of the subspace. How can I get it from the equations?
Find the dimension of the following subspaces of $\mathbb{R}^5$:
$$ U = \{(x_1,x_2,x_3,x_4,x_5) \ | \ 2x_1 - x_2 - x_3 = 0, x_4-3x_5=0 \}\\
V = \{(x_1,x_2,x_3,x_4,x_5) \ | \ 2x_1 - x_2 + x_3 + 4x_4 + 4x_5 = 0 \}.$$
Let $W$ be the subspace satisfying all 3 equations. Is it true that dim($W) = 5$?
 A: You can find the dimension by finding the dimension of the nullspace of the matrix
$$
\begin{bmatrix}
2& -1 & -1 & 0 & 0\\
0&0&0&1&-3
\end{bmatrix}
$$
The rank of this matrix is $2$, so the dimension of the nullspace is equal to $5-2 = 3$ by the rank+nullity theorem. 
A: The rank of the system of linear equations is equal to the codimension of the null space, by the rank-nullity theorem.
A: If $(x_1,x_2,x_3,x_4,x_5) \in U$, then $x_1 = \frac{1}{2}x_2 + \frac{1}{2}x_3$ and $x_4 = 3x_5$.
So
$$\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{pmatrix} = \begin{pmatrix}
\frac{1}{2}x_2 + \frac{1}{2}x_3\\
x_2 \\
x_3 \\
3x_5 \\
x_5
\end{pmatrix}= x_2\begin{pmatrix}
\frac{1}{2}\\
1\\
0\\
0\\
0\\
\end{pmatrix}+ x_3 \begin{pmatrix}
\frac{1}{2}\\
0\\
1\\
0\\
0\\
\end{pmatrix} + x_5 \begin{pmatrix}
0\\
0\\
0\\
3\\
1\\
\end{pmatrix}$$
So we see that these three vectors form a basis for $U$. Hence dim($U) = 3$.
Now if $(x_1,x_2,x_3,x_4,x_5) \in V$, then $x_2 = 2x_1 + x_3 + 4x_4 + 4x_5$.
So
$$\begin{pmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5\\
\end{pmatrix} = \begin{pmatrix}
x_1\\
2x_1 + x_3 + 4x_4 + 4x_5\\
x_3\\
x_4\\
x_5\\
\end{pmatrix}= x_1 \begin{pmatrix}
1\\
2\\
0\\
0\\
0\\
\end{pmatrix} + x_3 \begin{pmatrix}
0\\
1\\
1\\
0\\
0\\
\end{pmatrix} + x_4 \begin{pmatrix}
0\\
4\\
0\\
1\\
0\\
\end{pmatrix} + x_5 \begin{pmatrix}
0\\
4\\
0\\
0\\
1\\
\end{pmatrix}$$
So we see that these four vectors form a basis for $V$. Hence dim($V)=4$.
For $W$, rather than go through the same sort of computation, note that $W = U \cap V$. So $dim(W) \leq \min\{dim(U),dim(V)\} = 3$. So $dim(W) \neq 5$.
