<s>[INST] What is the dimension and base of the following vectors' sum and intersection? I have 2 Vector Subspaces of $\mathbb{R}^3$, namely $U = \operatorname{Span}(\begin{pmatrix} 2 \\ 5 \\ 9\end{pmatrix}, \begin{pmatrix} 0 \\ -1 \\ -3\end{pmatrix})$ and $W = \operatorname{Span}(\begin{pmatrix} -3 \\ 1 \\ 6\end{pmatrix}, \begin{pmatrix} 5 \\ 3 \\ 0\end{pmatrix})$. Now I want to find the dimension and base of $U + V$ and of $U \cap W$. This is what I have done so far:
Since the vectors of $U$ and $W$ are lineary independent, they have dimension $\dim(U) = \dim(W) = 2$. I can see that $U+W = \operatorname{Span}(\begin{pmatrix} -1 \\ 6 \\ 15\end{pmatrix}, \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix})$ and it is also lineary independent, so $\dim(U+W) = 2$.
This means that $\dim(U\cap W) = \dim(U) + \dim(W) - \dim(U+W) = 2 + 2 - 2 = 2$.
Now, going further, I want to find the base of $U\cap W$, so I construct the following matrix and reduce it:
$$\begin{pmatrix} 2 & 0 & 3 & -5 \\ 5 & -1 & -1 & -3 \\ 9 & -3 & -6 & 0 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & -1 \end{pmatrix}$$
So, this means that the two vector spaces are intersecting exactly when $$\lambda \begin{pmatrix} 2 \\ 5 \\ 9\end{pmatrix} + \lambda \begin{pmatrix} 0 \\ -1 \\ -3\end{pmatrix} = \lambda \begin{pmatrix} -1 \\ 6 \\ 15\end{pmatrix} + \lambda \begin{pmatrix} 5 \\ 2 \\ -3 \end{pmatrix}$$
So, a base of $U\cap W$ would be $\begin{pmatrix} 2 \\ 4 \\ 6\end{pmatrix}$
Now, $U \cap W $ has dimension 1. But earlier I found that it should be 2. Where is the error? Do I make some error when I calculate $U + W$? Can someone point me in the right direction?
 [/INST] You only formed the sum of the chosen basis elements of $U$ and $W$ instead of all vectors therein.
Your calculation for the intersection is correct, hence $\dim(U+V)=3$, i.e. $U+V=\Bbb R^3$, that is, every vector can be expressed as a sum $u+v$ with $u\in U$ and $v\in V$. 
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