Given two subspaces $L_1$ and $L_2$ of $\mathbb{R}^4$ as follows:
$$L_1:
\begin{array} \ x_1-x_2-x_4=0, \\ x_1+x_2+x_3+x_4=0, \\ x_2-x_3=0. \end{array} $$ and
$$L_2:
\begin{array} \ x_1-x_2+x_3-x_4=0, \\ x_1+x_2=0. \end{array} $$
Find the basis of $L_1+L_2$ and determine if the sum is direct.
I found out that $$L_1=\bigg\{\bigg(-\frac{1}{2}t, \ \ t, \ \ t,-\frac{3}{2}t\bigg) \ \ \bigg| \ \ t\in \mathbb{R}\bigg\}$$ and $$L_2=\bigg\{\bigg(\frac{-r+s}{2}, \ \ \frac{r-s}{2}, \ \ r, \ \ s \bigg)\ \ \bigg| \ \ r,s\in \mathbb{R}\bigg\}$$
Hence, the sum is
$$L_1+L_2=\bigg\{\bigg(\frac{-r+s}{2}-\frac{1}{2}t, \ \ \frac{r-s}{2}+t, \ \ r+t, \ \ s-\frac{3}{2}t \bigg) \ \ \bigg| \ \ r,s,t\in \mathbb{R}\bigg\}$$
Then, I checked that $L_1 \cap L_2 = \{0\}$. Is it enough to conclude the sum is direct?
Also, I could not find a basis for $L_1+L_2$.
Any help is appreciated.
Best Answer
Note that you should explicit the vectors spanning the sum a little more.
Note that sometimes could be easier think to $L_{1},L_{2}$ as solution of $$\begin{pmatrix}1 & -1 & 0 & -1 \\ 1 & 1 & 1 & 1 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\\end{pmatrix}\begin{pmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}$$ And $$\begin{pmatrix}1 & -1 &1 & -1 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4}\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix}$$
This representation could be useful in order to estimate apriori the dimension of those subspaces, since linear independance can be checked quite easily.
What we notice with this is representation is that we can study $L_{1},L_{2}$ as Kernel of those associated matrix, so the dimension can be studied through theorems or known formulas as the rank-nullity Theorem.
Additionally I think a good approach could be estimate the dimension of the intersection and determine a basis using Grassmann's identity
$$\dim(L_{1}+L_{2}) = \dim(L_{1})+\dim(L_{2})-\dim({L_{1} \cap L_{2}})$$
From here you can immediately observe that since $L_{1}$ has at least dimension $1$, but since we have a $3 \times 3$ invertible minor is at most one, thanks to the rank nullity theorem, and $L_{2}$ has at least dimension $2$ (the last two rows), and at most since we got a $2\times 2$ invertible minor), so the sum could be direct since our space is $\mathbb{R}^{4}$ but the sum of the dimension is $3$.
Since the intersection is a subspace of both $L_{1},L_{2}$ it has dimension $0$,$1$ since it has to be in particular a subspace of $L_{1}$.
To determine the exact dimension we can proceed as above, taking a basis of $L_{1},L_{2}$ taking the matrix spanned by those vector and calculate the rank : if the rank is $3$ they are linearly independant, hence the intersection is trivial and the sum is direct, otherwise the intersection is not trivial and has to coincide with $L_{1}$ in that case in order to find a basis you just have to take a vector in $w \in L_{2} : w \not\in L_{1}$ and extend a basis of $L_{1}$ with $w_{2}$ to get a basis for the sum.