To show: Direct sum of subspaces $M$ and $N$ in $R^2$ equals $R^2$, where $M=\{(x,y)\in R^2|2x+y=0 \}$ and $N=\{(x,y)\in R^2|x-y=0 \}$ If $M=\{(x,y)\in \mathbb{R}^2|2x+y=0 \}$ and $N=\{(x,y)\in \mathbb{R}^2|x-y=0 \}$. Show that $M+N=\mathbb{R}^2$
My Attempt:
$M+N=\{(x,y)\in \mathbb{R}^2|(x,y)=(x_1,y_1)+(x_2,y_2);\ (x_1,y_1)\in M,\ (x_2,y_2) \in N \}$ where
$x=x_1+x_2$ and $y=y_1+y_2$ since $(x_1,y_1) \in M$ and $(x_2,y_2) \in N$. Now
$2x_1+y_1=0$ and $x_2-y_2=0$ respectively
implies that $M+N=\{(x,y)\in \mathbb{R}^2|x=x_1+x_2,y=-2x_1+x_2) \}$.
How do I show that $M+N=\mathbb{R}^2$?
 A: Observe the bases of both the subspaces.
The basis of $M$ is
$\begin{pmatrix}
1 \\
-2 \\
\end{pmatrix}
$
and the basis of $N$ is
$\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}$.
Now consider $B = \Bigg(\begin{pmatrix}
1 \\
-2 \\
\end{pmatrix},
\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}\Bigg)$.
It is easily observable that vectors in $B$ are linearly independent (check it), and since $\dim(\mathbb{R}^2)=2$, we can conclude that $B$ is a basis of $\mathbb{R}^2$.
Thus any $v\in \mathbb{R}^2$ can be denoted as $v=a_1\begin{pmatrix}
1 \\
-2 \\
\end{pmatrix}+a_2\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}$.
Since $\begin{pmatrix}
1 \\
-2 \\
\end{pmatrix}$ and $\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}$ were basis of $M$ and $N$ respectively, we can denote $v$ as $v=m+n$, where $m\in M$ and $n\in N$.
Thus any $v\in \mathbb{R}^2$ can be denoted as $v=m+n$ for some $m\in M$ and $n\in N$, hence $M+N=\mathbb{R}^2$.
A: Quick proof: $M = \mbox{span}\{v\}$ and $N = \mbox{span}\{w\}$ are 1-D subspaces of $R^2$ where
$$
v = \left[ \begin{array}{c}
 1 \\
-2 \\
\end{array} \right] \ \mbox{and}
\ \
w = \left[ \begin{array}{c}
 1 \\
1 \\
\end{array} \right]
$$
Note that
$$
M \cap N = \{ \mathbf{0} \}
$$
and $\mbox{dim}(M \cap N) = 0.$
From Linear Algebra, we know that
$$
\mbox{dim}(M + N) = \mbox{dim}(M) + \mbox{dim}(N) - \mbox{dim}(M \cap N)
$$
or
$$
\mbox{dim}(M + N) = 1 + 1 - 0 = 2 = \mbox{dim}(R^2)
$$
Hence, we conclude that $M + N = R^2$.
