Direct sum and projection map for two dimensional vector

Question

$ U = \{(x,0) \ | \ x \in \Bbb R\} $ and $ W = \{(x,x) \ | \ x \in \Bbb R\} $

Prove that $ \Bbb R^2 = U \oplus W $

I am having trouble proving this, maybe because it is in a different form than I am used to.

Do I still need to show that this is a basis for $ \Bbb R^2$ ? Then how do you make the jump from $x$ as the only variable to $(x,y)$?

Answer 1

You can write the sets as $U = \{ (x,0) \, \mid \, x \in \mathbb{R} \}$ and $W = \{ (y,y) \, \mid \, y \in \mathbb{R} \}$. Then you just need to show that for any $(a,b) \in \mathbb{R}^2$, you can find $\textbf{unique}$ $(x,0) \in U$ and $(y,y) \in W$ so that $(a,b) = (x,0) + (y,y)$.

Answer 2

Let $v=(x_0,y_0)$ be an arbitrary. Then you can write it as $$v=u+w$$ where $u=(x_0-y_0,0) \in U$ and $w=(y_0,y_0) \in W$

Answer 3

In the plane draw the usual $x$- and $y$-axes. Draw another line making angle $\pi/4$ with $x$-axis, intersecting at the origin.

Now erase the $y$-axis. Mark points on both the lines at unit distance from the origin calling them vectors $u$ and $v$. Given a point P in the plane draw lines parallel to both the lines starting from P. Let them intersect those two lines at A and B.

Now by parallelogram law of addition, for the vector $OP$ we have $OP=OA+OB$. And this is same as $OP= au+bv$ where $a$ is the ratio of lengths of $OA$ to $u$ and $b$ is the ratio of $OB$ to $v$. This is unique so every vector is a unique linear combination of these two vectors. This answers your question geometrically.