Question

Let W = {(x, y) E R² | 2x + 3y = 0} Show
that W is a subspace of R2 . Find the
dimension of W. Also show that the cosets of
W are lines 2x + 3y + c = 0, where c belongs to R

Answer 1

Let $W = \{(x, y) \in \mathbb R^2\ |\ 2x + 3y = 0\}$. Let us show that $W$ is a subspace of $\mathbb R^2$. Let $(x_1,y_1),(x_2,y_2)\in W,\ a,b\in \mathbb R.$ Then $2x_1+3y_1=0, \ 2x_2+3y_2=0.$ Taking into account that $a(x_1,y_1)+b(x_2,y_2)=(ax_1+bx_2, ay_1+by_2)$ and $2(ax_1+bx_2)+3(ay_1+by_2)=2ax_1+2bx_2+3ay_1+3by_2=a(2x_1+3y_1)+b(2x_2+3y_2)=a\cdot 0+b\cdot 0=0,$ we conclude that $a(x_1,y_1)+b(x_2,y_2)\in W$, and thus $W$ is a subspace of $\mathbb R^2.$

For any $(x, y)\in W,$ $x=-1.5y$, and therefore, $(x,y)=(-1.5y,y)=y(-1.5,1)$. It follows that the vector $(-1.5,1)$ span $W$, and consequently, the dimension of $W$ is 1.

The set $v+W=\{v+w\ : w\in W\}$ is called a coset of $W$. Let $v=(a,b)$ be arbitrary element of $\mathbb R^2$. Denote by $c=-2a-3b$.Then $v+W=\{(a,b)+(x,y)\ | (x,y)\in W\}= \{(a+x,b+y)\ | 2x+3y=0\}=\{(x',y')\ | 2(x'-a)+3(y'-b)=0\}=\{(x',y')\ | 2x'+3y'-2a-3b=0\}=\{(x',y')\ | 2x'+3y'+c=0\}$

Therefore, the cosets of $W$ are lines $2x + 3y + c = 0$ , where $c\in\mathbb R.$