Answer to Question #203548 in Linear Algebra for Rohan

Question

How do we show that W= {(x,-3x,2x)|x€R} is a subspace of R³?Also find a basis for subspace U of R³ which satisfies R³=W⊕U?

Answer 1

Let us show that $W= \{(x,-3x,2x)|x\in\mathbb R\}$ is a subspace of $\mathbb R^3$. Let $\alpha, \beta\in\mathbb R,\ (x,-3x,2x), (y,-3y,2y)\in W.$ Then $\alpha (x,-3x,2x)+\beta (y,-3y,2y)= (\alpha x,-3\alpha x,2\alpha x)+(\beta y,-3\beta y,2\beta y)= (\alpha x+\beta y,-3\alpha x-3\beta y,2\alpha x+2\beta y)= (\alpha x+\beta y,-3(\alpha x+\beta y),2(\alpha x+\beta y))\in W.$

Therefore, $W= \{(x,-3x,2x)|x\in\mathbb R\}$ is a subspace of $\mathbb R^3$. This subspace is of dimention 1 generated by $w=(1,-3,2).$

Consider the vectors $v=(0,2,3)$ and $u=(-13,-3,2).$ Taking into account that for the dot products we have $w\cdot v=0-6+6=0, w\cdot u=-13+9+4=0,\ v\cdot u=-6+6=0,$ we conclude that the subspace $U=\langle v,u\rangle$ satisfies $\mathbb R^3=W⊕U$, and $U$ is of dimention 2 with basis consisting of vectors $v$ and $u.$