Answer to Question #336649 in Linear Algebra for dev

Question

Show that 𝑊 = {(𝑥, 4𝑥, 3𝑥) ∈ ℝ2 |𝑥 ∈ ℝ} is a subspace of ℝ 3 . Also find a basis for subspace 𝑈 of ℝ 3 which satisfies 𝑊 ⊕ 𝑈 = ℝ3

Answer 1

In order to prove that $W$ is a subspace of $\mathbb{R}^3$ we will use a subspace criteria. Namely, it is enough to prove that $u+w\in W$ for any two vectors $u,w\in W$ and $\alpha u\in W$ for any $\alpha\in{\mathbb{R}}$. Suppose that $u,w\in W$. It means that $u=(x,4x,3x)$ and $w=(y,4y,3y)$ where $x,y\in{\mathbb{R}}.$ We obtain: $u+w=(x+y,4(x+y),3(x+y))$. Denote $z:=x+y$. We receive: $(z,4z,3z)\in W$. Multiply a vector $u$ by $\alpha$: $\alpha u=(\alpha x,4\alpha x,3\alpha x)$. Denote $\alpha x=a$. We receive: $\alpha u=(a,4a,3a)\in W$. Thus, the criteria holds. Remind that $W\oplus U=\{W+U|W\cap U=\empty\}$. I.e., the direct sum consists of sums of vectors from $W$ and $U$. Consider a vector $(1,4,3)$. it belongs to $W$. The matrix $\left(\begin{array}{lll}1&4&3\\0&1&0\\0&0&1\end{array}\right)$ is non-degenerate. The determinant is $1$. It means that $(1,4,3)$, $(0,1,0)$,$(0,0,1)$ is a basis in $\mathbb{R}^3$. Set $U=\{(0,x,y)|x,y\in{\mathbb{R}} \}$. $W\cap U=\empty$, since $(1,4,3),(0,1,0),(0,0,1)$ are linearly independent. $W\oplus U=\mathbb{R}^3$, since $(1,4,3),(0,1,0)(0,0,1)$ is a basis.

Answer: $U=\{(0,x,y)|x,y\in{\mathbb{R}} \}$. The basis of $U$ is: $(0,1,0),(0,0,1).$