Question #336649

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


1
Expert's answer
2022-05-10T11:49:32-0400

In order to prove that WW is a subspace of R3\mathbb{R}^3 we will use a subspace criteria. Namely, it is enough to prove that u+wWu+w\in W for any two vectors u,wWu,w\in W and αuW\alpha u\in W for any αR\alpha\in{\mathbb{R}}. Suppose that u,wWu,w\in W. It means that u=(x,4x,3x)u=(x,4x,3x) and w=(y,4y,3y)w=(y,4y,3y) where x,yR.x,y\in{\mathbb{R}}. We obtain: u+w=(x+y,4(x+y),3(x+y))u+w=(x+y,4(x+y),3(x+y)). Denote z:=x+yz:=x+y. We receive: (z,4z,3z)W(z,4z,3z)\in W. Multiply a vector uu by α\alpha: αu=(αx,4αx,3αx)\alpha u=(\alpha x,4\alpha x,3\alpha x). Denote αx=a\alpha x=a. We receive: αu=(a,4a,3a)W\alpha u=(a,4a,3a)\in W. Thus, the criteria holds. Remind that WU={W+UWU=}W\oplus U=\{W+U|W\cap U=\empty\}. I.e., the direct sum consists of sums of vectors from WW and UU. Consider a vector (1,4,3)(1,4,3). it belongs to WW. The matrix (143010001)\left(\begin{array}{lll}1&4&3\\0&1&0\\0&0&1\end{array}\right) is non-degenerate. The determinant is 11. It means that (1,4,3)(1,4,3), (0,1,0)(0,1,0),(0,0,1)(0,0,1) is a basis in R3\mathbb{R}^3. Set U={(0,x,y)x,yR}U=\{(0,x,y)|x,y\in{\mathbb{R}} \}. WU=W\cap U=\empty, since (1,4,3),(0,1,0),(0,0,1)(1,4,3),(0,1,0),(0,0,1) are linearly independent. WU=R3W\oplus U=\mathbb{R}^3, since (1,4,3),(0,1,0)(0,0,1)(1,4,3),(0,1,0)(0,0,1) is a basis.

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


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