Question #162101

Let : 

V = R3, 

W = {(x1, X2, X3) ! x1 - x2 = .X3}. 

Show that W is a subspace of V. Further, 

find a basis for W, and hence, find the 

dimension of W.


1
Expert's answer
2021-02-12T10:32:28-0500

Let V=R3V = \mathbb R^3W={(x1,x2,x3) x1x2=x3}W = \{(x_1, x_2, x_3) |\ x_1 - x_2 = x_3\}

Let us show that WW is a subspace of VV. Let (x1,x2,x3),(y1,y2,y3)W,(x_1,x_2,x_3), (y_1,y_2,y_3)\in W, and  a,bR\ a, b\in\mathbb R. Then x1x2=x3x_1-x_2=x_3 and y1y2=y3y_1-y_2=y_3. It follows that (ax1+by1)(ax2+by2)=a(x1x2)+b(y1y2)=ax3+by3(ax_1+by_1)-(ax_2+by_2)=a(x_1-x_2)+b(y_1-y_2)=ax_3+by_3 , and therefore, a(x1,x2,x3)+b(y1,y2,y3)=(ax1+by1,ax2+by2,ax3+by3)Wa(x_1,x_2,x_3)+ b(y_1,y_2,y_3)= (ax_1+by_1,ax_2+by_2,ax_3+by_3)\in W. We conclude that WW is a subspace of VV.


Further, find a basis for WW . Since W={(x1,x2,x3)R3 x1x2=x3}={(x1,x2,x1x2)R3 x1,x2R}={x1(1,0,1)+x2(0,1,1)R3 x1,x2R},W = \{(x_1, x_2, x_3)\in\mathbb R^3 |\ x_1 - x_2 = x_3\}= \{(x_1, x_2, x_1-x_2)\in\mathbb R^3 |\ x_1, x_2\in\mathbb R\}= \{x_1(1, 0, 1)+x_2(0, 1, -1)\in\mathbb R^3 |\ x_1, x_2\in\mathbb R\},we conclude that linear independent vectors e1=(1,0,1)e_1=(1, 0, 1) and e2=(0,1,1)e_2=(0, 1, -1) form the basis of WW,

and hence dim(W)=2.\dim(W)=2.



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