Let V=R3, W={(x1,x2,x3)∣ x1−x2=x3}.
Let us show that W is a subspace of V. Let (x1,x2,x3),(y1,y2,y3)∈W, and a,b∈R. Then x1−x2=x3 and y1−y2=y3. It follows that (ax1+by1)−(ax2+by2)=a(x1−x2)+b(y1−y2)=ax3+by3 , and therefore, a(x1,x2,x3)+b(y1,y2,y3)=(ax1+by1,ax2+by2,ax3+by3)∈W. We conclude that W is a subspace of V.
Further, find a basis for W . Since W={(x1,x2,x3)∈R3∣ x1−x2=x3}={(x1,x2,x1−x2)∈R3∣ x1,x2∈R}={x1(1,0,1)+x2(0,1,−1)∈R3∣ x1,x2∈R},we conclude that linear independent vectors e1=(1,0,1) and e2=(0,1,−1) form the basis of W,
and hence dim(W)=2.
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