Answer to Question #162101 in Linear Algebra for Sourav Mondal

Question

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.

Answer 1

Let $V = \mathbb R^3$,  $W = \{(x_1, x_2, x_3) |\ x_1 - x_2 = x_3\}$. 

Let us show that $W$ is a subspace of $V$. Let $(x_1,x_2,x_3), (y_1,y_2,y_3)\in W,$ and $\ a, b\in\mathbb R$. Then $x_1-x_2=x_3$ and $y_1-y_2=y_3$. It follows that $(ax_1+by_1)-(ax_2+by_2)=a(x_1-x_2)+b(y_1-y_2)=ax_3+by_3$ , and therefore, $a(x_1,x_2,x_3)+ b(y_1,y_2,y_3)= (ax_1+by_1,ax_2+by_2,ax_3+by_3)\in W$. We conclude that $W$ is a subspace of $V$.

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

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