Answer to Question #229275 in Linear Algebra for Dhruv rawat

Question

Let V= R^3

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 us show that $W=\{(x_1, x_2, x_3)| x_1-x_2 =x_3\}$ is a subspace of $V$. Let $(x_1, x_2, x_3),(y_1, y_2, y_3)\in W,\ a\in \R.$ Then $x_1-x_2=x_3,\ y_1-y_2=y_3.$ It follows that $a(x_1-x_2)=ax_3,$ and hence $ax_1-ax_2=ax_3.$ We conclude that $a(x_1, x_2, x_3)\in W.$ It follows also that $x_1-x_2+y_1-y_2=x_3+y_3,$ and thus $(x_1+y_1)-(x_2+y_2)=x_3+y_3.$ Therefore, $(x_1, x_2, x_3)+(y_1, y_2, y_3)\in W.$ We conclude that $W$ is a subspace of $V.$

Taking into account that

$W=\{(x_1, x_2, x_3)\in\R^3| x_1-x_2 =x_3\}=\{(x_2+x_3, x_2, x_3)| x_2, x_3\in\R\}\\ =\{x_2(1, 1, 0)+x_3(1, 0, 1)| x_2, x_3\in\R\},$

we conclude that $\{(1, 1, 0),(1, 0, 1)\}$ is a basis of $W,$ and hence $dim(W)=2.$