Question

Let A = {(x, y, z) E R^3ㅣx - y + z = 0} where E represents "an element of" and R the set of real numbers
(a) Show that A is a vector space
(b) Give the dimensions and basis of A.

Answer 1

Let $A = \{(x, y, z) \in\mathbb R^3ㅣx - y + z = 0\}$

(a) Let us show that $A$ is a vector space. For this it is sufficient to show that $A$ is a subspace of the vector space $\mathbb R^3$. Let $(x_1,y_1,z_1),(x_2,y_2,z_2)\in A,\ \ \alpha\in \mathbb R$. Then $x_1-y_1+z_1=0$ and $x_2-y_2+z_2=0$. Since $(x_1+x_2)-(y_1+y_2)+(z_1+z_2)=(x_1-y_1+z_1)+(x_2-y_2+z_2)=0+0=0$ and $\alpha x_1-\alpha y_1+\alpha z_1=\alpha( x_1-y_1+z_1)=\alpha\cdot 0=0$, we conclude that $(x_1,y_1,z_1)+(x_2,y_2,z_2)=(x_1+x_2,y_1+y_2,z_1+z_2)\in A$ and $\alpha (x_1,y_1,z_1)=(\alpha x_1,\alpha y_1,\alpha z_1)\in A$. Therefore, $A$ is a vector space.

(b) Let us give the dimensions and basis of $A$. Since $A = \{(x, y, z) \in\mathbb R^3ㅣx - y + z = 0\}= \{(x, x+z, z) \in\mathbb R^3ㅣx, z \in\mathbb R\}= \{(x, x, 0)+(0, z, z) \in\mathbb R^3ㅣx, z \in\mathbb R\}= \{x(1, 1, 0)+z(0, 1, 1) \in\mathbb R^3ㅣx, z \in\mathbb R\},$

we conclude that the lenearly independent vectors $(1,1,0)$ and $(0,1,1)$ form a basis of $A$. Consequently, the dimension of $A$ is equal to 2.