Answer to Question #156280 in Linear Algebra for Yolande

Question

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

Answer 1

(a) We remind that a vector space is a set closed under finite vector addition and scalar multipication.

Since $A\subset{\mathbb{R}}^3$ and ${\mathbb{R}}^3$ is a vector space, it is enough to check that $A$ is a subspace. Assume that $u=(x_1,y_1,z_1),v=(x_2,y_2,z_2)\in A.$ It means that $x_1-y_1+z_1=0$ and $x_2-y_2+z_2=0$. We get: $x_1+x_2-(y_1+y_2)+z_1+z_2=0$. Thus, $u+v\in A$. In addition, for any $\alpha\in{\mathbb{C}}$ we have: $\alpha u\in A$ ( it holds, because $\alpha x_1-\alpha y_1+\alpha z_1=0$ ).Thus, $A$ is a vector space.

(b) We remind that a basis is a maximum number of independen vectors. It is obvious that vectors $u=(1,1,0)$ and $v=(0,1,1)$ belong to $A$ and they are independent. We can present any vector $w$ that belongs to $A$ in the form: $w=(x,x+z,z)$. it is obvious that $xu+zv=w$. Thus, the dimension of the basis of $A$ is 2 and it consists of vectors $u=(1,1,0)$ and $v=(0,1,1)$.