# Extend a set to a basis for the kernel of a linear transformation

Calculate the kernel of the linear transformation $$T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$$ given by:

$$T(x_1,...,x_6) = (x_1-x_2+2x_4-3x_5+x_6,2x_1-x_2-x_3+3x_4-4x_5+4x_6)$$

By definition, this we have that $$ker(T) = \{(x_1,...x_6) \in \mathbb{R}^6\ |$$ $$x_1-x_2+2x_4-3x_5+x_6=0$$ $$\text{and}$$ $$2x_1-x_2-x_3+3x_4-4x_5+4x_6=0 \}$$

Somebody helped show me to calculate that $$\ker(T)=\{(a-b+c-3d,a+b-2c-2d,a,b,c,d)\mid a,b,c,d\in\Bbb R\}.$$

My question is then this: how do I extend this set of 2 vectors: $$\{(1,0,1,1,1,0),(0,-1,0,1,1,0) \}$$ to a basis for $$Ker(T)$$?

Thank you

Those two vectors have two things in common, besides being elements of $$\ker(T)$$:
1. the sixth coordinate is equal to $$0$$;
So, add to them an element of $$\ker(T)$$ for which the first condition holds but not the second one (such as $$(-1,1,0,1,0,0)$$) and an element of $$\ker(T)$$ for which the second condition holds but not the first one (such as $$(-3,-2,0,0,0,1)$$) and then you will have $$4$$ linearly independent elements of $$\ker(T)$$, that is, you will have a basis of $$\ker(T)$$.