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

Question

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 $ $\text{and}$ $2x_1-x_2-x_3+3x_4-4x_5+4x_6=0 \}$

In this thread: Calculate the kernel of the linear transformation $T: \mathbb{R}^6 \rightarrow \mathbb{R}^2$

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

Answer 1

Those two vectors have two things in common, besides being elements of $\ker(T)$:

1. the sixth coordinate is equal to $0$;
2. the fourth and the fifth coordinates are equal.

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 $\ker(T)$.