Consider the following 4 vectors in

, , ,

These vectors are linearly independent and span . Thus the vectors form a basis of called the usual basis of .

The standard unit vectors are linearly independent for if we writeIf someone could please explain the steps on how to tackle this problem I would be appreciative of that.

Secondly, the question later reads

Find the basis of the kernel of T

and

Find the standard basis of the kernel of T.

If anyone could clarify what is the difference between the two and how the calculation procedures are different I would also appreciate that.

Thank you very much.

The usual or standar basis in component form then we obtain , which implies the coeficients are zeroes. Furthermore the vectors span because an arbitrary vector can be expressed as

We call the usual or standard basis for .

Edit: I think in your problem to find a basis of the kernel of the map T you must set T(v)=0 where v=(x,y,z,t) and then set corresponding components equal to each other to form a homogeneous system whose solution space is the kernel of T.