can someone define in childrens language what the rank, nullity, and basis for the range are?

Question

I'm supposed to find the rank of T, the nullity of T and a basis for the range of T.
T is defined as such:
T(X1, X2, X3, X4) = (0, X1, X2, X3)

Answer 1

OK, let's review the basic concepts.

A basis for a vector space is a set of linearly independent vectors that span the vector space.

There is a theorem that says that if there is a basis for a vector space with a finite number of elements n, then every basis for that vector space has n elements. This number of basis elements is called the dimension of the space. (There are such things as infinite-dimensional vector spaces, but you don't need to worry about them - at least not yet.)

If T is a linear transformation from a finite-dimensional vector space X to a finite-dimensional vector space Y, we can prove a number of things:
- The range of T, which is {Tx: x ∈ X}, is a subspace of Y; its dimension is called the rank of T, and is less than or equal to the dimension of X.
- The kernel of T, which is {x ∈ X: Tx = 0}, is a subspace of X. Its dimension is called the nullity of T.
- The rank and the nullity of T add up to the dimension of X (this is called the Rank-Nullity theorem).

So, in this particular case:
T(X1, X2, X3, X4) = (0, X1, X2, X3)

The range is {(0, X1, X2, X3): X1, X2, X3 ∈ R} which fairly obviously has a basis
{(0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1)}
since we can write
(0, X1, X2, X3) = X1(0, 1, 0, 0) + X2(0, 0, 1, 0) + X3(0, 0, 0, 1) (so they span the range of T), and this can only be 0 if X1, X2 and X3 are all 0 (so they are linearly independent).
Since there are three elements in the basis, the range of T has dimension 3 and therefore T has rank 3.

The nullity is the dimension of the kernel (which is the set of vectors that T maps to 0). In this case the kernel is {(0, 0, 0, X4): X4 ∈ R} and has basis {(0, 0, 0, 1)}, so it is one-dimensional and therefore the nullity is 1. Note that the rank and nullity of T add up to 4, as required.