"T is one-to-one if and only if the equation T(x) = 0 has only the trivial solution" is correct and means that the vector "x" is null. (For example, x=(0,0,0,0))

So, to verify this statement, you have to solve a system of equations:

T(x) = 0

where the unknowns will be the components of "x".

For example:

T(x)=(x1+x2,x1)

T is defined: T:R2->R2. The system to be solved is:

(1) x1+x2=0

(2) x1=0

From (2), (1) is 0+x2=0 and x2=0. In this case, x=(0.0). This is the trivial solution.

Well, now, suppose that N(x)=(x1-x2,x1-x2). Then, you have:

(1) x1-x2=0

(2) x1-x2=0

So x1=x2. And x=(x1,x1). There you have multiple solutions to the system. Just asign a value for x1. Then, N is not one-to-one.

In the case of a 3x4-matrix, you will have a system of 3 equations of 4 unknowns. You have to resolve:

(1) equation1 = 0

(2) equation2 = 0

(3) equation3 = 0

If you find more than one solution (you will), then it is not one-to-one.

It is a simple help. If you want to go in-deep, there are multiple theorems to solve this problem in a different manner.

Hope this helps.

Federico.