### One-to-One

##### 2. Kernel and uniqueness

The concept of range has been introduced in the discussion of existence/onto. For general transformations, however, there is no similar concepts associated with uniqueness/one-to-one. For a linear transformation T(x): RnRm, on the other hand, the definition of uniqueness

T(x) = T(x') ⇒ x = x'.

is equivalent to

T(x - x') = 0x - x' = 0.

This leads to the following definition.

kernelT = {xRn: T(x) = 0}

The discussion before the definition leads to the following conclusion.

A linear transformation T is one-to-one ⇔ kernelT = {0}

We remark that the conclusion is comparable to the earlier observation that the following are equivalent

1. Solution of Ax = b is unique
2. The homogeneous system Ax = 0 has only the trivial solution.

We also note that if T is given by a matrix A, then kernelT is all solutions of Ax = 0.

Example The system in this example has the vector (2, 2, 6) as the unique solution. Since the uniqueness is independent of the right side, we conclude that the linear transformation

 T[ x1 ] = [ 3x1 + x2 - x3 ] : R3 → R3 x2 x1 - x2 + x3 x3 2x1 + 2x2 + x3

is one-to-one.

In contrast, this system has infinitely many solutions. The corresponding linear transformation

 T[ x1 ] = [ x1 - x2 + 3 x3 ] : R3 → R3 x2 3x1 + x2 + x3 x3 x1 + x2 - x3

is not one-to-one.

Finally, in this example we have a system with no solution. In general, the inconsistency of the system alone would not tell us whether the corresponding linear transformation

 T[ x1 ] = [ 3x1 + x2 - x3 ] : R3 → R3 x2 x1 - x2 + x3 x3 2x1 - x2 + x3

is one-to-one or not. More computation or argument is needed. In this case, fortunately, we do not need to do any computation. By applying the basic principle of linear algebra, since 3 (number of variables/columns) = 3 (number of equations/rows), the system cannot have unique solution for any right side. Therefore T is not one-to-one.

Example To determine whether the linear transformation

 T1[ x1 ] x2 x3 x4 x5 x6
=
 [ x1 + 3x2 + 2x3 + x5 ] - x1 - x2 - x3 + x4 + x6 4x2 + 2x3 + 4x4 + 3x5 + 3x6 x1 + 3x2 + 2x3 - 2x4
: R6R4

is one-to-one or not, we find the row echelon forms of the corresponding matrix. In this example, the matrix corresponding to T1

 [ 1 3 2 0 1 0 ] -1 -1 -1 1 0 1 0 4 2 4 3 3 1 3 2 -2 0 0

has been simplified, by row operations, to

 [ 1 3 2 0 1 0 ] 0 2 1 -1 0 1 0 0 0 2 1 0 0 0 0 0 0 1

Since not all columns are pivot, we do not have uniqueness, and the transformation T is not one-to-one.

In fact, we do not even need to make any computation in order to show T is not one-to-one, since the solutions to the corresponding 4 equations in 6 variables cannot be unique (see the numerical consequence of uniqueness).

In order to modify the transformation to become one-to-one, we need to make sure to have all rows pivot. Since the pivots columns are 1, 2, 4, 6, we delete x3 and x5 to get

 T2[ x1 ] x2 x3 x4
=
 [ x1 + 3x2 ] - x1 - x2 + x4 + x6 4x2 + 4x4 + 3x6 x1 + 3x2 - 2x4
: R4R4

The row echelon form of the corresponding matrix is

 [ 1 3 0 0 ] 0 2 -1 1 0 0 2 0 0 0 0 1

with all columns pivot, the transformation T2 is one-to-one.

As in the case of onto, one-to-one linear transformations also has numerical consequences. Expanding on the argument in the last example, we have

A linear transformation T: RnRm is one-to-one ⇒ mn