The following example should convince you that the expression ** Ax** is linear:

` A`(

A more general proof can be found here.

= [A |
a |
b |
] |

c |
d |

The corresponding linear transformation is

[T |
x_{1} |
] = [ | ax_{1} + bx_{2} |
] : R^{2}
→ R^{2} |

x_{2} |
cx_{1} + dx_{2} |

We have

[T |
x_{1} + y_{1} |
] = [ | a(x_{1} + y_{1})
+ b(x_{2} + y_{2}) |
] = [ | ax_{1} + bx_{2} |
] + [ | ay_{1} + by_{2} |
] = [T |
x_{1} |
] + [T |
y_{1} |
] |

x_{2} + y_{2} |
c(x_{1} + y_{1})
+ d(x_{2} + y_{2}) |
cx_{1} + dx_{2} |
cy_{1} + dy_{2} |
x_{2} |
y_{2} |

and (we use `k` instead of `c` as the scalar to avoid
confusion in notations)

[T |
kx_{1} |
] = [ | akx_{1} + bkx_{2} |
] = [ | k(ax_{1} + bx_{2}) |
] = k[ |
ay_{1} + by_{2} |
] = k[T |
x_{1} |
] |

kx_{2} |
ckx_{1} + dkx_{2} |
k(cx_{1} + dx_{2}) |
cy_{1} + dy_{2} |
x_{2} |

You are strongly encouraged to verify the linearity for matrices of other sizes.

The discussion suggests us to introduce the following concept.

A transformation ` T`:

` T`(

By taking `c` = 0 in the definition, we have

` T`(

Moreover, by combining the two properties in the definition, we see that linear transformations preserve linear combinations.

` T`(

Example
Suppose ` T`:

` T`(

Then by

(5, -3) = 5 (1, 0) - 3(0,1) = 5**e**_{1} - 3**e**_{2},

(-2, 4) = -2**e**_{1} + 4**e**_{2},

we have

` T`(5, -3) =
5

` T`(-2, 4) =
-2

In general, by

(`x`_{1}, `x`_{2}) =
`x`_{1}**e**_{1} + `x`_{2}**e**_{2},

we have

` T`(

We remark that ` T` is a matrix transformation given by

[ | 1 | 3 | ] =
[(Te_{1}) (Te_{2})] |

2 | 4 |

Example The transformation

` S`(

is not a linear transformation because

` S`(

The transformation

` T`(

does satisfy ` T`(

` T`(1, 0) +

This is different from

` T`((1, 0) + (0, 1)) =

Therefore ` T` is not linear.

Since the transformations are not linear, they are not matrix transformations (see this example).

The linearity of ** Ax** helps us to construct solutions of systems of linear equations.
The key idea is that, if

Example The vectors ` x` = (1, 1, 2) and

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 6 |

x_{1} |
- x_{2} |
+ x_{3} |
= | -2 |

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 1 |

Note that the two systems have the same left side. The following systems

x_{1} |
- x_{2} |
+ x_{3} |
= | 0 |

3x_{1} |
+ x_{2} |
- x_{3} |
= | 4 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 7 |

x_{1} |
- x_{2} |
+ x_{3} |
= | 1 |

3x_{1} |
+ x_{2} |
- x_{3} |
= | 1 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 3 |

x_{1} |
- x_{2} |
+ x_{3} |
= | 8 |

3x_{1} |
+ x_{2} |
- x_{3} |
= | -4 |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 3 |

still have the same left side but different right sides. By

(0, 4, 7) = (2, 2, 6) + (-2, 2 1),

(1, 1, 3) = 1/2 (2, 2, 6),

(8, -4, 3) = (2, 2, 6) - 3(-2, 2 1),

we see that

` x` +

1/2

are the respective solutions of the three new systems.

[Extra: Matrix transformations are linear]