The composition of two linear transformations is also linear. The geometrical reason is clear: if parallelograms and scalings are preserved by each transformation, then they are preserved by the combined transformation. See here for a rigorous proof.

Since linear transformations between euclidean spaces are equivalent to matrices, the composition of linear transformations should have corresponding operation on matrices.

Let ` A` and

Example Consider linear transformations

` T`:

By substituting the formulae of `y`_{1}, `y`_{2} in terms of
`x`_{1}, `x`_{2} into the formulae of `z`_{1}, `z`_{2} in terms of
`y`_{1}, `y`_{2}, we find the composition
` TS`:

`z`_{1}
= `a`_{11}(`b`_{11}`x`_{1} + `b`_{12}`x`_{2})
+ `a`_{12}(`b`_{21}`x`_{1} + `b`_{22}`x`_{2})
= (`a`_{11}`b`_{11} + `a`_{12}`b`_{21})`x`_{1}
+ (`a`_{11}`b`_{12} + `a`_{12}`b`_{22})`x`_{2}

`z`_{2}
= `a`_{21}(`b`_{11}`x`_{1} + `b`_{12}`x`_{2})
+ `a`_{22}(`b`_{21}`x`_{1} + `b`_{22}`x`_{2})
= (`a`_{21}`b`_{11} + `a`_{22}`b`_{21})`x`_{1}
+ (`a`_{21}`b`_{12} + `a`_{22}`b`_{22})`x`_{2}

In terms of the product of matrices, we have

[ | a_{11} |
a_{12} |
][ | b_{11} |
b_{12} |
] = [ | a_{11}b_{11} + a_{12}b_{21} |
a_{11}b_{12} + a_{12}b_{22} |
] |

a_{21} |
a_{22} |
b_{21} |
b_{22} |
a_{21}b_{11} + a_{22}b_{21} |
a_{21}b_{12} + a_{22}b_{22} |

Note that in the example above, the first entry
`a`_{11}`b`_{11} + `a`_{12}`b`_{21}
of the matrix ** AB** is obtained by "multiplying"
the first row (

The result ** AB** is an

= [A |
1 | 2 | ] |

3 | 4 |

= [B |
-1 | 0 | 2 | ] |

0 | 1 | 3 |

= [C |
0 | 1 | -1 | ] |

2 | 0 | 3 | ||

-1 | 1 | 2 |

= [D |
1 | 2 | ] |

2 | 3 | ||

3 | 4 |

Then

= [AB |
1×(-1) + 2×0 | 1×0 + 2×1 | 1×2 + 2×3 | ] = [ | -1 | 2 | 8 | ] |

3×(-1) + 4×0 | 3×0 + 4×1 | 3×2 + 4×3 | -3 | 4 | 18 |

A^{2} = = [AA |
1×1 + 2×3 | 1×2 + 2×4 | ] = [ | 7 | 10 | ] |

3×1 + 4×3 | 3×2 + 4×4 | 15 | 22 |

= [DA |
1×1 + 2×3 | 1×2 + 2×4 | ] = [ | 7 | 10 | ] |

2×1 + 3×3 | 2×2 + 3×4 | 11 | 16 | |||

3×1 + 4×3 | 3×2 + 4×4 | 15 | 22 |

= [BC |
(-1)×0 + 0×2 + 2×(-1) | (-1)×1 + 0×0 + 2×1 | (-1)×(-1) + 0×3 + 2×2 | ] = [ | -2 | 1 | 5 | ] |

0×0 + 1×2 + 3×(-1) | 0×1 + 1×0 + 3×1 | 0×(-1) + 1×3 + 3×2 | -1 | 3 | 9 |

()AB = [C |
(-1)×0 + 2×2 + 8×(-1) | (-1)×1 + 2×0 + 8×1 | (-1)×(-1) + 2×3 + 8×2 | ] = [ | -4 | 7 | 23 | ] |

(-3)×0 + 4×2 + 18×(-1) | (-3)×1 + 4×0 + 18×1 | (-3)×(-1) + 4×3 + 18×2 | -10 | 15 | 51 |

(A) = [BC |
1×(-2) + 2×(-1) | 1×1 + 2×3 | 1×5 + 2×9 | ] = [ | -4 | 7 | 23 | ] |

3×(-2) + 4×(-1) | 3×1 + 4×3 | 3×5 + 4×9 | -10 | 15 | 51 |

On the other hand, the products such as ** AD**,

Example According to an earlier
example,
the matrix for the rotation by angle `α` is

= [R_{α} |
cosα |
- sinα |
] |

sinα |
cosα |

Since rotation by `β` and then rotation by `α` is the same as
rotation by `α` + `β`, we must have
` R_{α}R_{β}` =

[ | cosα |
- sinα |
] [ | cosβ |
- sinβ |
] = [ | cos(α + β) |
- sin(α + β) |
] |

sinα |
cosα |
sinβ |
cosβ |
sin(α + β) |
cos(α + β) |

If we compute matrix multiplication on the left side in the usual way, we get the following familiar formulae in trigonometry.

cos`α`cos`β` - sin`α`sin`β` = cos(`α + β`)

sin`α`cos`β` + cos`α`sin`β` = sin(`α + β`)

[Extra: Composition of linear transformations is linear]