Доведення: доведемо наприклад перше твердження. Нехай у деякому базисі (
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,
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,
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),
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(
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,
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,
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),
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(
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,
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,
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). Тоді за означенням координат вектора
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=
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+
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+
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,
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=
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+
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+
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.
Отже,
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+
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=
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+
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+
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+
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+
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+
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= (
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+
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)
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+ (
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+
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)
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+ (
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+
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)
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.
Звідси випливає, що координати вектора
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+
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відповідно дорівнюють
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+ +
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,
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+
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,
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+
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, що й треба було довести.
Аналогічно доводяться й інші властивості.
Теорема (2-га ознака колінеарності двох векторів): для того, щоб два вектори
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(
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,
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,
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),
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(
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,
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,
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) задані в деякому базисі (
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,
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,
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), були колінеарними, необхідно і достатньо, щоб їх координати були пропорційними.
Доведення: якщо
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=
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, то твердження очевидне. Припустимо, що
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.
1. Необхідність. Нехай
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||
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. Тоді існує таке число λ, що
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= λ
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, звідки випиває, що
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= λ
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,
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= λ
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,
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= λ
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;
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= λ.
Отже, якщо вектори колінеарні, то їх координати пропорційні.
2. Достатність. Нехай
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= λ, тоді
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= λ
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,
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= λ
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,
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= λ
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. Помноживши ці рівності на вектори
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,
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,
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відповідно, дістанемо
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= λ
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,
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= λ
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,
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= λ
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. Додавши ці рівності дістанемо
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+
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+
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= λ
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+ λ
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+ λ
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або
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+
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+
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= λ(
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+
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+
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), тобто
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= λ
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||
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. Теорему доведено.