This quote was added by
user91542

If it is given that V is a subset of a vector space W, then showing that V is a subspace means we must only verify vector space properties three and four for V. That is to say, V must contain the zero vector, and a negative vector for each vector in V. All other vector space properties are implied by the fact that V is a subset of W.
2.5 out of 5
based on 34 ratings.

Unfortunately, this is wrong. What you probably meant is that for all v, w ∈ V, we have v - w ∈ V. What you wrote is not enough. For example, take V = {0, 1, -1} as a subset of say ℚ (regarded as a vector space over itself).

Also the terminology "negative vector" is slightly imprecise. Better to say "opposite".

Finally, saying "properties 3 and 4" doesn't really make much sense. I guess those refer to your textbook, but the numbering is not universal, so most people wouldn't know which properties you're talking about.

Also the terminology "negative vector" is slightly imprecise. Better to say "opposite".

Finally, saying "properties 3 and 4" doesn't really make much sense. I guess those refer to your textbook, but the numbering is not universal, so most people wouldn't know which properties you're talking about.

random math lesson amid typing practice absolutely loved it

Hell yeah.

wuut?

Name | WPM | Accuracy |
---|---|---|

hippogriffo | 140.50 | 97.4% |

hunterz1200 | 124.44 | 95.2% |

user81230 | 122.95 | 97.4% |

venerated | 122.70 | 97.4% |

user697099 | 119.98 | 95.4% |

user939249 | 119.67 | 95.5% |

69buttpractice | 117.74 | 97.7% |

venerated | 117.05 | 99.4% |

Name | WPM | Accuracy |
---|---|---|

bagdat2004 | 68.36 | 94.4% |

similarmotion | 69.64 | 92.5% |

2001or2 | 96.94 | 88.4% |

rishikkshah | 64.42 | 89.0% |

matrixx | 67.52 | 98.0% |

hummer350 | 68.18 | 95.7% |

user82395 | 48.31 | 95.7% |

ccandll | 55.26 | 96.3% |