# User:IssaRice/Linear algebra/Properties of a list of vectors and their images

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Let and be two finite-dimensional vector spaces, let be a linear transformation, and let be a list of vectors ( is not necessarily equal to or ).

Given the above background setting, we can now place various conditions on the list and the transformation to conclude various things about the list . For instance, the first row below says that if is a linearly independent list and is an injective map, then is also linearly independent.

Condition on | Condition on | Conclusion about | Notes |
---|---|---|---|

Linearly independent | Injective | Linearly independent | |

Spans | Surjective | Spans | |

Basis for | Bijective | Basis for | Combine previous two rows |

Basis for | No condition | No conclusion | Even though we have placed a strong condition on , if we don't place any conditions on , then there is not much we can conclude about . If sends everything to zero, then we just get , which neither spans nor is it a linearly independent list. |

No condition | Bijective | No conclusion | Even though we have placed a strong condition on , if we don't place conditions on the list of vectors, then there is not much we can say about the list of the images. For instance, if , if is the identity map, and if our list is , then the image is also , which is linearly dependent and does not span . |

Orthonormal | Is an isometry | Orthonormal | |

Eigenvectors of corresponding to distinct non-zero eigenvalues | No condition | Linearly independent; eigenvectors of corresponding to distinct eigenvalues | Eigenvectors corresponding to distinct eigenvalues are linearly independent, so is a linearly independent list. Since each is an eigenvector, we see that is the list , where each is the corresponding eigenvalue. Since by assumption none of the eigenvalues are zero, the new list is also linearly independent. |