# $\mathrm{ker}L=\{0\}$ if and only if $L$ is injective

###### Theorem.

A linear map between vector spaces^{} is injective if and only if its kernel is $\mathrm{\{}\mathrm{0}\mathrm{\}}$.

###### Proof.

Let $L:V\to W$ be a linear map. Suppose $L$ is injective, and $L(v)=0$ for some vector $v\in V$. Also, $L(0)=0$ because $L$ is linear. Then $L(v)=L(0)$, so $v=0$. On the other hand, suppose $\mathrm{ker}L=\{0\}$, and $L(v)=L({v}^{\prime})$ for vectors $v,{v}^{\prime}\in V$. Hence $L(v-{v}^{\prime})=L(v)-L({v}^{\prime})=0$ because $L$ is linear. Therefore, $v-{v}^{\prime}$ is in $\mathrm{ker}L=\{0\}$, which means that $v-{v}^{\prime}$ must be $0$. ∎

Title | $\mathrm{ker}L=\{0\}$ if and only if $L$ is injective |
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Canonical name | operatornamekerL0IfAndOnlyIfLIsInjective |

Date of creation | 2013-03-22 14:44:46 |

Last modified on | 2013-03-22 14:44:46 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 11 |

Author | Mathprof (13753) |

Entry type | Theorem |

Classification | msc 15A04 |