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Is there a relationship between the null space $N(T)$ of a linear transformation $T$ and whether or not it is invertible?

For example, if you know $N(T) \neq \{0\}$, can you be sure it's not an invertible transformation?

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If you know the null space is nonzero, then it certainly isn't invertible, since more than one element maps to zero (so to what element would the inverse map zero?). –  Alan Guo Dec 9 '12 at 18:31
    
Yes that's true - I didn't think of that! –  Imray Dec 9 '12 at 18:35
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Yes, what is true is that $T:V\to W$ is injective if and only if $N(T)=\{0\}$. Thus, if you know that $N(T)\ne\{0\}$ you know that $T$ can't be injective, and so particularly can't be invertible. Note that $N(T)=\{0\}$ is not sufficient for invertibility as the map $T:\mathbb{R}^3\to\mathbb{R}^4:(x,y,z)\mapsto (x,y,z,0)$ shows.

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A linear map is invertible iff it is bijective, and injective iff its kernel is trivial.

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