I was struck with the following question: Is there a linear map that's injective, but not surjective? I know full well the difference between the concepts, but I'll explain why I have this question.(adsbygoogle = window.adsbygoogle || []).push({});

Given two finite spaces [itex]V[/itex] and [itex]W[/itex] and a transformation [itex]T: V→W[/itex] represented by a matrix [itex]\textbf{A}[/itex], we know that [itex]T[/itex] is invertible if and only if there is a one-to-one mapping between spaces, which means, [itex]\textbf{A}[/itex] is invertible.

On the other hand, we know that [itex]T[/itex] is surjective if its image space can generate [itex]W[/itex]. The image space is related with the column space of [itex]\textbf{A}[/itex], so let's consider the system

[itex]\textbf{A v} = \textbf{w}[/itex]

where [itex]\textbf{v}[/itex] and [itex]\textbf{w}[/itex] are the coordinates of an element of [itex]V[/itex] and [itex]W[/itex], respectively. If [itex]\textbf{A}[/itex] is invertible than this system is always possible, meaning we can obtain any element of [itex]W[/itex] from an element of [itex]V[/itex], so [itex]T[/itex] must also be surjective.

Perhaps I didn't formulate this right and I may be doing some mistake somewhere, but it's just a simple curiosity.

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# Injective and Surjective linear transformations

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