/sci/ - Science & Math

>>11455795
Generally we have an exact sequence [math]0\rightarrow \operatorname{ker}A \rightarrow V \xrightarrow{A} W \rightarrow \operatorname{coker}A \rightarrow 0[/math], which tells you [math]W = \operatorname{im}A \oplus \operatorname{coker}A[/math]. On the image [math]\operatorname{im}A \subset W[/math], [math]A[/math] is certainly invertible so it just suffices to show that [math]\operatorname{coker}A = 0[/math]. Now since [math]\operatorname{im}A = \operatorname{Span}_{j\leq \operatorname{dim}W}\{\operatorname{Col}_jA\}[/math] is the span of the column space, any [math]w\in W/\operatorname{im}A[/math] gets you a vector linearly independent to the [math]\operatorname{dim}W[/math]-columns. However this cannot happen if [math]\{\operatorname{Col}_jA\}_{j\leq \operatorname{dim}W}[/math] is linearly independent, unless [math]w = 0[/math]. This means that [math]W/\operatorname{im}A = \operatorname{coker}A = 0[/math].