Exact Sequences

Observation.

We use rank nullity to show that for a short exact sequence of vector spaces,

$$0\to U\stackrel{T}{\to }V\stackrel{S}{\to }W\to 0$$

the alternating sum of the dimensions of the spaces is 0, i.e. $-dim(U)+dim(V)-dim(w)=0$. The exactness of the sequence implies that $T$ is injective and $S$ is surjective. The former implies that $dim(U)=dim(range(T))$. The latter implies that $dim(V)=dim(null(S))+dim(W))$. Again, the exactness of the sequence implies that $range(T)=null(S)$. Thus, $dim(U)=dim(V)-dim(W)$, as desired.

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It can be shown in general that the alternating sum of the dimensions of spaces in an exact sequence is 0. I’ve been told this is useful when dealing with infinite-dimensions.