Talk:Rank–nullity theorem

[edit] Rank theorem

Why does "Rank theorem" redirect to this article?128.135.239.238 (talk) 02:20, 23 April 2008 (UTC)

[edit] Proof

Can't u show all the process of proving the relation : rank(T)+Nullity(T)=Dim(V)

-- I've posted a proof of this (DriveOnTheAutobahn/128.253.69.185)

The proof (notation mostly), should be modified so that it works for infinite dimensional vector spaces, at least, this should be mentioned, since it is in the introduction. Lewallen (talk) 00:20, 28 February 2009 (UTC)

[edit] Infinite-dimensional

Surely V can't be infinite dimensional? How then would it's rank be well defined? —Preceding unsigned comment added by 152.78.171.187 (talk) 20:58, 1 June 2009 (UTC)

Of course, $V$ can be infinite dimensional. $rank(T)$ is then still well defined in the same old way: $rank(T) = \dim(im(T)))$ Still $im(T)$ is a subspace of $W$ and as such it is a vector space and has a dimension $\dim(im(T))$ , which is the cardinality of any of its bases. The theorem still holds true since the addition of cardinalities of disjoint sets is just the cardinality of the union.