If you have a linear transformation going from V,W where V,W are finite-dimensional vector spaces. Does the dim(range(T)) being equal to the dimension of the target space imply that the linear transformation is onto. If it does do you mind explaining why.
[Math] Does rank(T) = dim(W) imply linear transformation is onto.
linear algebralinear-transformationsvector-spaces
Best Answer
rank $T$ is the dimension of the range of $T$. So the range of $T$ is a subspace of $W$ that has the same dimension as $W$, hence the image is $W$, or in other words, $T$ is onto.
This argument relies on the fact that a subspace $U$ of $W$ that has the same dimension as $W$ is $W$ (call this dimension $n$). We can prove this easily.
Take a basis for $U$, it has $n$ elements, since the basis is a linearly independent subset of $W$ it can be extended to a basis of $W$, of course all the basis for $W$ have $n$ elements, so the extension consists on adding nothing, therefore the basis for $U$ is a basis for $W$ and hence $U=W$ as desired.