Linear Algebra, finite-dimensional null space and range

• Sep 24th 2008, 05:13 PM
cutmeout
Linear Algebra, finite-dimensional null space and range
Hi everyone,

I'm trying to show that given a Linear Map T from V to V and that the range and null space of T are both finite-dimensional, then V must be finite-dimensional.

My start at the solution is as follows:

Because the range and null space are f.d., we write a basis for them and note that since the range is a subset of V, this basis is in V. Applying the map hasn't seemed to help much, although I strongly suspect the fact that the map maps back into the domain is key to this problem. Could you point me in the right direction please? (No blatant solutions, however please; I'd just like a hint.)

Best,

Alex
• Sep 24th 2008, 05:59 PM
ThePerfectHacker
Quote:

Originally Posted by cutmeout
Hi everyone,

I'm trying to show that given a Linear Map T from V to V and that the range and null space of T are both finite-dimensional, then V must be finite-dimensional.

Hint: Rank-Nullity theorem.
• Sep 24th 2008, 06:01 PM
cutmeout
Quote:

Originally Posted by ThePerfectHacker
Hint: Rank-Nullity theorem.

I definitely thought that first, but the Rank-Nullity theorem assumes V is finite-dimensional, so using it directly is out. However, my Professor told me somehting in the proof of it would be helpful, but I've yet to find that.
• Sep 25th 2008, 08:01 AM
ThePerfectHacker
Quote:

Originally Posted by cutmeout
I'm trying to show that given a Linear Map T from V to V and that the range and null space of T are both finite-dimensional, then V must be finite-dimensional.

Let $\displaystyle T:V\to V$ be a linear transformation. Assume that $\displaystyle \text{dim}(\ker (T)) > 0$ (you need to treat the case $\displaystyle =0$ seperately). We know that this dimension is finite (by hypothesis) and so let $\displaystyle B=\{ \bold{v}_1,...,\bold{v}_m\}$ be a basis for $\displaystyle \ker (T)$. Assume that $\displaystyle V$ is infinite-dimensional. Then we can add vectors $\displaystyle \bold{v}_{m+1}, ... , \bold{v}_n$ to $\displaystyle B$ so that all $\displaystyle n$ vectors are linearly independent, note that there is no bound on $\displaystyle n$ since we are assuming $\displaystyle V$ is infinite-dimensional. The claim is that $\displaystyle \{ T(\bold{v}_{m+1}), ... , T(\bold{v}_n) \}$ is linearly independent - look at proof of Rank Nullity Theorem, it is similar. Thus, it means we can make this set as large as we please by keeping it linearly independent, but, then it means the range of $\displaystyle T$ cannot be finite dimensional since there is a limit to the number of linearly independent elements.