Having a problem proving $\operatorname{rank}T = m$.

Question

So, I have managed to do parts a,b and the first half of c, However I am struggling with showing that $\operatorname{rank} T = m$.

Any tips are really appreciated.

Thanks

Answer 1

So I will attempt to prove what you need without invoking the Rank-Nullity Theorem, as I feel that the whole point of this exercise is to prove it. If you already have access to it, then I think you can be more direct in your argumentation.

You are given an $n$-dimensional real vector space $V$, an $m$-dimensional subspace $U$, and you are told there is an inner product at your disposal. You are also given a basis $\{u_1,\cdots,u_m\}$ of $U$.

Now, choose a vector in $V\setminus U$, say $w_1$. Define $v_1=w_1-\sum u_i\frac{\langle u_i,w_1\rangle}{\langle u_i,u_i\rangle}$. Then it is clear that $v_1\neq \vec{0}$ (since that would imply that $w_1\in U$), and by construction $v_1\in U^\perp$. Now, choose a vector in $V\setminus \text{span}(U,\{v_1\})$, say $w_2\ldots$

Basically, repeat the arguemntation that gives you the Gram-Schmidt process, but use it to create new basis vectors that are orthogonal to everything you have so far. This process ends because $V$ is finite dimensional, and you will be left with a basis of $V$ of the form $\{u_1,\ldots,u_m,v_1,\ldots,v_{n-m}\}$ with precisely the properties you need.

If you have the Rank-Nullity Theorem you are then done, as you have pinned down the dimensions of the ambient space and the kernel. If you don't have that theorem, then you need to argue about the dimensions of the image of your transformation $T$, but that is probably even easier than what I have done here; the dimension of the image is the size of a basis for the image, and you need only show that the transformation is a surjection.

EDIT: I don't think I got enough sleep last night, I appear to have wandered a lot in my answer, and accidentally answered part (d) as well.