I"m not sure if I'm understanding the notion of a linear extension as explained in my linear algebra text so some of my hw is difficult to approach. There's a linear extension theorem for bases and this is my attempt at explaining to myself. Suppose I have simple vector space:

$\displaystyle U={(x,y,z)| x=z}$

The vector $\displaystyle (x,y,z)$ is in $\displaystyle R^3$, but the dimension of $\displaystyle U$ is only $\displaystyle 2$, which we can discern from $\displaystyle (a,b,a)=a(1,0,1)+b(0,1,0)$. So only two vectors in the basis for U. So even though it is in $\displaystyle R^3$, it still has a dimension of $\displaystyle 2$.

Does the idea of a linear extension have to do with the difference between the dimension of U compared to the dimension of $\displaystyle R^3$? Basically, I can find a basis vector and add it to the 2-dimensional basis of U and obtain a basis with 3 vectors so that the dimension of U is now the same as $\displaystyle R^3$? And this modified "U" set of vectors is my "linear extension?"

So when we speak of the linear extension $\displaystyle T:V->W$ of a linear transformation$\displaystyle T(A_i)=B_i$ with $\displaystyle i=1,2,...,n$, the former is just a completed form of the latter? And by "competed" I mean there are fewer vectors in $\displaystyle T(A_i)=B_i$ than the vectors $\displaystyle {A_1,...,A_n}\in V$ and $\displaystyle {B_1,...,B_n}\in W$ if $\displaystyle i<n$.