Subspaces and Dimension of sum of two subspaces. A simple question.i am in the initial phase of learning linear algebra. Need your help.
I have made it through my own understanding. Wanted to know if i am thinking right. 
Lets take the subspace {(a,b,c,d,e,0,0,0,0,0,0) $ \forall $ a,b,c,d,e $ \in $ R} . What i wanted to know is whether we can regard this subspace as the vector space $ R^5 $ ??? as we can get any vector in $ R^5 $ using this subspace as all the other coordinates in the remaining dimensions are "0"...so, we can say that :
{(a,b,c,d,e,0,0,0,0,0,0) $ \forall $ a,b,c,d,e $ \in $ R} is a subspace of $ R^{11} $ or it is the vector space $ R^5 $ ....we can say either way right?? 
as for another example lets take the subspace {(a,b,1) $ \forall $ a,b $ \in $ R}. Here we can say that it is subspace of $ R^3 $ but not the vector space $ R^2 $ as the third coordinate is "1" here and not "0"..had it been "0" we could have said it is subspace of $ R^3 $ or it is the vector space $R^2$..am i thinking it the right way???
my second doubt is :
we know dim(U1+U2) = dim(U1)+dim(U2)-dim(U1 $\cap$ U2)  (here dim denotes dimension)
i have gone through the proof and its fine.What i also thought of it is as follows : 
Let U1 = {(a,b,c,0,0,0) $\forall$ a,b,c $\in$ R} 
let U2 = {(0,0,d,e,f,g) $\forall$ d,e,f,g $\in$ R}
Now,when we will take the dimension of U1+U2, we will get the 3rd coordinate of the vector space (U1+U2) as common from both the subspaces, thus all we have to do is write the dimension as :
dim(U1+U2) = 3+4-1 = 6. I guess i am right on this one.If not, kindly point out my mistake.
Besides, as i mentioned in my first doubt, can we write here U1 is the subspace of $R^6$ or it is the vector space $R^3$?? Similarly U2 is the subspace of $R^6$ or it is the vector space $R^4$??? In case of U2, the coordinates are any real number at 3rd,4th,5th and 6th place..if i am thinking right in my first doubt,can we here also say that U2 is the vector space $R^4$?? coz what i think where the coordinates are real (other than "0" ofcourse)   the place where the coordinates are placed should not bother us ... 
Thanks for your patience...one final question..
If whatever i have said upto now is right, then lets take the subspace :
{(0,a,0,0,b,0,0,0,c) $\forall$ a,b,c $\in$ R} 
My question is the same.Can we say it is a subspace of $R^9$ or that it is the vector space $R^3$ ???
So, these are the things i wanna clear out. i know its simple ones. I have just started learning linear algebra. and the way i am thinking is not explained in the book. So, needed help from you guys.Thanks for taking time to read and Thanks for any help in advance.
 A: $U:=\{(a,b,0)\mid a,b\in\Bbb R\}$ is a (linear) subspace of $\Bbb R^3$, that's right.
It is not equal to $\Bbb R^2$, because its elements are certain triples of real numbers, whereas the elements of $\Bbb R^2$ are pairs of real numbers.
However, $U$ has dimension $2$, and as such, is isomorphic to $\Bbb R^2$, in other words, they can be identified by the mapping $(a,b,0)\mapsto (a,b)$, which is a linear bijection (isomorphism) between $U$ and $\Bbb R^2$.
Also, $\{(a,b,1)\mid a,b\in\Bbb R\}$ is not a linear subspace of $\Bbb R^3$, because each linear subspace must contain the zero (now $(0,0,0)$). However, it is a 'shifted subspace', so called affine subspace.
The subspaces of the $3$ dimensional space are: the origin itself (0d), all the lines through the origin (1d), all the planes through the origin (2d), and the space itself (3d).
A: You've done the sum dimension just right.
As for $U_2$, for example, you can easily say that $U_2$ is isomorphic to (roughly, "behaves the same as") $\Bbb R^4$. You should be cautious, though, about saying that it is $\Bbb R^4$, since there are many subspaces of $\Bbb R^6$ isomorphic to $\Bbb R^4$.
Only one thing is well and truly incorrect: $\{(a,b,1):a,b\in\Bbb R\}$ is not a linear subspace of $\Bbb R^3,$ as (for one thing) it isn't closed under vector addition.
