@BalarkaSen I'm having trouble visualizing how the dimension of quotient space $V/U$ is $\text{dim}(V)-\text{dim}(U)$

Suppose we consider $V$ as $\Bbb R^2$

And $U$ as the line having slope $2$ and passing through origin

@BalarkaSen Huh, okay. I guess it would be helpful if we take the above example for visualization purpose

I'll try that out, though

Could you check if my definition of quotient space is correct, first of all:

Suppose $U$ is a subspace of $V$, then the quotient space $V/U$ is the set of all affine subsets of $V$ parallel to $U$ (i.e. the set of all lines of slope 2, in this case....right?)

So, you're saying that the set of all lines of slope $2$ is isomorphic to the line perpendicular to the line of slope $2$? Or am I doing something wrong?

Ah, I think it's just this: Consider a map $P:V\to(V/U)$. Then $\text{null}(P)=U$, since if you take any element of $U$, it produces the zero vector in $V/U$ after being acted upon by the linear map $P$.

Then we use $dim(null(P))+dim(range(P))=dim(V)$

The isomorphism part is easy to show after we show that they have the same dimension i.e. 1

By mapping the basis vector

Alright, I will try