Linear Algebra intuition behind subspaces Hello I am trying to understand better the intuition behind subspaces. I am aware that the subspace of a vector space must have zero(go through the origin), has to be closed by addition, and has to be closed by multiplication. The aspect I fail to understand is why must it equal zero/go through the origin. If anyone has any analogy for subspaces I'd also love to hear it thank you.
 A: A subspace has to be a vector space and therefore must be closed under scalar multiplication. There is always a $0$ value in the field of scalars, so $0\cdot\vec v$, which is the origin $\vec0$, has to be in the subspace.
You might be familiar with thinking of Euclidean vector spaces like $\mathbb R^3$, where the nontrivial subspaces are lines or planes through the origin. The collection of lines or planes that don’t necessarily go through the origin are called “affine spaces” or “affine subspaces,”* but those are not vector spaces if they don’t go through the origin.
*Like sometimes happens with mathematical definitions more than in English, the adjective “affine” is not restrictive; affine subspaces are not necessarily subspaces.
A: Closed by multiplication $\implies$ zero must be in the subspace $U\subseteq V$, as stated at the answer above. Also, a subspace is, in particular, a vector space and vector spaces must have a neutral element: the vector $e\in U$ for which $u+e=u$ for all vectors $u\in U$. And by unicity of the neutral element of the "bigger" vector space $V$, the element $e\in U$ with that property must be the same element as the neutral of $V$, that is: zero.
Indeed, you only prove that zero is in $U$ ($U$ is the set you want to be a subspace) to conclude that $U\not = \emptyset$, because although $\emptyset$ is vacuosly closed by addition and multiplication, it doesn't have a neutral element (hence is not a vector space). So, on proving that $U$ is a subspace you can show that $0\in U$ or that $c \in U$ for some convenient constant vector (and show those "close properties")... Again, what really matters is that $U\not = \emptyset$.
