A subspace is a vector space which is a subset of a vector space, using the same addition and scalar multiplication operations. Intuitively, a subspace of \mathbb{R}^{2} are all the lines through the origin as well as \{0\}; a subspace of \mathbb{R}^{3} are all the planes through the origin as well as \{0\}, etc. etc. constituents vector space V A subset U \subset V which is itself a vector space requirements You check if U is a subspace of V by checking IFF the following three conditions: additive identity: 0 \in U closed under the same addition as in V: u,w \in U: u+w \in U closed under scalar multiplication as in V: a \in \mathbb{F} and u \in U means au \in U Yes, by only checking three you can prove everything else. additional information simplified check for subspace commutativity, associativity, distributivity These properties are inherited from V as they hold for every element in V so they will hold for U \subset V. additive inverse Because scalar multiplication is defined, and we proved in Axler 1.B that -1v=-v (proof: v+(-1)v = (1+(-1))v = 0v = 0). multiplicative identity Its still 1. \blacksquare finite-dimensional subspaces Every subspace of a finite-dimensional vector space is a finite-dimensional vector space. We prove this result again via induction. base case If U=\{0\}, we know U is finite-dimensional and are done. If not, take some v_1 \in U and create a list with only v_1 thus far; the invariant here is that the list is linearly independent as we see that a list containing this one element as indeed linearly independent. case j If the linearly independent list we created v_1, \dots v_{j-1} spans U, we are done. We have created a finite list which spans U, making U finite-dimensional. If not, that means that we can pick some u \in U that cannot be written as a linear combination of the invariantly linearly independent vectors v_1, \dots v_{j-1}. We append u to the list, naming it v_{j}. As v_{j} cannot be written as a linear combination of the original list, appending it to the list doesn’t make the list dependent. This means that the list is still linearly independent. induction Therefore, we have constructed a list of increasing length that is linearly independent. By the fact that length of linearly-independent list \leq length of spanning list, and the fact that the spanning list of V has finite length (it is given that V is a finite-dimensional vector space), the increasingly longer linearly independent list—building upwards to eventually span U in finite length.