If S, T &subset. V are subspaces of V, and S &subseteq. T, then
Proof: Let #{~v}_1 ... #{~v}_{~n} be a basis for S (dim (S) = ~n)
If S, T &subset. V are subspaces of V, their ~#{join} or ~#{sum} is
S + T = &set. #{~s} + #{~t} | #{~s} &in. S, #{~t} &in. T &xset.
S + T is a subspace of V, and _ S &subseteq. S + T, _ T &subseteq. S + T.
If S, T &subset. V are subspaces of V, their ~{#{intersection}} is
S &intersect. T = &set. #{~v} | #{~v} &in. S and #{~v} &in. T &xset.
S &intersect. T is a subspace of V, and _ S &intersect. T &subseteq. S , _ S &intersect. T &subseteq. T.
#{Lemma}: dim (S + T) + dim (S &intersect. T) = dim (S) + dim (T)
If S, T &subset. V are subspaces of V, and S &intersect. T = \{#0\}, then their intersection or sum is sometimes written as _ S &oplus. T , and is known as their #~{direct sum} _ I.e. this symbol not only represents the space _ S + T , _ but also indicates that their intersection is the trivial vector space.
If F is a field, define F^{~n} = &set. (~a_1, ... ,~a_{~n}) | ~a_{~i} &in. F^{~n} &xset., (i.e. the Cartesian Product of F with itself ~n times), with the operations
Then F^{~n} is a vector space over F.
dim (F^{~n}) = ~n, _ _ where &set.#e_{~i}&xset._{~i = 1, ... ~n} _ is a basis for F^{~n}, _ #e_{~i} = (0 ... 1 ... 0), _ _ [~i^{th} element = 1].
If V is a vector space over F with basis #{~u}_1, ... ,#{~u}_{~n}, then the map &pi.#: V --> F^{~n}, given by #{~u} (= &sum. λ_{~i}#{~u}_{~i}) -> (λ_1 ... λ_{~n}), is bijective, a linear map [i.e. it is well-defined and preserves linear independence]. Furthermore &pi.^{ -1 } is also a bijective linear map.