Linear Algebra 2

Let {V} be an finite-dimensional vector space over a field {F}. A basis for {V} is a set {S\subset V} such that

  • {S} is linearly independent, and
  • {\mathrm{Span}(S) = V}.

For example, take {V=F^n}, that is, the set of {n}-tuples of elements of {F}. Then the standard basis for {F^n} is {\{e_1, e_2, \ldots, e_n\}} where {e_1 = (1,0,\ldots,0)}, {e_2=(0,1,0,\ldots,0)}, {\ldots}, {e_n=(0,\ldots,0, 1)}. It is clear that this set is linearly independent, as

\displaystyle \sum_{i=1}^n c_ie_i = 0

implies that {c_1\cdot1=0} for each {i}, and hence {c_1=\cdots=c_n=0}. If {v=(v_1, \ldots, v_n)\in F^n}, then

\displaystyle v = \sum_{i=1}^n v_ie_i.

A nonempty set {S\subset V} is said to be a subspace of {V} if {S} is itself a vector space. Since any subset of {V} inherits the operations of addition and scalar multiplication from {V}, the associative and distributive laws follow automatically. Therefore {S} is a subspace of {V} if and only if

  • {S} is closed under addition, i.e. {u,v\in S} implies {u+v\in S}, and
  • {S} is closed under scalar multiplication, i.e. {v\in S}, {c\in F} implies {cv\in S}.

Note that the second condition implies that {0_V\in S}, as for any {v\in S}, {0_V= 0v\in S} (this is a necessary condition for {(S,+)} to be a subgroup of {(V,+)}). For an example of a subspace, let {V=F^n}, and

\displaystyle S= \{(v_1,v_2,\ldots,v_n)\in F^n : v_1=0\}.

For if {u,v\in S} then

\displaystyle u+v = (0,u_1,\ldots,u_n)+(0,v_1,\ldots,v_n) = (0,u_2+v_2,\ldots,u_n+v_n)\in S,

and if {c\in F} then

\displaystyle cv = c(0,v_2,\ldots, v_n) = (0,cv_2,\ldots,cv_n)\in S.

One last definition: if {V}, {W} are vector spaces over a field {F}, a function {T:V\rightarrow W} is called a linear operator if

  • For all {u,v\in V}, {T(u+v) = Tu + Tv}, and
  • For all {v\in V} and {c\in F}, {T(cv) = cTv}.

If in addition {T} is a bijection, then {T} is called an isomorphism of vector spaces. The next post on linear algebra will show that all finite-dimensional vector spaces of the same dimension are isomorphic.

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