# 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.