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Definition 15. Let U be a subspace of Rn . A set {X1 , X2 , · · · , Xk } of vectors of Rn is
called a basis of U if:
1. {X1 , X2 , · · · , Xk } is linearly independent.
2. span{X1 , X2 , · · · , Xk } = Rn .
Theorem 32. Fundamental Theorem: Let U = span{X1 , X2 , · · · , Xm } be a subspace of
Rn Let {Y1 , Y2 , · · · , Yk } be a linearly independent subset of U . Then, k ≤ m.
Proof. Let k > m. Y1 ∈ U . Thus, Y1 = a1 X1 + · · · + am Xm . Y1 6= 0, thus one of ai must
be non-zero. Say, a1 6= 0, then X1 ∈ span{Y1 , X2 , · · · , Xm }. By continuing this process,
we can replace Xi with Yi and thus U = span{Y1 , Y2 , · · · , Ym }. Thus, Ym+1 ∈ U is a linear
combination of {Y1 , Y2 , · · · , Ym }. This is contradiction, with the fact that {Y1 , Y2 , · · · , Yk }
is a linearly independent set.
Remark 11. Any two bases have the same number of elements. This number is called the
dimension of U and we denote it by dimU .
Remark 12. Rn = span{E1 , · · · , En }. Thus, by Fundamental Theorem any linearly independent set has at most n vectors.
• dim Rn = n
• Dimension of the line through origin in R3 = 1
• We define the dimension of zero subspace to be zero:
dim{0} = 0.
Theorem 33. {X1 , X2 , · · · , Xk } is a linearly independent set of vectors in Rn and Y is
not in span{X1 , X2 , · · · , Xk }. Then, {Y, X1 , X2 , · · · , Xk } is also linearly independent.
Proof. Let
bY + a1 X1 + · · · + ak Xk = 0.
Then, b = 0 (otherwise, Y is in span{X1 , · · · , Xk }). Therefore,
a1 X1 + · · · + ak Xk = 0.
{X1 , X2 , · · · , Xk } is linearly independent, therefore
a1 = a2 = · · · = ak = 0.
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Theorem 34. U 6= {0} is any subspace of Rn . U has a basis and dim U ≤ n.
Proof. U 6= {0}, therefore there is X1 6= 0 ∈ U . If U = span{X1 }, then {X1 } is a basis
for U . If U 6= span{X1 }, pick X2 ∈ U which is not in span{X1 }. Then, by Theorem
33, {X1 , X2 } is linearly independent. If U = span{X1 , X2 }, then {X1 , X2 } is a basis for
U . If U 6= span{X1 , X2 } enlarge the linearly independent set again. By Fundamental
Theorem, we have to stop this process after finite steps. Thus, U = span{X1 , X2 , · · · , Xk }
and {X1 , X2 , · · · , Xk } is a linearly independent set.
Theorem 35. Let U and V denote subspaces of Rn . If U ⊆ V and dim U = dim V , then
U =V.
Proof. Let {X1 , · · · , Xk } be a basis for U . If U 6= V , then there are vectors Y1 , · · · , Yt ∈ V
which are not in U = span{X1 , · · · , Xk } and {X1 , · · · , Xk , Y1 , · · · , Yt } is a basis for V .
This, is contradiction with the fact that dim U = dim V .
Definition 16. Let A be an m × n matrix. Denote the rows of A by R1 , · · · , Rm and the
columns of A by C1 , · · · , Cn . The row space of A is defined by
T
rowA = span{R1T , · · · , Rm
},
and the column space of A is defined by
colA = span{C1 , · · · , Cn }.
To find a basis for rowA find the reduce form R. Then the non-zero rows are a basis for
rowA.
To find a basis for colA, find the reduce form R. If the leading ones are in columns j1 , · · · , jr ,
then {Cj1 , · · · , Cjr } is a basis for colA. Recall that
ImA = {Y ∈ Rm |AX = Y for some X inRn }
thus, ImA = span{C1 , · · · , Cn } = colA. To find a basis for
nullA = {X ∈ Rn |AX = 0},
solve the system AX = 0 and find the basic solutions X1 , · · · , Xk . A basis for nullA is
{X1 , · · · , Xk }.