The Span of the set of vectors
Definition 1
Let \(\mathcal { S } = \left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \dots , \mathbf { u } _ { k } \right\}\) is a set of vectors from \(\mathcal{R^n}\), the span of \(\mathcal{S}\) is all linear combinations in \(\mathcal{R^n}\), the set is denoted by span \(\mathcal{S}\), or span \(\left\{ \mathbf { u } _ { 1 } , \mathbf { u } _ { 2 } , \dots , \mathbf { u } _ { k } \right\}\)
\(\vec v \in \text{span } \mathcal{S} \Longleftrightarrow \vec v \text{ can be some linear combination by the vectors from $\mathcal{S}$} \Longleftrightarrow \mathbf{A}\vec x = \vec v\text{ is consistent where $\mathbf{A} = [\vec u_1~ \cdots~\vec u_n]$}\)
Definition 2
If \(\mathcal{ V}\) is a set of vectors from \(\mathcal{R^n}\) and span \(\mathcal{S}\) = \(\mathcal{V}\), then we can say that \(\mathcal{S}\) generates \(\mathcal{V}\) or that \(\mathcal{S}\) is a generating set of \(\mathcal{V}\),
Example 1
Let \(\mathcal { S } = \left\{ \left[ \begin{array} { l } { 1 } \\ { 0 } \\ { 0 } \end{array} \right] , \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 0 } \end{array} \right] , \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] , \left[ \begin{array} { r } { 1 } \\ { - 2 } \\ { - 1 } \end{array} \right] \right\}\), show that span \(\mathcal{S} = \mathcal{R^3}\)
prove:
Since the \ (\ text {span} \ mathcal {S} \ subset \ mathcal {R} ^ 3 \) is obviously established, we need to prove that \ (\ mathcal {R} ^ 3 \ subset \ text {span} \ mathcal {S} \) , i.e., \ (\ mathcal {R ^ 3 } \) either a vector \ (\ vec v \) belonging \ (\ text {span} \ mathcal {S} \) , and further, we want to prove the equation \ (\ mathbf {a} x = v \) is solvability (or the named consistent) \ ((\ FORALL \ VEC V \ in \ ^ R & lt mathcal {}. 3) \) , where
\ [ \ mathbf {A} = \ left [\ begin {array} {rrrr} {1} & {1} & {1} & {1} \\ {0} & {1} & {1} & {- 2} \\ {0} & {0}
& {1} & {- 1} \ end {array} \ right] \] the matrix A into row echelon form simplest R & lt
\ [\ R & lt mathbf {} = \ left [ \ begin {array} {cccc} {1} & {0} & {0} & {3} \\ {0} & {1} & {0} & {-1} \\ {0} & {0} & {1} &{-1}\end{array}\right] \]
It can be seen, the augmented matrix \ ([\ mathbf {A} ~ b] \) into \ ([\ mathbf {R} ~ c] \) then, regardless of where \ (\ vec c \) is the number, linear equations must be solved, i.e., regardless of the augmented matrix \ ([\ mathbf {a} ~ b] \) is the number of linear equations must be solvable, and the pair of \ (\ mathcal {R ^ 3 } \) any vector \ (\ vec v \) are true, then the \ (\ mathcal {R & lt ^. 3} \ Subset \ text {span} \ mathcal {S} \) , conclusions can be of the above-mentioned evidenced the span \ (\ mathcal {S} = \ ^ R & lt mathcal. 3} {\) .
Theorem 1.6
The following statements about an \(m \times n\) matrix \(\mathbf{A}\) are equivalent
(a) The span of columns of \(\mathbf{A}\)is \(\mathcal{R^m}\)
(b) The equation \(\mathbf{A}x = b\) has at least one solution (\(Ax = b\) is consistent) for each b in \(\mathcal{R^m}\)
(c) The rank of \(\mathbf{A}\) is m, the numer of rows in \(\mathcal{A}\)
(d) The reduced row echelon form of \(\mathcal{A}\) has m nonzero rows
(e) There is a pivot in each row of \(\mathcal{A}\)
Prove only (b) \ (\ Longleftrightarrow \) (c), the other can be directly launched by the definition of
(1) (b) \(\Rightarrow\) (c)
Suppose (c) is not satisfied, i.e., Rank ( \ (\ mathbf {A} \ ) ) <m, the matrix \ (\ mathbf {A} \ ) then into the most simple row echelon form matrix R, the last row must be nonzero OK, we take m-th element is a unit vector 1 \ (\ vec e_m \) and \ (\ mathbf {R} \ ) consisting of a matrix \ ([\ mathbf {R & lt} ~ e_m] \) , then \ ([ \ mathbf {R} ~ e_m] \) is mismatched (Theorem 1.5), since the \ (\ mathbf {a} \ ) becomes \ (\ mathbf {R} \ ) is substantially the line change is reversible, we \ ([\ mathbf {R} ~ e_m] \) an inverse transform is applied to this series of line change, can be obtained augmented matrix \ ([\ mathbf {a} ~ B] \) , the \ ([\ mathbf {R} ~ e_m] \ ) is not matched, we can get \ ([\ mathbf {R} ~ e_m] \) is not matched, and \ (\ VEC B \ in \ mathbf {R & lt ^ m} \) , the premise and (b) contrary to assumption does not hold, that is, (c) is true, (b) \ (\ Rightarrow \) (c) is proved
(2) (c) \(\Rightarrow\) (b)
Seen from (C) \ ([\ mathbf {A} ~ B] \) of \ ([\ mathbf {R} ~ c] \) after, \ ([\ mathbf {R} ~ c] \) is not present only in the last column there are non-zero elements of the line, from the previous theorems for 1.5, \ (\ mathbf {a} B = X \) is the consistency, i.e., (b) the establishment, (b) \ (\ Rightarrow \) (C) proved.
In summary, (B) \ (\ Longleftrightarrow \) (C)
\ ([1 ~ -1] ^ {T}, [0 ~ 1] ^ {T} \) may Zhang \ (\ R & lt mathcal {} ^ 2 \) , and further $ [1 ~ -1] ^ { T }, [~ 0. 1] ^ {T}, [2 ~. 3] T ^ {} \ (may span \) \ R & lt mathcal {} ^ 2 \ (, how it gives the minimum of a finite set \) \ mathcal {S} $, such that span \ (\ mathcal {S} = \ mathcal {R & lt} ^ m \) , if it is not the minimum, \ (\ mathcal {S} \) excess vector \ (~ \ vec v ~ \) what is the nature.
Theorem 1.7
Let \(\mathcal{S} = \lbrace \vec u_1, \cdots , \vec u_n\rbrace\), prove that
\ [Span ~ \ lbrace \ vec u_1, \ cdots, \ vec u_n \ rbrace = span ~ \ lbrace \ vec u_1, \ cdots, \ vec u_n, \ vec v \ rbrace \ Longleftrightarrow \ vec v \ in span ~ \ mathcal {S} \]
proof:
sufficiency:
\ (\ VEC V \ in span ~ \ lbrace \ VEC U_1, \ cdots, \ VEC U_n, \ VEC V \ rbrace \) , and \ (span ~ \ lbrace \ vec u_1 , \ cdots, \ VEC U_n \ rbrace = span ~ \ lbrace \ VEC U_1, \ cdots, \ VEC U_n, \ VEC V \ rbrace \) , there are \ (\ vec v \ in span ~ \ lbrace \ vec u_1, \ cdots, \ VEC U_n \ rbrace span ~ = ~ \ mathcal {S} \) , adequacy proved.
Necessity:
\ (\ VEC V \ ~ in ~ span \ mathcal {S} \) then \ (\ vec v \) is a linear combination of vectors in the set S, may wish to set
\ [\ vec v = c_1 \ vec u_1 + c_2 \ vec u_2 + \ cdots
+ c_n \ vec u_n \] the span { \ (\ mathcal {S} \ Cup {\ VEC V} \) } to any one vector \ (\ vec \ alpha = l_1 \ vec U_1 + L_2 \ VEC U_2 + \ cdots + L_n \ VEC U_n + c_1 and \ VEC U_1 + c_2 \ VEC U_2 + \ cdots + C_N \ VEC U_n \) , finishing available
\ [\ vec \ alpha = ( l_1 + c_1) \ vec u_1 + (l_2 + c_2
) \ vec u_2 + \ cdots + (l_n + c_n) \ vec u_n \ in ~ span ~ \ mathcal {S} \] i.e. span { \ (\ mathcal {S} \ Cup {\ V} VEC \) } \ (\ Subset \) span \ (\ mathcal {S} \) , on the other hand, span \ (\ mathcal {S} \) \ (\ Subset \) span { \ (\ mathcal { S} \ cup {\ vec v } \)} Is obvious, therefore span \ (\ mathcal {S} \) = span { \ (\ mathcal {S} \ {Cup \ VEC V} \) }, the necessity is proved.
Remark
Rank Solution and coefficient matrix equations \ (rank (\ mathbf {A }) \) and the augmented matrix rank \ (rank ([\ mathbf { A} ~ b]) \) Relationship
Set \ (rank (\ mathbf {A }) = m, rank ([\ mathbf {A} ~ b]) = n, \ mathbf {A} \) the number of columns is \ (K ~ \) .
(1) n> m linear equations has no solution (the presence \ (0a_1 + \ cdots 0a_n = D + \ NEQ 0 \) )
(2) n = m solvable linear equations, 1. N = m = unique nonzero solution k 2. n = m <k numerous solutions
When \ (~ b = 0 ~ \ ) when, \ (\ mathbf {A} B = X \ rightarrow \ mathbf {A} = 0 X \) is the homogeneous linear equation, at least a zero solution.
(1) n <k linear equations numerous solutions
(2) n = k linear equations has a unique solution zero
Problem
(1) matrix \ (\ mathbf {A} \ ) after primary transformation into the most simple step type matrix row \ (\ R & lt mathbf {} \) , provided \ (\ mathcal {A} \ ) is represented by \ (\ mathbf {a} \ ) set row vector composition, \ (\ mathcal {R & lt} \) represented by (\ mathbf {R} \) \ set of row vectors of the composition, there are
\ [span ~ \ mathcal {A} = span ~ \ mathcal {R} \] $ \ qquad (span ~ \ lbrace \ vec u_1, \ vec u_2, \ cdots, \ vec u_n \ rbrace = span ~ \ mathcal {A}) $ \]
Proof: For simplicity, we first consider the elementary transformation occurs in the \ (\ mathbf {A} \ ) between the first and second lines
case I: Matrix \ (\ mathcal {A} \ ) occurs exchange line
is obviously, after the exchange of a collection of row vectors consisting of the elements are not changed, it will not change span
case II: a matrix is multiplied by scalar k \ (\ mathcal {a} \ ) a first row ( \ (K \ NEQ 0 \) )
\ [\ FORALL \ Alpha \ in span ~ \ mathcal {A} \ Rightarrow \ Alpha = c_1 and \ VEC U_1 + c_2 \ VEC U_2 + \ cdots + C_N \ VEC U_n = \ dfrac { c_1} {k} k \ vec u_1 + c_2 \ vec u_2 + \ cdots + c_n \ vec u_n \ in span \ lbrace k \ vec u_1, \ vec u_2, \ cdots, \ vec u_n \ rbrace \\ \ forall \ alpha \ in span \ lbrace k \ vec u_1, \ vec u_2, \ cdots, \ vec u_n \ rbrace \ Rightarrow \ alpha = m_1 k \ vec u_1 + \ cdots + m_n \ vec u_n \ in span ~ \ mathcal {A} \ ]
we can get
\ [~ Span \ mathcal {A} \ Longleftrightarrow span ~ \ lbrace K \ VEC U_1, \ U_2 VEC, \ cdots, \ VEC U_n \ rbrace \]
Case III: Matrix \ (\ mathbf {A} \ ) first It is applied to the second row after row by K Scalar
\ [\ FORALL \ Alpha \ span in ~ \ mathcal {A} \ Rightarrow \ Alpha = c_1 and \ U_1 VEC + c_2 \ U_2 VEC + \ + cdots C_N \ VEC = U_n (c_1-c_2k) \ vec u_1 + c_2 (k \ vec u_1 + \ vec u_2) + \ cdots + c_n \ vec u_n \ in span \ lbrace \ vec u_1, \ vec u_2 + k \ vec u_1, \ cdots, \ vec u_n \ rbrace \\ \ forall \ alpha \ in span \ lbrace \ vec u_1, \ vec u_2 + k \ vec u_1, \ cdots, \ vec u_n \ rbrace \ Rightarrow \ alpha = (m_1 + km_2) \ vec u_1 + \ cdots + m_n \ vec u_n \ in span ~ \ mathcal {A} \]
We can get
\ [span ~ \ mathcal {A
} \ Longleftrightarrow span ~ \ lbrace \ vec u_1, \ vec u_2 + k \ vec u_1, \ cdots, \ vec u_n \ rbrace \] i.e. matrix \ (\ mathbf {A} \) after several primary transformation matrix of \ (\ mathbf {B} \ ) have \ (~ span \ mathcal {a} \ ~ Longleftrightarrow span \ mathcal {B} \) , and so on, we can get
\ [ span ~ \ mathcal {a} \ Longleftrightarrow span ~ \ mathcal {R & lt} \]
(2) the title of the \ (\ mathbf {a} \ ) column vector and \ (\ mathbf {R} \ ) column vector whether the space spanned by an equivalent
否，反例\(A = \left[ \begin{array} { l l } { 1 } & { 0 } \\ { 1 } & { 0 } \end{array} \right]\)，\(R = \left[ \begin{array} { l l } { 1 } & { 0 } \\ { 0 } & { 0 } \end{array} \right]\)