Rank and Nullity
Rank (Rank) matrix and the zero degree (nullity) (from the perspective of linear equations)
Definition:
Rank: Minimal line non-zero number of rows stepped
Zero degree: n - Rank (A) (n is the number of columns of the matrix)
Relationship between the amount set forth below and two linear equations from several aspects
Remark:
Reduced row augmented matrix form Echelon \ ([\ R & lt mathbf {C} ~] \) of the row corresponding to a non-zero principal components, which are the main element 1, a row corresponds to a respective nonzero element row 1 is column vector, such as a main element on the i-th row corresponds to \ (\ vec e_i = [0,0 , \ cdots, 1 (~ i_ {th} ~ entry), \ cdots, 0, 0] ^ {T} \) , which is
Rank \ (\ Longleftrightarrow \) unit number column vector
nullity \ (\ Longleftrightarrow \) non-primary element where the number of columns = n - rank (A)
On the other hand, a base corresponding to non-zero row variable, so they can be said to rank = number of basic variables, such is the case, can be derived
Rank \ (\ Longleftrightarrow \) unit column number vector \ (\ Longleftrightarrow \) Basic solution vector variable number
nullity \ (\ Longleftrightarrow \) the number of non-primary element of the column of \ (\ Longleftrightarrow \) solution vector in the number of free variable
Property
For the case of linear equation class
\ (^ {O}. 1 ~ nullity = 0 \) , i.e. all the underlying variables, the equation has a unique solution (unique solution)
\ (2 ^ {O} ~ nullity> 0 \) , there is a free variable, there are numerous solutions of equations (infinitely many solutions)
Theoroem 1.5
The following several conditions are equivalent
(1) linear equations are \ (A \ vec x = \ vec b \) is the consistency of the (consistent)
(2) constant vector \ (\ vec b \) is the solution vector \ (\ vec x \) the linear combination of the components
(3) the augmented matrix form Echelon Row reduced \ ([\ R & lt mathbf {C} ~] \) last row does not exist \ ([0,0, \ cdots, 0, D] \) ( \ (D \ neq 0 \) form)
According to definition, (1) with (2) is equivalent to
Proof (1) \ (\ Longleftrightarrow \) (3)
Proband (1) -> (3)
Hypothesis (3) does not hold, then there exists such a line of the form, we can draw
[[0, 0,\cdots, 0] [x_1, x_2, \cdots, x_n]^{T} = 0 x_1 + 0x_2 + \cdots + 0x_n = d \neq 0 ]
We can not come to the solution vector, the equations are mismatched (inconsistent) In other words, the premise contradicts the assumption does not hold, that is, (3) established
Another card (3) -> (1)
If we do not exist can be determined by back substitution method \ ([\ mathbf {R} ~ c] \) of the solution is naturally \ ([\ mathbf {A} ~ b] \) of the solution, i.e., (1 ) established