1. 5.5 Row Space, Column Space, and Nullspace

2. Row Space, Column Space, and Nullspace
Definition:
For an mxn matrix
the vectors
in Rn formed from the rows of A are called the row vectors of A, and the vectors
in Rm formed from the columns of A are called the column vectors of A

3. Row Space, Column Space, and Nullspace
Definition:
If A is an mxn matrix, then the subspace of Rn spanned by the row vectors of A is called the row space of A,
and the subspace of Rm spanned by the column vectors is called the column space of A.
The solution space of the homogeneous system of equations Ax=0, which is a subspace of Rn, is called the nullspace of A.
Theorem 5.5.1: A system of linar equations Ax=b is consistent iff b is in the column space of A

4. example Show that b in column space of A!
The solution by G.E. X1 = 2, X2 = -1, X3 = 3, the system is consistent, b is in the column space of A

5. Row Space, Column Space, and Nullspace
Theorem 5.5.2: If x0 denotes any single solution of a consistent linear system Ax=b, and if v1,v2,...,vk form a basis for the nullspace of A, that is, the solution space of the homogeneous system Ax=0, then every solution of Ax=b can be expressed in the form
and, conversely, for all choices of scalars c1,c2,...,ck, the vector x in this formula is a solution of Ax=b.

6. General and Particular Solutions
Terminology:
Vector x0 is called a particularly solution of Ax=b.
The expression x0+c1v1+c2v2+...+ckvk is called the general solution of Ax=b.
The expression c1v1+c2v2+...+ckvk is called the general solution of Ax=0.

7. Bases for Row Spaces, Column Spaces, and Nullspaces
Theorem 5.5.3: Elementary row operations do not change the nullspace of a matrix
Theorem 5.5.4: Elementary row operations do not change the row space of a matrix
Theorem 5.5.5: If A and B are row equivalent matrices, then:
A given set of column vectors of A is linearly independent iff the corresponding column vectors of B are linearly independent.
A given set of column vectors of A forms a basis for the column space of A iff the corresponding column vectors of B form a basis for the column space of B.

8. Bases for Row Spaces, Column Spaces, and Nullspaces
Theorem: If a matrix R is in row-echelon form, then the row vectors with the leading 1’s (i.e., the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading 1’s of the row vectors form a basis for the column space of R.
Example: [Bases for Row and Column Spaces]
The matrix R is in row-echelon form, while the vectors r
form a basis for the row space of R

9. Bases for Row Spaces, Column Spaces, and Nullspaces
and the vectors
form a basis for the column space of R
Example: [Bases for Row and Column Spaces]
Find bases for the row and column spaces of

10. Bases for Row Spaces, Column Spaces, and Nullspaces
The basis vectors are
The first, third, and fifth columns of R contain the leading 1’s of the row vectors that form a basis for the column space of R.
Thus the corresponding column vectors of A, form a basis for the column space of A

11. Bases for Row Spaces, Column Spaces, and Nullspaces
Example: [Basis and Linear Combinations]
Find a subset of the vectors
v1=(1,-2,0,3), v2=(2,-5,-3,6), v3=(0,1,3,0),
v4=(2,-1,4,-7), v5=(5,-8,1,2) that forms a basis for the space spanned by these vectors.
Express each vector not in the basis as a linear combination of the basis vector
Solution:

12. Bases for Row Spaces, Column Spaces, and Nullspaces
Basis for the column space of matrix vectors w is {w1,w2,w4} and consequently basis for the column space of matrix vectors v is {v1,v2,v4}.
Expressing w3 and w5 as linear combinations of the basis vectors w1,w2, and w4 (dependency equations).
w3 = 2w1 - w2
w5 = w1 + w2 + w4
The corresponding relationships are
v3 = 2v1 – v2
v5 = v1 + v2 + v4

13. Bases for Row Spaces, Column Spaces, and Nullspaces
Given a set of vectors S={v1,v2,...,vk) in Rn, the following procedure produces a subset of these vectors that forms a basis for span(S) and expresses those vectors of S that are not in the basis as linear combinations of the basis vectors.
Step 1. Form the matrix A having v1,v2,...,vk as its column vectors.
Step 2. Reduce the matrix A to its reduced row-echelon form R, and let w1,w2,...,wk be the column vectors of R.
Step 3. Identify the columns that contain the leading 1’s in R. The corresponding column vectors of A are the basis vectors for span(S).
Step 4. Express each column vector of R that does not contain a leading 1 as a linear combination of preceding column vectors that do contain leading 1’s.

14. 5.6 Rank and Nullity

15. Four Fundamental Matrix Spaces
Row space of A, Column space of A
Nullspace of A, Nullspace of AT
Relationships between the dimensions of these four vector spaces.

16. Row and Column Spaces have Equal Dimensions
Theorem 5.6.1: If A is any matrix, then the row space and column space of A have the same dimension.
The common dimension of the row space and column space of a matrix A is called the rank of A and is denoted by rank(A); the dimension of the nullspace of A is called the nullity of A and is denoted by nullity(A).

17. Row and Column Spaces have Equal Dimensions
Example: [Rank and Nullity of a 4x6 Matrix]
Find the rank and nullity
of the matrix
Solution:
The reduced row-echeclon
form of A is
rank(A) = 2 and the corresponding system will be

18. Row and Column Spaces Have Equal Dimensions
The general solution of the system is

19. Row and Column Spaces Have Equal Dimensions
Nullity(A)=4

20. Row and Column Spaces Have Equal Dimensions
Theorem 5.6.2: If A is any matrix, then rank(A) = rank(AT).
Theorem 5.6.3: [Dimension Theorem for Matrices]
If A is a matrix with n columns, then
rank(A) + nullity(A) = n
Theorem 5.6.4: If A is an mxn matrix, then:
Rank(A) = the number of leading variables in the solution of Ax = 0.
Nullity(A) = the number of parameters in the general solution of Ax = 0.

21. Row and Column Spaces Have Equal Dimensions
A is an mxn matrix of rank r
Fundamental Space
Dimension
Row space of A r
Column space of A
Nullspace of A
n-r
Nullspace of AT
m-r

22. Maximum Value for Rank A is an mxn matrix: rank(A) ≤ min(m,n)
where min(m,n) denotes the smaller of the numbers m and n if m≠n or their common value if m=n.

23. Linear Systems of m Equations in n Unknowns
Theorem 5.6.5: [The Consistency Theorem]
If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent.
Ax = b is consistent
b is in the column space of A.
The coefficient matrix A and the augmented matrix [A|b] have the same rank.

24. Linear Systems of m Equations in n Unknowns
Theorem:
If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent.
Ax = b is consistent for every mx1 matrix b.
The column vectors of A span Rm.
Rank(A) = m
A linear system with more equations than unknowns is called an overdetermined linear system. The system cannot be consistent for every possible b.

25. Linear Systems of m Equations in n Unknowns
Example: [Overdetermined System]
The system is consistent
iff b1, b2, b3, b4, and b5
satisfy the conditions

26. Linear Systems of m Equations in n Unknowns
Theorem 5.6.7: If Ax=b is a consistent linear system of m equations in n unknowns, and if A has rank r, then the general solution of the system contains n-r parameters.
Theorem 5.6.8: If A is an mxn matrix, then the following are equivalent.
Ax=0 has only the trivial solution.
The column vectors of A are linearly independent.
Ax=b has at most one solution (none or one) for every mx1 matrix b.
A linear system with more unknowns than equations is called an underdetermined linear system.
Underdetermined linear system is consistent if its solution has at least one parameter → has infinitely many solution.

27. Summary Theorem 5.6.9: [Equivalent Statements]
If A is an nxn matrix, and if TA:Rn→Rn is multiplication by A, then the following are equivalent.
A is invertible
Ax=0 has only the trivial solution
The reduced row-echelon form of A is In.
A is expressible as a product of elementary matrices.
Ax=b is consistent for every nx1 matrix b
Ax=b has exactly one solution for every nx1 matrix b
Det(A)≠0

28. Summary The range of TA is Rn TA is one-to-one
The column vectors of A are linearly independent
The row vectors of A are linearly independent
The column vectors of A span Rn
The row vectors of A span Rn
The column vectors of A form a basis for Rn
The row vectors of A form a basis for Rn
A has rank n
A has nullity 0