Section 5 5 row space column space and nullsapce
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Section 5.5 Row Space, Column Space, and Nullsapce. Definition. For an mxn matrix The vectors in R n formed from the rows of A are called the row vectors of A,. Row Vectors and Column Vectors. And the vectors

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Section 5 5 row space column space and nullsapce
Section 5.5 Row Space, Column Space, and Nullsapce

Definition. For an mxn matrix

The vectors

in Rn formed from the rows of A are called the row vectors of A,


Row vectors and column vectors
Row Vectors and Column Vectors

And the vectors

In Rn formed from the columns of A are called the column vectors of A.

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 of A 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 nullsapce of A.


Nullspace
Nullspace

Theorem

Elementary row operations do not change the nullspace of a matrix.

Example

Find a basis for the nullspace of

The nullspace of A is the solution space of the homogeneous system Ax=0.

2x1 + 2x2 – x3 + x5 = 0

-x1 + x2 + 2x3 – 3x4 + x5 = 0

x1 + x2 – 2x3 – x5 = 0

x3+ x4 + x5 = 0.


Nullspace cont
Nullspace Cont.

Then by the previous example, we know

Form a basis for this space.


Theorems
Theorems

Theorem

Elementary row operations do not change the row space of a matrix.

Note: Elementary row operations DO change the column space of a matrix.

However, we have the following theorem

Theorem

If A and B are row equivalent matrices, then

  • A given set of column vectors of A is linearly independent if and only if 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 if and only if the corresponding column vectors of B form a basis for the column space of B.


Theorems cont
Theorems Cont.

Theorem

If a matrix R is in row-echelon form, then the row vectors with the leading 1’s (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
Example

Example

Find bases for the row and column spaces of

Solution. Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis for the row space of any row-echelon form of A.


Example1
Example

By Theorem, the nonzero row vectors of R form a basis for the row space of R and hence form a basis for the row space of A. These basis vectors are

Note that A ad R may have different column spaces, but from Theorem that if we can find a set of column vectors of R that forms a basisi for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A.


Example2
Example

Note

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.


Section 5 6 rank and nullity
Section 5.6 Rank and Nullity

The four fundamental matrix spaces associated with A are:

Row space of A column space of A

Nullspace of A nullspace of AT

Theorem

If A is any matrix, then the row space and column space of A have the same dimension.

Definition

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).


Example3
Example

Example

Find the rank and nullity of the matrix

Solution. The reduced row-echelon form of A is


Example4
Example

Since there are two nonzero rows (or, equivalently, two leading 1’s), the row space and column space are both two-dimensional, so rank(A)=2. To find the nullity of A, we must find the dimension of the solution space of the linear system Ax=0. This system can be solved by reducing the augmented matrix to reduced row-echelon form. The corresponding system of equations will be

X1-4x3-28x4-37x5+13x6=0

X2-2x3-12x4-16x5+5x6=0

Solving for the leading variables, we have

x1=4x3+28x4+37x5-13x6

X2=2x3+12x4 +16x5-5x6

It follows that the general solution of the system is


Example cont
Example Cont.

x1=4r+28s+37t-13u

X2=2r+12s+16t-5u

X3=r

X4=s

X5=t

X6=u

Equivalently,

Because the four vectors on the right side of the equation form a basis for the solution space, nullity(A)=4.


Theorems1
Theorems

Theorem

If A is any matrix, then rank(A)=rank(AT).

Theorem (Dimension Theorem for Matrices)

If A is a matrix with n columns, then

Rank(A)+nullity(A=n

Theorem

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.


Theorems2
Theorems

Theorem (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.

    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..


Theorems3
Theorems

Theorem

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

If A is an mxn matrix, then the following are equivalent.

  • Ax=0 has only the trivial solution.

  • The column vectors of A are linear independent.

  • Ax=b has at most one solution (none or one) for every mx1 matrix b.


Theorems4
Theorems

Theorem (Equivalent Statements)

I 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 expressed 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.

  • The range of TA is Rn.

  • TA is one-to-one.

  • The column vectors of A are linearly independent.


Theorem cont
Theorem Cont.

(k) 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.


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