null space, rank and nullity theorem
TRANSCRIPT
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Row Space, Column Space, Null Space And
Rank Nullity Theorem
A part of active learning assignment.An initiative by GTU.
Prepared by, Ronak Machhi (130410109042) (130410109044) (130410109046) (130410109048) (130410109050)
Guided by:Prof. Harshad R. PatelProf. J. B. Patel
Vector Calculus and Linear Algebra
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Outline Row Space Column Space Null space Rank And Nullity
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Definition
11 12 1
21 22 2
1 2
1 11 12 1
2 21 22 2
For an m n matrix
...
... A=
: : :
...
the vectors
[ ]
[ ]
:
n
n
m m mn
n
n
a a a
a a a
a a a
a a a
a a a
r
r
. . .
. . .
m 1 2
n
11 12 1
21 22 21 2
1 2
[ ]
in R formed form the rows of Aare called the row vectors of A, and the vectors
, , ..., : : :
in
m m mn
n
nn
m m mn
a a a
a a a
a a a
a a a
r
c c c
. . .
m R formed from the columns of A are called the column vectors of A.
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Example 1Row and Column Vectors in a 2×3 Matrix
Let
The row vectors of A are
and the column vectors of A are
2 1 0
3 1 4A
1 22 1 0 and 3 1 4 r r
2 1 0, , and
3 1 4
1 2 3c c c
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Definition If A is an m×n 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 equation Ax=0, which is a subspace of Rn, is called the nullsapce of A.
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Theorem
Elementary row operations do not change the nullspace of a matrix.
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Example 4 Basis for Nullspace Find a basis for the nullspace of
Solution.The nullspace of A is the solution space of the
homogeneous system
After solution we find that the vectors
Form a basis for this space.
2 2 1 0 1
1 1 2 3 1
1 1 2 0 1
0 0 1 1 1
A
1 2 3 5
1 2 3 4 5
1 2 3 5
3 4 5
2 2 0
2 3 0
2 0
0
x x x x
x x x x x
x x x x
x x x
1 2
1 1
1 0
and 0 1
0 0
0 1
v v
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Theorem
Elementary row operations do not change the row space of a matrix.
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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.
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Example 5Bases for Row and Column Spaces
1
2
3
The matrix
1 2 5 0 3
0 1 3 0 0
0 0 0 1 0
0 0 0 0 0
is in row-echelon form. From Theorem 5.5.6 the vectors
[1 -2 5 0 3]
[0 1 3 0 0]
[0 0 0 1 0]
form a
R
r
r
r
1 2 4
basis for the row space of R, and the vectors
1 2 0
0 1 0 , ,
0 0 1
0 0 0
form a basis for the column space of R.
c c c
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Example 6Bases for Row and Column Spaces (1/2)
Find bases for the row and column spaces of
Solution.Reducing A to row-echelon form we obtain
By Theorem 5.5.6 the nonzero row vectors of R form a basis for row space of R, and hence form a basis for the row space of A. These basis vectors are
1 3 4 2 5 4
2 6 9 1 8 2
2 6 9 1 9 7
1 3 4 2 5 4
A
1 3 4 2 5 4
0 0 1 3 2 6
0 0 0 0 1 5
0 0 0 0 0 0
R
1
2
3
r 1 3 4 2 5 4
r 0 0 1 3 2 6
r 0 0 0 0 1 5
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Example 6Bases for Row and Column Spaces (2/2)
We can find a set of column vectors of R that forms a basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A.
The first, third, and fifth columns of R contain the leading 1’s of the row vector , so
Form a basis for the column space of R; thus
Form a basis for the column space of A.
1 4 5
0 1 2, ,
0 0 1
0 0 0
' ' '1 5 5c c c
1 4 5
2 9 8, ,
2 9 9
1 4 5
1 3 5c c c
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Example 7Basis for a Vector Space Using Row Operations (1/2) Find a basis for the space spanned by the vectorsv1=(1, -2, 0, 0, 3), v2=(2, -5, -3, -2, 6), v3=(0, 5, 15, 10, 0),
v4=(2, 6, 18, 8, 6)Solution.Except for a variation in notation, the space spanned by
these vectors is row space of the matrix
Reducing this matrix to row-echelon form we obtain
1 2 0 0 3
2 5 3 2 6
0 5 15 10 0
2 6 18 8 6
1 2 0 0 3
0 1 3 2 0
0 0 1 1 0
0 0 0 0 0
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Example 7Basis for a Vector Space Using Row Operations (2/2)
The nonzero row vectors in this matrix are
w1=(1, -2, 0, 0, 3), w2=(0, 1, 3, 2, 0), w3=(0, 0, 1, 1, 0)
These vectors form a basis for the row space and consequently form a basis for the subspace of R5 spanned by v1, v2, v3, and v4.
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Example 8Basis for the Row Space of a Matrix (1/2) Find a basis for the row space of
Consisting entirely of row vectors from A.Solution.Transposing A yields
1 2 0 0 3
2 5 3 2 6
0 5 15 10 0
2 6 18 8 6
A
1 2 0 2
2 5 5 6
0 3 15 18
0 2 10 8
3 6 0 6
TA
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Example 8Basis for the Row Space of a Matrix (2/2)
Reducing this matrix to row-echelon form yields
The first, second, and fourth columns contain the leading 1’s, so the corresponding column vectors in AT form a basis for the column space of ; these are
Transposing again and adjusting the notation appropriately yields the basis vectors
r1=[1 -2 0 0 3], r2=[2 -5 -3 -2 6], and r3=[2 -5 -3 -2 6]
for the row space of A.
1 2 0 2
0 1 5 10
0 0 0 1
0 0 0 0
0 0 0 0
1 2 2
2 5 6
, , and 0 3 18
0 2 8
3 6 6
1 2 4c c c
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Rank And Nullity
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Theorem
If A is any matrix, then the row space and column space of A have the same dimension.
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Definition
The common dimension of the row 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).
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Example 1Rank and Nullity of a 4×6 Matrix (1/2)
Find the rank and nullity of the matrix
Solution.The reduced row-echelon form of A is
Since there are two nonzero rows, the row space and column space are both two-dimensional, so rank(A)=2.
1 2 0 4 5 3
3 7 2 0 1 4
2 5 2 4 6 1
4 9 2 4 4 7
A
1 0 4 28 37 13
0 1 2 12 16 5
0 0 0 0 0 0
0 0 0 0 0 0
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Example 1Rank and Nullity of a 4×6 Matrix (2/2)
The corresponding system of equations will be x1-4x3-28x4-37x5+13x6=0
x2-2x3-12x4-16x5+5x6=0It follows that the general solution of the system is x1=4r+28s+37t-13u; x2=2r+12s+16t-5u; x3=r;
x4=s; x5=t; x6=uOr
So that nullity(A)=4.
1
2
3
4
5
6
4 28 37 13
2 12 16 5
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
x
x
xr s t u
x
x
x
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Theorem
If A is any matrix, then rank(A)= rank(AT).
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Theorem 5.6.3Dimension Theorem for Matrices
If A is a matrix with n columns, then
rank(A)+nullity(A)=n
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Theorem
If A is an m×n matrix, then:a) rank(A)=the number of leading
variables in the solution of Ax=0.b) nullity(A)=the number of
parameters in the general solution of Ax=0.
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Example 2The Sum of Rank and Nullity The matrix
has 6 columns, so rank(A)+nullity(A)=6This is consistent with Example 1, where we
should showed that rank(A)=2 and nullity(A)=4
1 2 0 4 5 3
3 7 2 0 1 4
2 5 2 4 6 1
4 9 2 4 4 7
A
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References:
Textbook of Vector Calculus and Linear Algebra (Atul Publications)
Note: All the excerpts taken from the respective book were used only for academic purpose. NO commercial purpose was entertained.
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Thank you.