3.5 Rank and nullity of a matrix

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We can now de¯ne three important integers associated with a matrix.

DEFINITION 3.5.1 Let A 2 Mm£n(F). Then

(a) column rankA =dimC(A);

(b) row rankA =dimR(A);

(c) nullityA =dimN(A).

We will now see that the reduced row{echelon form B of a matrix A allows

us to exhibit bases for the row space, column space and null space of A.

Moreover, an examination of the number of elements in each of these bases

will immediately result in the following theorem:

THEOREM 3.5.1 Let A 2 Mm£n(F). Then

(a) column rankA =row rankA;

(b) column rankA+nullityA = n.

Finding a basis for R(A): The r non{zero rows of B form a basis for R(A)

and hence row rankA = r.

For we have seen earlier that R(A) = R(B). Also

R(B) = hB1¤; : : : ;Bm¤i

= hB1¤; : : : ;Br¤; 0 : : : ; 0i

= hB1¤; : : : ;Br¤i:

The linear independence of the non{zero rows of B is proved as follows: Let

the leading entries of rows 1; : : : ; r of B occur in columns c1; : : : ; cr. Suppose

that

x1B1¤ + ¢ ¢ ¢ + xrBr¤ = 0:

Then equating components c1; : : : ; cr of both sides of the last equation, gives

x1 = 0; : : : ; xr = 0, in view of the fact that B is in reduced row{ echelon

form.

Finding a basis for C(A): The r columns A¤c1 ; : : : ;A¤cr form a basis for

C(A) and hence column rank A = r. For it is clear that columns c1; : : : ; cr

of B form the left{to{right basis for C(B) and consequently from parts (b)

and (c) of Theorem 3.3.5, it follows that columns c1; : : : ; cr of A form the

left{to{right basis for C(A).

3.5. RANK AND NULLITY OF A MATRIX 65

Finding a basis for N(A): For notational simplicity, let us suppose that c1 =

1; : : : ; cr = r. Then B has the form

B =

2

66666666664

1 0 ¢ ¢ ¢ 0 b1r+1 ¢ ¢ ¢ b1n

0 1 ¢ ¢ ¢ 0 b2r+1 ¢ ¢ ¢ b2n

...

...

¢ ¢ ¢

...

...

¢ ¢ ¢

...

0 0 ¢ ¢ ¢ 1 brr+1 ¢ ¢ ¢ brn

0 0 ¢ ¢ ¢ 0 0 ¢ ¢ ¢ 0

...

...

¢ ¢ ¢

...

...

¢ ¢ ¢

...

0 0 ¢ ¢ ¢ 0 0 ¢ ¢ ¢ 0

3

77777777775

:

Then N(B) and hence N(A) are determined by the equations

x1 = (¡b1r+1)xr+1 + ¢ ¢ ¢ + (¡b1n)xn

...

xr = (¡brr+1)xr+1 + ¢ ¢ ¢ + (¡brn)xn;

where xr+1; : : : ; xn are arbitrary elements of F. Hence the general vector X

in N(A) is given by

2

666666664

x1

...

xr

xr+1

...

xn

3

777777775

= xr+1

2

666666664

¡b1r+1

...

¡brr+1

1

...

0

3

777777775

+ ¢ ¢ ¢ + xn

2

666666664

¡bn

... ¡

b

r

n

0

...

1

3

777777775

(3.2)

= xr+1X1 + ¢ ¢ ¢ + xnXn¡r:

Hence N(A) is spanned by X1; : : : ;Xn¡r, as xr+1; : : : ; xn are arbitrary. Also

X1; : : : ;Xn¡r are linearly independent. For equating the right hand side of

equation 3.2 to 0 and then equating components r + 1; : : : ; n of both sides

of the resulting equation, gives xr+1 = 0; : : : ; xn = 0.

Consequently X1; : : : ;Xn¡r form a basis for N(A).

Theorem 3.5.1 now follows. For we have

row rankA = dimR(A) = r

column rankA = dimC(A) = r:

Hence

row rankA = column rank A:

66 CHAPTER 3. SUBSPACES

Also

column rankA + nullityA = r + dimN(A) = r + (n ¡ r) = n:

DEFINITION 3.5.2 The common value of column rankA and row rankA

is called the rank of A and is denoted by rankA.

EXAMPLE 3.5.1 Given that the reduced row{echelon form of

A =

2

4

1 1 5 1 4

2 ¡1 1 2 2

3 0 6 0 ¡3

3

5

equal to

B =

2

4

1 0 2 0 ¡1

0 1 3 0 2

0 0 0 1 3

3

5 ;

¯nd bases for R(A); C(A) and N(A).

Solution. [1; 0; 2; 0; ¡1]; [0; 1; 3; 0; 2] and [0; 0; 0; 1; 3] form a basis for

R(A). Also

A¤1 =

2

4

1

2

3

3

5 ; A¤2 =

2

4

1

¡1

0

3

5 ; A¤4 =

2

4

1

2

0

3

5

form a basis for C(A).

Finally N(A) is given by

2

66664

x1

x2

x3

x4

x5

3

77775

=

2

66664

¡2x3 + x5

¡3x3 ¡ 2x5

x3

¡3x5

x5

3

77775

= x3

2

66664

¡2

¡3

1

0

0

3

77775

+ x5

2

66664

1

¡2

0

¡3

1

3

77775

= x3X1 + x5X2;

where x3 and x5 are arbitrary. Hence X1 and X2 form a basis for N(A).

Here rankA = 3 and nullityA = 2.

EXAMPLE 3.5.2 Let A =

·

1 2

2 4

¸

. Then B =

·

1 2

0 0

¸

is the reduced

row{echelon form of A.

3.6. PROBLEMS 67

Hence [1; 2] is a basis for R(A) and

·

1

2

¸

is a basis for C(A). Also N(A)

is given by the equation x1 = ¡2x2, where x2 is arbitrary. Then

·

x1

x2

¸

=

·

¡2x2

x2

¸

= x2

·

¡2

1

¸

and hence

·

¡2

1

¸

is a basis for N(A).

Here rankA = 1 and nullityA = 1.

EXAMPLE 3.5.3 Let A =

·

1 2

3 4

¸

. Then B =

·

1 0

0 1

¸

is the reduced

row{echelon form of A.

Hence [1; 0]; [0; 1] form a basis for R(A) while [1; 3]; [2; 4] form a basis

for C(A). Also N(A) = f0g.

Here rankA = 2 and nullityA = 0.

We conclude this introduction to vector spaces with a result of great

theoretical importance.

THEOREM 3.5.2 Every linearly independent family of vectors in a sub-

space S can be extended to a basis of S.

Proof. Suppose S has basis X1; : : : ;Xm and that Y1; : : : ; Yr is a linearly

independent family of vectors in S. Then

S = hX1; : : : ;Xmi = hY1; : : : ; Yr; X1; : : : ;Xmi;

as each of Y1; : : : ; Yr is a linear combination of X1; : : : ;Xm.

Then applying the left{to{right algorithm to the second spanning family

for S will yield a basis for S which includes Y1; : : : ; Yr.

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