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General linear mappings

4
...
11 � IR
...
Since vector spaces encompass the essential
properties of IR
...
This also turns out to be extremely
useful and important in many real-world applications
...
, a function L : V

Linear Mapping

satisfies the linearity properties
L1 Lx
(

y)

W is a linear mapping if it

�

L(x)

= + L(y)
+= tL(x)

L2 L(tx)

for all x, y E V and t ER If W

=

V, then Lmay be called a linear operator
...


+ L(y)

Moreover, we have that

EXAMPLE 1

Let L : M(2, 2)

�

P2 be defined by L

d

d)x

(b

([; �]) = + + +

2
x
...


[;; �:], [;� �� ]
( [�: �:]+[�� ��])=L([��::�� ��::��])
= + + + + +
+ +
= + + +
+ + + +

Solution: For any

E M(2, 2) and t E JR
...


EXAMPLE2

Let M : P3

�

P3 be defined by M(ao

linear operator
...


a1

2a2x
...


Xi+
X
+
X
3
2
]
L =
X
[[�:ll [6 2


...
L

1s

Remark
Since a linear mapping

L :V

-

Wis just a function, we define operations (addition,

scalar multiplication, and composition ) and the concept of invertibility on these more
general linear mappings in the same way as we did for linear mappings

L

:

]Rn

-

JR
...


Definition
Range

Let

V

and Wbe vector spaces over JR
...
We write

={x EV I L(x) =Ow}

Null(L)

EXAMPLE3

Let

L

: M(2, 2)

P2
L([; �])=c +(b + d)x+ax2
...

a, b, d
l+x+x2=L([� !])=c+(b+d)x+ax2
c= b+d = a =
+x+x2 E
A
L(A)= +x+x2 A=[ � �]
...
Observe that we get a system of linear
equations that is consistent with infinitely many solutions
...

c-a

is in the range of

L
...
Row

reducing the corresponding augmented matrix gives

-

1
1
0

0
-1
1

Henee, the system is inconsistent, so

Theorem 1

1

0
0

�

1

[;]

0
-1
0

1
0

1
3

is not in the span of L
...
Then

( 1 ) L(Ov) = Ow
(2) Null(L) is a subspace of V
(3) Range(L) is a subspace of W

The proof is left as Problem D 1
...


EXAMPLES

[ � ]·

Determine a basis for the range and nullspace of the linear mapping l : P1
defined by L(a+bx) =

a- 2b

Solution: If a+bx

E Null(L), then we have

]
[
�

=

l(a+bx) =

a- 2b

[ �]

�

IR
...
Thus, the only polynomial in the nullspace
of l is the zero polynomial
...


Any vector y in the range of l has the form

y

=

�
I l [�l [ 1
{ [�l [J]}
...
Hence C is a basis for the range of l
...

Solution: If[: �] Null(L), then 0 =L([: �]) =(b + c) + (c - d)x2, sob + c =0
and c - d =0
...

Any polynomial in the range of L has the form (b+c)+(c-d)x2
...
Also, {1, x2} is clearly linearly independent and hence a basis for
2
Range(L)
...
This result
reminds us of the Rank Theorem (Theorem 3
...
8)
...

Definition
Rank of
a Linear Mapping

Definition
Nullity of
a Linear Mapping

Let V and W be vector spaces over R The rank of a linear mapping L : V
the dimension of the range of L:
rank(L) = dim ( Range(L))
Let V and W be vector spaces over R The nullity of a linear mapping L
the dimension of the nullspace of L:
nullity(L) =dim (Null(L))

:

V

�

W

is

�

W

is

Theorem 2

[Rank-Nullity Theorem]
Let V and W be vector spaces over IR'
...
Then,
rank(L)

+

nullity(L)

=

n

Proof: The idea of the proof is to assume that a basis for the nullspace of L contains
k vectors and show that we can then construct a basis for the range of L that contains
n k vectors
...
,vd be a basis for Null(L), so that nullity(L) = k
...
3
...
,u,,
such that {v1,
...
,Un} is a basis for V
...
Then w = L(x) for some x
But any x

E

·

V
...
,vbuk+1,
...
,tn such that x
tk+I Uk+! +
+ t11 u11• Then,
·

E

=

t1V1

+
...
,L(un)}
...
Is it linearly

independent? We consider

By the linearity of L, this is equivalent to

If this is true, then tk+I Uk+!

+

·

·

·

+ t11u,, is a vector in the nullspace of L
...
, db we have

But this is impossible unless all

t;

and d; are zero, because {v1,
...
,u11} is

a basis for V and hence linearly independent
...
Hence it is a
basis for Range(L) containing n - k vectors
...


+

nullity(L)

=

(n - k) + k

=

=

n

n

-

k and

P3
([� :])=cx+(a+b)x3

Determine the rank and nullity of L

EXAMPLE7

L

:

M(2,2)

defined by

�

and verify the Rank-Nullity Theorem
...

Clearly a basis for the range of L is
since this set is linea�ly independent
and spans the range
...

Then, as predicted by the Rank-Nullity Theorem, we have
=

{x, x3},

=

rank(L)

+

nullity(L)

=+ =
2

2

=

4

dim M(2,2)

PROBLEMS 4
...


IR3 IR2
x x 2 , x 3)
(X1 +X2, X\ +X2+X3)
P,
m J = (a+ b)+
(a+b+c)x
�IR
([� :])=a+d

(a) L :

(b) L

�

,

R'

�

(c) tr: M(2,2)

defined by L(

1,

=

defined by L

defined by tr

(Taking the trace of a matrix is a linear opera­
tion
...


� IR
ad-be
P
�
P
2
2
(a-b)+(b+c)x2
T: IR2
]
[�l :2
M([� �])= [� �]

(a) det : M(2,2)
(b) L :
(c)

defined by

�

([� :]) =
L(a + bx + cx2) =
r([��])=

defined by det

M(2,2) defined by

(d) M : M(2,2)

�

M(2,2) defined by

[-

A3 For each of the following, determine whether the

-2Xt

given vector y is in the range of the given linear
mapping L : V � W
...

(a) L : JR3 � JR4 defined by L

([�:ll

Xt +X3

y=

X2 - X4

[=� -�]

A4 Find a basis for the range and nullspace of the fol­

0
0

=

-

2 x2 - 2x3 - 2x4

lowing linear mappings and verify the Rank-Nullity
Theorem
...
3 � IR
...


([�ll

(a+b+c)x
(c) tr : M(2,2) � IR
...

2
(a) L: P2 � JR3 defined by /Aa+bx+cx ) =

(

(b) L : P2 � Pt defined by L a +bx+cx
b+ 2cx

(c) L: M(2,2) �IR
...
defined by L(a+bx+cx +dx3) =
a
b
c
d

(d) Let JD be the subspace of M(2,2) of diagonal

]� )
!])

linear
...
defined by M(a + bx+ cx ) =
2
b - 4ac

(e)

N : JR3 �

[

Xl

M(2,2) defined by

N

� X3 � ]

mm

=

2

(d) L : M(2,2) � M(2, 2) defined by L (A ) =

[ � ;]

(e) T

:

A

M(2,2) � P2 defined by T

2
a+ (ab)x+ (abc)x

)
([ ]
a

b

c

d

--

B3 For each of the following, determine whether the
given vector y is in the range of the given linear
mapping L : V -- W
...


L([::Jl =
l· [ �1

:\ � �:

2

y=

=l+x- x2

[

--

JR
...
defined by L

(d) Let D be the subspace of M(2,2) of diagonal
matrices; L: D -- P defined by
2
a+(a+b)x+bx2

[:=� �=�]

([��])=

(f) L : M(2,2) -- P defined by L
2

y=

([; :]) =

(a+b +c + d)x2

(e) Let T denote the subspace of 2 x 2 upper­
triangular matrices; L : T -- P defined by
2

L([� �])= -

( a- c) +(a - 2b)x+

] [-

-- M(2, 2) defined by T(A)=AT

(g) T : M(2,2)

[ ;: : ��

(h) L: P

2

M(2,2) defined by L(a+bx+cx2) =

--

-�: = �]

_

(-2a+b + c)x2, y= 2 +x- x2
L : P -- M(2,2) defined by L(a+bx+cx2)
2
-a - 2c
2b - c
2
2
=
y
0 -2
-2a+2c -2b - c '

[

-- P1 defined by L a+bx+cx2

(e) L: M(2,2) -- M(2,2) defined by

(d) L: JR
...


(b) L : P

1
-2x1 + x - 2x3
2
(b) L : P -- P defined by L(a + bx + cx2) =
2
2
(-a
2b) + (2a + c)x + (-2a + b - 2c)x2,

-

lowing linear mappings and verify the Rank-Nullity

(a) L • P,

R3 defined by
2

B4 Find a basis for the range and nullspace of the fol­

]

=

Conceptual Problems
Dl Prove Theorem 1
...

(a) V = R 3, W = P • L
2

L

([�ll

([�ll

= x', L

([�]l

= 2x
,

rank(L)= 2, and L

•

,vk} is a linearly independent set

in V
...


Range(L)

{O} and

=

W be a linear mapping
...


DS Let U, V andW be finite-dimensional vector spaces
--

U and M: U -- W be linear

mappings
...


(b) Prove that rank(M

o

L) $ rank(L)
...
• •

=

--

over JR
...
Prove that if {L(v1),

in V but {L(v1),

let L : V

=
{[� �J
...
rn �]}
=
=
([� �]) �
JR
...
, vd is linearly independent

= I +x+x2

(b) V = P , W = M(2, 2); Null(L)
2

then {v1,

,

of the ranks
...
Define the left shift L : S -- S by

and let L : V -- Ube a linear mapping and M :

L(x1, x2,x3,
...
) and the right shift

U -- Ube a linear operator such that Null(M) =

R : S -- S by R(x1,X2,X3,
...

Then it is easy to verify that L and R are linear
...
x
...
It is important in this example that S is

Ou}
...

D7 Let S denote the set of all infinite sequences of
real numbers
...
,x11,•••)
...
Then S is a

infinite-dimensional

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