Chapter 1
Linear Mappings
Definition 1.1. Let V and W be vector spaces over a field F. A function :VW is called a
linear mapping iff
(a)
(u,vV) (u+v)= (u)+ (v), and
(b)
(vV)(pF) (pv)=p(v)
Proposition 1.1. A function :VW is a linear mapping iff
(c)
(p,qF) (u,vV) (pu+qv) = p(u)+q(v)
Proof. () Suppose  is a linear mapping. Then (pu+qv) = (pu)+ (qv) = p(u) + q(v), by
(a) and (b) in that order.
() To prove (a) we simply put p=q=1 in (c) and to prove (b) we put q=0 and u=v. Then (pv) =
=(pv+) = (pv+0v) = p(v)+0(v) = p(v)+ = p(v).
Example 1.1. V=W=Rn[x], (f(x))=f ’(x). Differentiation is obviously a linear mapping
Example 1.2. :R3R4, (x,y,z)=(x+y,2xz,x+y+z,y). We have (x,y,z)+(a,b,c)) =
(x+a,y+b,z+c) = ((x+a)+(y+b), 2(x+a)(z+c), (x+a)+(y+b)+(z+c), (y+b)) = (x+y+a+b,
2xz+2ac, x+y+z+a+b+c, y+b) = (x,y,z) + (a,b,c) and (p(x,y,z))= (px,py,pz) =
(px+py,2pxpz,px+py+pz,py) = =(p(x+y),p(2xz),p(x+y+z),py) = p(x,y,z). Hence  is a linear
mapping.
Example 1.3. :R3R2, (x,y,z)=(x+y1,xz).  is not a linear mapping as (+) = () =
(0,0,0)= (0+01,00) = (1,0), while ()+()=(1,0)+(1,0) = (2,0). This is enough to
show that  is not linear, but let us note that  does not satisfy the second condition, too, as
(2)=()=(1,0) and 2()=2(1,0)=(2,0).
Example 1.4. V=2{1,2,…,n}, the space of all subsets of {1,2, … ,n}, over the field Z2 (see Example
2 in Chapter 4), W= Z2, (A)=|A|mod2. There are two cases to be considered while verifying the
second condition, namely p=0 or p=1. In the case p=0 we have (0A)= ()=0 and
0(A)=0(|A|mod2)=0. In the second case (1A)= (A) = |A|mod2 = 1(|A|mod2)=1(A), so the
second condition holds. To verify the first condition let us consider (AB) = |(AB)|mod2 =
|(AB)(AB)|mod2. Since AB is a subset of AB we have |(AB)(AB|=|AB||AB|.
From the inclusion-exclusion principle we have |AB| = |A|+|B||AB|. Hence (AB) =
(|A|+|B|2|AB|)mod2 = (|A|+|B|)mod2 = (|A|mod2 + |B|mod2)mod2 = (A)(B).
We will now define two important parameters of a linear mapping, rank and nullity.
Definition 1.2. Let :VW be a linear mapping. The image of  is the set im = (V) and the
kernel of  is the set ker={vV: (v)=W}
Proposition 1.2. im is a subspace of W and ker is a subspace of V.
Proof. Let w1,w2im. There exist v1,v2V such that (vi)=wi, i=1,2. Then pw1+qw2 = p(v1)+
q(v2) = (pv1+qv2)im, hence im is a subspace of W. For every v1 and v2 from ker we have
(av1+bv2) = a(v1)+b(v2) = a+b = , so ker is a subspace of V.
Definition 1.3. rank()=dim(im), nullity()=dim(ker)
Example 1.5. Let V=W=Rn[x] and (f(x))=f ’(x). Then ker=R0[x], the subspace consisting of
all constant polynomials and im=Rn1[x]. Obviously, rank()=n and nullity()=1.
Proposition 1.3. For every set {v1,v2, … ,vn}V, and for every linear mapping :VW,
(span(v1,v2, … ,vn))=span((v1),(v2), … , (v2)).
n
Proof. w(span(v1,v2, … ,vn)) iff there exist scalars a1,a2, … an such that w=  ( a i v i ) =
i 1
n
 a  (v ) . That means w(span(v1,v2, … ,vn)) iff wspan((v1),(v2), … , (v2)).
i 1
i
i
1 2 3
 x


Example 1.6. Let :R R , (X) = AX, where A  2 3 4 and X   y  . In the traditional
1 3 5
 z 
3
3
notation, (x,y,z)=(x+2y+3z,2x+3y+4z,x+3y+5z). To find the kernel of  we must solve the
matrix equation AX=0, in other words we must solve the system of linear equation
 x  2 y  3z  0

2 x  3 y  4 z  0 This system is equivalent to
 x  3 y  5z  0

 x  2 y  3z  0

  y  2 z  0 and then to

y  2z  0

x


z 0
y  2z  0
which means x=z, y=2z and z ranges over R. Hence every vector from ker has the form
(z,2z,z) = z(1,2,1) i.e. ker=span((1,2,1)) and nullity()=1.
To find im we notice that, by Proposition 1.3 im=span((1,0,0), (0,1,0), (0,0,1)) =
span((1,2,1),(2,3,3),(3,4,5)). If the three spanning vectors were linearly independent then rank
would be equal to 3, but (1,2,1)=2(2,3,3)(3,4,5) hence, by Theorem 4.5, ??? im = span((2,3,3),
(3,4,5)). Since the last two vectors are obviously linearly independent, rank()=2.
Theorem 1.1. Let S={v1,v2, … ,vn} be a basis for V and let f be a function, mapping S into
W, i.e. f:{v1,v2,…,vn}W. Then there exists exactly one linear mapping :VW such that
for each i=1,2, … ,n (vi)=f(vi)
Proof. We must prove two things, that there exists such linear mapping  and that it is unique. To
n
do the first,. let us take any vector v from V. Since S is a basis for V, v   a i v i for some scalars
i 1
n
a1,a2, … an. We define  (v)   a i f (v i ) . By Theorem 4.6<?>, the scalars a1,a2, … an are
i 1
uniquely determined by v, hence  is a function. Is  a linear mapping? Let us take another
n
vector uV. There exist scalars b1,b2, … ,bn such that u   bi v i . Let us take any two scalars
i 1
p,qF and consider
n
n
i 1
i 1
(pu+qv) =  ( p( bi v i )  q( a i v i )) =
n
=  ( ( pbi  qa i )v i ) =
by properties of vector addition and scaling
i 1
n
=
 ( pb
i
i 1
n
=
 ( pb
i
i 1
 qa i ) f (v i ) =
by the definition of 
f (v i )  qa i f (v i )) =
n
n
i 1
i 1
= p  bi f (vi )  q  ai f (vi ) =
by properties of vector addition and scaling
by properties of vector addition and scaling
= p(u)+ q(v) by the definition of . Hence, by Proposition 1.1,  is a linear mapping. To
prove the uniqueness of  let us suppose that there is another linear mapping , such that
n
(vi)=f(vi) for i=1,2, … ,n. Consider arbitrarily chosen vector v   a i v i . By linearity of ,
i 1
(v)=
n
n
 a  (v )   a
i 1
i
i
i 1
i
f (vi ) = (v).
Theorem 1.1 says that every linear mapping from V into W is uniquely determined by its
values on vectors from a basis of V, or, equivalently, that every function defined on a basis is
uniquely extendable to a linear mapping.
Theorem 1.2. rank()+nullity()=dimV
Proof. Let {v1,v2, … ,vn} be a basis for ker. Then there exist vectors w1,w2, … ,wk such that
{v1,v2, … ,vn,w1,w2, … ,wk} is a basis for V. It is enough to show that {(w1), (w1), … , (wk)}
is a basis for im. By Proposition 1.3 it is enough to show that {(w1), (w1), … , (wk)} is
k
linearly independent. Let
k
 a  ( w ) = . By linearity of  we have  =  a  ( w ) =
i
i 1
k
 ( ai wi ) , i.e.
i 1
i
i 1
i
k
 ai wi  ker  . Hence, for some scalars b1,b2, … ,bn we have
i 1
k
This implies that
n
 a w   (b )v
i 1
i
i
j 1
j
j
i
k
n
i 1
j 1
 ai wi   b j v j .
  , hence all coefficients ai and bj are equal to 0, in
particular all ai-s are zeroes.
Theorem 1.3. A linear mapping :VW is one-to-one iff  preserves linear independence, i.e.
for every linearly independent subset S of V, (S) is a linearly independent subset of W.
n
Proof. () Suppose S={v1,v2, … ,vn} is a linearly independent subset of V and
 a  (v ) =W.
i 1
We must prove that a1= … =an=0. Since  is a linear mapping we have
= . Since (V)=W and  is a 1-1 function we have
n
a v
i 1
of S, we have a1= … =an=0.
i
i
i
i
n
n
i 1
i 1
 a i  (v i ) =  (  a i v i )
 V and, by linear independence
n
() Suppose S={v1,v2, … ,vn} is a basis of V and for some v and u from V, v   a i v i and
i 1
n
u   bi v i , and (v)=(u). Then W = (v)(u) = (vu) =
i 1
n
 (a
i 1
i
 bi ) (v i ) . Since 
preserves linear independence we obtain that ai=bi for i=1,2, …,n and that means that u=v.
Theorem 1.4. A linear mapping :VW is one-to-one iff ker={}
Proof. () From Theorem 1.3 and Proposition 1.3 it follows that for every basis {v1,v2, … ,vn}
of V {(v1),(v2), … ,(vn)} is a basis for im, hence dim(ker)=ndim(im)=0 and that means
that ker={}.
() If ker{} then there exists v0 such that (v)=0. Since (0)=0,  is not one-to-one.