Linear Functions: Review and Summary
1
Basic Definitions
A function L : Rn → Rm is linear if it preserves vector addition and scalar
product, that is,
L(~x + ~y ) = L(~x) + L(~y ) for all ~x, ~y ∈ Rn and
L(t~x) = tL(~x) for all ~x ∈ Rn , t ∈ R.
Taking t = 0 in the second property, we see that L must take ~0 ∈ Rn to
~0 ∈ Rm . The matrix of L is the m × n matrix whose columns are the images
under L of the standard basis vectors
 
 
 
1
0
0
1
0
0
 
 
 
0
0
0
 
 
 
ê1 =  ..  , ê2 =  ..  , . . . , ên =  ..  ∈ Rn .
.
.
.
 
 
 
0
0
0
0
0
1
Matrix multiplication is defined so that if L : Rn → Rm has matrix [L] and
T : Rm → Rk has matrix [T ], then the matrix of the composition T ◦ L :
Rn → Rk is the matrix product [T ][L].
The range of L is the the set L(Rn ) = {L(~x)| ~x ∈ Rn }: it is a subspace of
m
R . The kernel of L is the set ker L = {~x ∈ Rn | L(~x) = ~0}: it is a subspace
of Rn . The rank of L is the the dimension of the range of L, and the nullity
of L is the dimension of the kernel of L. The rank-nullity theorem says that
rank L + nullity L = n.
˜ obtained by
The rank of L is the number of leading 1’s in the matrix [L]
row-reducing the matrix [L] of L. A basis for the range of L can be obtained
1
by choosing rank L linearly independent columns of [L]. Since
˜ x = ~0},
ker L = {~x ∈ Rn | [L]~x = ~0} = {~x ∈ Rn | [L]~
˜ x = ~0.
nullity L=dim ker L is the number of free variables in the solution of [L]~
2
Inverses
Any function f : A → B is one-to-one if f (x) 6= f (y) whenever x 6= y in A,
and onto if for every z ∈ B there is some x ∈ A with f (x) = z. The function
f has an inverse (that is, a function f −1 : B → A such that f ◦ f −1 is the
identity function on B and f −1 ◦ f is the identity function on A) if and only
if f is one-to-one and onto. A linear function L : Rn → Rm is one-to-one
exactly when nullity L = 0, and is onto if and only if rank L = m. Thus,
L has an inverse only if nullity L = 0 and rank L = m: by the rank-nullity
theorem, this can only happen if n = m = rank L. In that case, the inverse
[L]−1 of the matrix [L] of L is the matrix of the linear function L−1 . Your
calculator will compute matrix inverses: a useful fact for hand calculation is
that a 2 × 2 matrix
a b
c d
has an inverse if and only if ad − bc 6= 0, in which case the inverse matrix is
1
d −b
.
ad − bc −c a
3
Projections
If W ⊂ Rn is a subspace of Rn , the function PW that takes x ∈ Rn to its
orthogonal projection onto W is a linear function from Rn to itself. The function PW has range W and kernel W ⊥ = {~x ∈ Rn | ~x · ~z = 0 for all ~z ∈ W }, so
by the rank-nullity theorem dim W + dim W ⊥ = n. If {~y1 , . . . , ~yk } is a (not
necessarily orthogonal) basis for W , then we can form an n × k matrix Y
whose columns are the vectors ~y1 , . . . , ~yn , and if PW ~x = a1 ~y1 +· · ·+ak ~yk = Y ~a,
(where ~a is the column vector containing the ai ), then ~a = (Y T Y )−1 Y T ~x and
PW ~x = Y ~a = Y (Y T Y )−1 Y T ~x, where T means transpose. Note that
2
PW
= Y (Y T Y )−1 Y T Y (Y T Y )−1 Y T = Y (Y T Y )−1 Y T = PW .
2
4
Exercises
1. Suppose f : Rn → Rm is linear.
a. If V is a subspace of Rn , prove that f (V ) = {f (~x)| ~x ∈ Rn } is a
subspace of Rm .
b. For ~y ∈ f (Rn ), prove that {~x ∈ Rn | f (~x) = ~y } is an affine set.
2. Let G : R3 → R3 be the linear function with matrix


1 2 3
4 5 6 .
7 8 9
a. Find the rank and nullity of G.
b. Give bases for the subspaces ker G and G(R3 ) of R3 .
3. Let W be the subspace of R4 with basis
   
1
−1 


   

−1 ,  2  .
 0  −1





−1
1


1
3

Find PW 
−1, and also work out the matrix of PW .
0
3