AM 10 Prof.

Daniele Venturi

Lecture 7: Linear transformations

Let V and W be two vector spaces over a field K. We say that a transformation

F : V 7→ W (1)

is linear if
1. F (u + v) = F (u) + F (v) ∀u, v ∈ V ,
2. F (cu) = cF (u) ∀u ∈ V, ∀c ∈ K.
Conditions 1. and 2. imply that

F (au + bv) = aF (u) + bF (v) ∀u, v ∈ V, ∀a, b ∈ K. (2)

Let us discuss a few examples of linear and nonlinear transformations.

• Example 1: Let V = W = R, K = R. The transformation

F :R→R
x → sin(x)

is nonlinear. In fact, sin(x + y) 6= sin(x) + sin(y) for arbitrary x and y in R.

• Example 2: Let V = C (1) (R) (vector space of real-valued continuously differentiable functions),
W = C (0) (R) (vector space of real-valued continuous functions), K = R. The transformation

F : C 1 (R) → C 0 (R)
df (x)
f (x) →
dx
is linear. In fact, we have
df df (x) dg(x)
(af (x) + bg(x)) = a +b ∀f, g ∈ C (1) (R), ∀a, b ∈ R. (3)
dx dx dx

• Example 3: The transformation

F : R3 → R2
 
x1  
x2  → x1 − x2
2x1 + x2 − x3
x3

is linear. In fact, we have

        
x1 y1   x1 y1
a(x1 − x2 ) + b(y1 − y2 )
F a x2 + b y 2
     = = aF  x2  +bF  y2  .
a(2x1 + x2 − x3 ) + b(2y1 + y2 − y3 )
x3 y3 x3 y3

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• Example 4: The transformation

F : R3 → R2
 
x1  
x2  → x 1 + x 2 + 1
(4)
x3 + x1
x3
is not linear. In fact,
       
x1 y1 x1 y1
F x2  + y2  6= F x2  + F y2  .
x3 y3 x3 y3
Transformations of the form (4) is called affine transformations. Affine transformations are
obtained by adding a constant vector to a linear transformation. For the transformation (4)
we have    
x1   x1  
1 1 0  x2  + 1 .
F x2  → (5)
1 0 0 0
x3 x3 |{z}
| {z } constant
linear transformation

• Example 5: The transformation1

Tr : Mn×n (R) → R
Xn
A→ aii (trace of the matrix A) (6)
k=1

is linear. In fact,
Tr(aA + bB) = aTr(A) + bTr(B). (7)

Hereafter we show that the composition of two linear transformation is a linear transformation.

Theorem 1. Let U , V , and W be vector spaces. Consider the linear transformations F : U → V

and G : V → W . Then G(F (u)) : U → W is a linear transformation.

Proof. If F and G are linear transformations then

G(F (au + bv)) = G(aF (u) + bF (v)) = aG(F (u)) + bG(F (v)). (8)
Hence, the composition of F and G is a linear transformation.

Injective, surjective and invertible transformations. Let V and W be two sets (not neces-
sarily vector spaces). Consider the following (linear or nonlinear) transformation
F :U →V (9)
1
The trace of a square matrix is defined to be the sum of all diagonal entries of A.

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1. We say that F is injective or one-to-one if:

for all u, v ∈ V F (u) = F (v) ⇒ u=v (10)

2. We say that F is surjective or onto if

for all w ∈ W there exists u ∈ V such that F (u) = w (11)

Note that there may be more than one element in V that is mapped onto w. In the figure
above, two elements u and v are mapped onto the same element w.
3. We say that F is invertible2 if is is one-to-one and onto (injective and surjective). The following
diagram sketches an invertible transformation between the sets V and W .

2
Invertible transformations are often called bijections or bijective transformations.

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Example 6: The nonlinear transformation

F :R→R
x → sin(x)

is not injective nor surjective on the real line.

In fact, there are multiple points on the x axis with the same value of sin(x). For example,

sin(1) = sin(1 + kπ) k ∈ Z. (12)

Hence the function is not injective.The function sin(x) is also not surjective in R, as there is no
x ∈ R such that sin(x) = 2. However, if we restrict the domain and range of F as follows
h π πi
F : − , → [−1, 1]
2 2
x → sin(x)

then F is invertible, since it is injective and surjective. The inverse function is denoted by sin−1 (x)
or arcsin(x)

Example 7: The linear transformation

F : R2 → R2
      
x1 1 2 x1 x1 + 2x2
→ = ,
x2 −1 1 x2 −x1 + x2
| {z } | {z } | {z }
x A x

is one-to-one and onto. In fact it is easy to show Ax = Ay implies x = y (inejctivity), and that for
each y ∈ R there exits x ∈ R2 such that Ax = y. Therefore the transformation F is invertible. The
inverse transformation is defined by the inverse matrix A−1

F −1 : R2 → R2
      
x1 1 1 −2 x1 1 x1 − 2x2
→ = .
x2 3 1 1 x2 3 x1 + x2
| {z } | {z } | {z }
x A−1 x

By applying F −1 (x) to F (x) = Ax we obtain

F −1 (F (x)) = A−1 (Ax) = x. (13)

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Definition. Let V , W be vector spaces, F : V → W a linear transformation. If F is invertible

then we say that F is an isomorphism between V and W . If there exists an isomorphism between
the vector spaces V and W (i.e., an invertible linear trasformation) then we say that V and W are
isomorphic.

Theorem 2. Let V be a vector space of dimension n over a field K. Then V is isomorphic to K n .

Proof. Let v1 , . . . , vn ∈ V be a basis of F . Any vector v ∈ V can be represented uniquely relative

to the basis as
v = x1 v1 + · · · + xn vn xi ∈ K. (14)
The transformation

F : V → Kn (15)
 
x1
 .. 
v→ .  (16)
xn

is linear, one-to-one and onto. These properties follow immediately from the definition of basis
(surjectivity), and from the fact that the coordinates of v ∈ V relative to a basis are unique
(injectivity). Hence, (16) defines a bijection between V and K n . This means that V is isomorphic
to K n .

Example 8: The space of polynomials of degree at most 4 with real coefficients, i.e., P4 (R), is
isomorphic to R5 . In fact, if we set up a basis for P5 (R), i.e., a set of 5 linearly independent
polynomials of degree at most 4, e.g.,

p4 (x) = x4 − 3x, p3 (x) = x3 , p2 (x) = x3 + x2 + 1, p1 (x) = x − x3 , p0 (x) = x2 + 1, (17)

then we see that each polynomial in p ∈ P4 (R) is uniquely identified by 5 real coefficients x0 , . . . , x4 :

p(x) = x4 p4 (x) + x3 p3 (x) + x2 p2 (x) + x1 p1 (x) + x0 p0 (x). (18)

Hence, there exists a bijection between R5 and the space of polynomials P4 (R). In other words,
P4 (R) and R5 are isomorphic.

Example 9: The vector space of 3 × 3 symmetric matrices with real coefficient is isomorphic to
R6 .

Since the inverse of an isomorphism is an isomorphism we have that all vector spaces of dimension
n over some field K are isomorphic to one another. For example, the vector space of polynomials
of degree at most 3 is isomorphic to the vector space of 2 × 2 matrices with real coefficients.

Theorem 3. The set of all linear mappings between two vector spaces V and W is a vector space.
Such a space is denoted by L(V, W ).

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Nullspace and range of a linear transformation. Let V , W be vector spaces. Consider the
linear transformation
F : V → W. (19)
• The nullspace (or kernel3 ) of F is the set vectors in V that are mapped into 0W (zero vector
of W ), i.e.,

N (F ) = {v ∈ V such that F (v) = 0W } (nullspace of F ). (20)

Clearly, since F is linear we have that the element 0V is always mapped onto 0W . Therefore,
0V is always in the nullspace of F .
• The range of F is the set of vectors w in W such that w is the image of some v ∈ V under F ,
i.e., there exists v ∈ V such that F (v) = w.

R(F ) = {F (v) ∈ W such that v ∈ V } (21)

Let us determine the nullspace and the range of simple linear transformations.

Example 10: Consider the following linear transformation

F : R3 → R2
   
x1     x1
x2  → x1 + x2 + x3 = 1 1 1 x2  (22)
x3 0 0 1
x3 | {z } x3
A

3
The nullspace/kernel of a linear transformation F is often denoted as ker(F ).

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The nullspace of F is the set of vectors in R3 that mapped onto the zero vector of R2 . Hence, the
nullspace of F is defined by the following homogeneous linear system of equations
( (
x1 + x2 + x3 = 0 x1 = −x2
⇒ (23)
x3 = 0 x3 = 0

Note that the nullspace of F is a vector subspace of R3 (line passing through the origin). The range
of F is the image of R3 under F . Such range coincides with column space of the matrix A, i.e., the
span of the columns of A. In fact,
       
x1 + x2 + x3 1 1 1
= x1 + x2 + x3 . (24)
x3 0 0 1

Hence,          
1 1 1 1 1
R(F ) = span , , = span , = R2 (25)
0 0 1 0 1

Theorem 4. Let V , W be vector spaces, F : V → W linear. Then N (F ) is a vector subspace of

V and R(F ) is a vector subspace of W .

Proof. Let u, v ∈ N (F ). Clearly, u + v is in N (F ). In fact, since F is linear we have F (u + v) =

F (u) + F (v) = 0W . Thus, u + v is in N (F ). Moreover, 0V ∈ N (F ) and cu ∈ N (F ) for all u ∈ N (F )
and all c ∈ K. This implies that N (F ) is a vector subspace of V . To prove that R(F ) is a vector
subspace of W , let w, s ∈ R(F ). This means that there exist u, v ∈ V such that F (u) = w and
F (v) = s. Obviously, (w + s) ∈ R(F ). In fact, by using the linearity of F we have F (u + v) = w + s,
and therefore w + s ∈ R(F ). Also, 0W is R(F ) and cu ∈ R(F ) for all u ∈ R(F ). Thus, R(F ) is a
vector subspace of W .

The nullspace and the range of linear transformation characterize the injectivity and surjectivity of
the transformation. In particular we have the following two theorems.

Theorem 5. Let V , W be vector spaces, F : V → W a linear transformation. Then F is injective

(one-to-one) if and only if N (F ) = {0V }, i.e., the if nullspace of F reduces the singleton {0V }.

Proof. To prove the theorem we need to prove two statements:

1. F is injective ⇒ N (F ) = {0V },

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2. F is injective ⇐ N (F ) = {0V }.
⇒ Suppose that F is one-to-one. We want to show that this implies N (F ) = {0V }. To this end,
let v ∈ N (F ), i.e., F (v) = 0W . Clearly v = 0V is mapped onto 0W , i.e., 0V ∈ N (F ). The
assumption that F is one-to-one rules out the existence of any other element in V mapped
onto 0W . In other workds, 0V is the only element of V mapped into 0W . Hence, if F is
one-to-one then N (F ) = {0V }.
⇐ Conversely, let us assume that N (F ) = {0V }. We want to show that this implies that F is
one-to-one. To this end, suppose there are two elements u, v ∈ V such that F (u) = F (v). By
using the linearity of F we have F (u − v) = 0W , i.e., (u − v) ∈ N (F ). Since, by assumption,
the only element in the nullspace of F is 0V we have that u − v = 0V , i.e., u = v. In other
words, N (F ) = {0V } implies that F is one-to-one.

Theorem 6. Let V , W be vector spaces, F : V → W linear. Then F is surjective (onto) if and

only if dim(R(F )) = dim(W ).

Proof. As before, to prove the theorem we need to prove two statements:

1. F is surjective ⇒ dim(R(F )) = dim(W ),
2. F is surjective ⇐ dim(R(F )) = dim(W ).
Let F be surjective (onto), i.e., ∀w ∈ W there exists (at least one) v ∈ V such that F (v) =
w. This means that R(F ) = W and therefore dim(R(F )) = dim(W ). Conversely, suppose that
dim(R(F )) = dim(W ). Then R(F ) is a vector subspace of W with the same dimension. This
implies that R(F ) = W , i.e., F is surjective (onto).

Next we present a very important theorem for linear transformations between vector spaces.

Theorem 7. Let F : V → W be any linear transformation between two vector spaces V and W .
Then
dim(V ) = dim(N (F )) + dim(R(F )). (26)

Proof. If R(F ) = 0W the statement is trivial since the entire V is mapped to the 0W . This
implies N (F ) = V , and of course dim(N (F )) = dim(V ). Consider now dim(R(F )) = s > 0
and let {w1 , . . . , ws } be a basis of R(F ). Then there exist s elements v1 , . . . , vs ∈ V such that
F (v1 ) = w1 , . . . , F (vs ) = ws . Suppose dim(N (F )) = q and let {u1 , . . . , uq } be a basis for N (F ).
We would like to show that {u1 , . . . , uq , v1 , . . . , vs } is a basis of V 4 . To this end, pick an arbitrary
v ∈ V . Then, there exists x1 , . . . , xs ∈ K such that F (v) = x1 w1 + . . . + xs ws (since w1 , . . . , ws is a
basis for R(F )). Recalling that F (v1 ) = w1 , ..., F (vs ) = ws
F (v) = x1 F (v1 ) + . . . + xs F (vs )
= F (x1 v1 + . . . + xs vs ).
4
Note that if {u1 , . . . , uq , v1 , . . . , vs } is a basis of V then dim(V ) = q +s, where q = dim(N (F ) and s = dim(R(F )).

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By using the linearity of F we obtain

F (v − x1 v1 − · · · − xs vs ) = 0W ⇒ (v − x1 v1 − · · · − xs vs ) ∈ N (F ).

At this point we represent (v − x1 v1 − · · · − xs vs ) relative to the basis of N (F )

v − x1 v1 − . . . − xs vs = y1 u1 + . . . + yq uq ,

i.e.,
v = x1 v1 + . . . + xs vs + y1 u1 + . . . + yq uq .
This shows that V = span{v1 , . . . , vs , u1 , . . . , uq }, i.e., that V is generated by {v1 , . . . , vs , u1 , . . . , uq }.
To prove the theorem it remains to prove that the the vectors {v1 , . . . , vs , u1 , . . . , uq } are linearly
independent. In this way we can claim that n = s + q, i.e., dim(V ) = dim(N (F )) + dim(R(F )).
To this end, consider the linear combination

x1 v1 + . . . + xs vs + y1 u1 + . . . + yq uq = 0V . (27)

By applying F and recalling that F (ui ) = 0W (ui ∈ N (F )) we obtain

x1 w1 + . . . + xs ws = 0W ⇒ x1 , . . . , xs = 0. (28)

In fact {w1 , . . . , ws } is a basis for R(F ) and therefore wi are linearly independent. Substituting this
result back into (27) yields

x1 v1 + . . . + xs vs = 0V ⇒ x1 , . . . , x s = 0 (29)

since {v1 , . . . , vs } is a basis for N (F ). Equations (28), (29) and (27) allow us to conclude that
{v1 , . . . , vs , u1 , . . . , uq } are linearly independent. Moreover the vectors {v1 , . . . , vs , u1 , . . . , uq } gen-
erate V , and therefore they are a basis for V . This implies that

dim(V ) = s + q = dim(N (F )) + dim(R(F )). (30)

Matrix rank theorem. Consider the linear transformation

F : Rn → Rm
    
x1 a11 · · · a1n x1
 ..   .. ... .
..   ... 
 . → . (31)
 

xn am1 · · · amn xn
| {z } | {z } | {z }
x A x

We know that range of F coincides with the column space of A. Also the dimension of the column
space is the rank of the matrix A. Therefore from equation (26) it follows that

n = dim(N (A)) + rank(A). (32)

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Matrix associated with a linear transformation Let V and W be finite-dimensional vector

spaces, and let
F :V →W (33)
an arbitrary linear transformation. In this section we show how to represent F in terms of a matrix.
To this end, suppose that

BV = {v1 , . . . , vn } → basis of V , dim(V ) = n,

BW = {w1 , . . . , wm } → basis of W , dim(W ) = m.

The transformation F is uniquely determined by the image of the basis BV under F , i.e.,

{v1 , . . . , vn } → {F (v1 ), . . . , F (vn )}. (34)

Clearly, for all i = 1, . . . , n we have that F (vi ) ∈ R(F ) ⊆ W . Therefore, each F (vi ) can be
represented in terms of the basis BW as

F (v1 ) = a11 w1 + · · · + am1 wm

..
. . (35)

F (vn ) = a1n w1 + · · · + amn wm


Note that aij is the i-th component of F (vj ) relative to the basis {w1 , . . . , wm }. The matrix asso-
ciated with the linear transformation F depends bases BV and BW and it is defined as
 
a11 · · · a1n
ABBW (F ) =  ... .. ..  . (36)

V
. . 
am1 · · · amn

Example 11: Let V and W be vector spaces of dimension dim(V ) = 2 and dim(W ) = 3, respectively.
We consider the following bases in V and W :

BV = {v1 , v2 }, BW = {w1 , w2 , w3 }. (37)

Relative to such bases, suppose that F is defined as

(
F (v1 ) = w1 − 2w2 − w3
. (38)
F (v2 ) = w1 + w2 + w3

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Then the matrix representing F is

 
1 1
ABBW
V
(F ) = −2 1 . (39)
−1 1

If v = x1 v1 + x2 v2 is an arbitrary vector in V then

F (v) =x1 F (v1 ) + x2 F (v2 )

=x1 (w1 − 2w2 − w3 ) + x2 (w1 + w2 + w3 )
= (x1 + x2 ) w1 + (x2 − 2x1 ) w2 + (x2 − x1 ) w3 (40)
| {z } | {z } | {z }
y1 y2 y3

Note that the coordinates of F (v) relative to the basis BW , i.e., {y1 , y2 , y3 } are given by the standard
matrix-vector product    
y1 1 1  
y2  = −2 1 x1 (41)
x2
y3 −1 1

As we shall see hereafter this is a general result. In fact, consider an arbitrary element v ∈ V , and
represent it in terms of the basis BV

v = x1 v1 + · · · + xn vn . (42)

By applying F and taking (35) into account we obtain

F (v) =x1 F (v1 ) + · · · + xn F (vn )

=x1 (a11 w1 + · · · + am1 wm ) + · · · + xn (a1n w1 + · · · + amn wm )
= (a11 x1 + · · · + a1n xn ) w1 + · · · + (am1 x1 + · · · + amn xn ) wn . (43)

At this point we define the following two column vectors

   
x1 y1
 ..   .. 
[v]BV =  .  , [F (v)]BW =  .  (44)
xn ym

representing the coordinates of v and F (v) relative to BV and BW , respectively. With this notation,
we see from (43) and (36) that
[F (v)]BW = ABBWV
(F )[v]BV . (45)
Therefore, the coordinates of F (v) relative to BW are obtained by taking the matrix-vector product
between the matrix ABBWV
(F ) and the coordinates of v relative to BV .

Change of basis transformation Consider the following two bases in the vector space V

B1 = {u1 , . . . , un }
B2 = {v1 , . . . , vn }

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Obviously, we can express any element in B1 as a linear combination of elements in B2 and vice
versa. For example, 
F (v1 ) = α11 u1 + · · · + αn1 um

..
. . (46)

F (vn ) = α1n u1 + · · · + αnn un


The matrix associated with the linear transformation “change of basis from B1 to B2 ” is
 
α11 · · · α1n
MBB12 =  ... . . . ..  . (47)

. 
αn1 · · · αnn

Such a matrix is invertible and it allows us to transform the coordinates of any vector v ∈ V from
those relative to B1 to those relative to B2 , i.e.,

[v]B2 = MBB12 [v]B1 . (48)

Moreover, we have

−1
−1
[v]B1 = MBB21 [v]B2 = MBB12 [v]B2 which implies MBB21 = MBB12 . (49)

The change of basis transformation can be also used to represent a linear transformation F : V → W
relative to different bases in V and W . To show this, let

B1 , B2 → Bases of V , dim(V ) = n,
B3 , B4 → Bases of W , dim(V ) = m.

We have,

[F (v)]B4 = MBB34 [F (v)]B3 = MBB34 ABB32 [v]B2 = MBB34 ABB32 MBB12 [v]B1 , (50)
| {z }
B
AB4
1

i.e.,
ABB41 = MBB34 ABB32 MBB12 . (51)
The matrix ABB41 represents the linear transformation F relative to the bases B1 (basis of V ) and B4
(basis of W ). Similarly, ABB32 represents the linear transformation F relative to the bases B2 (basis
of V ) and B3 (basis of W ).

Example 12: (Change of basis in R2 ) Consider the following bases of R2

   
1 0
B1 = {e1 , e2 }, e1 = , e2 = (canonical basis of R2 ),
0 1
   
1 1
B2 = {v1 , v2 }, v1 = , v2 = .
1 2

Define the change of basis transformation

F : R2 → R2 (52)

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as        
1 1 0 1
F = , F = . (53)
0 1 1 2
Clearly, (
v1 = e 1 + e 2
. (54)
v2 = e1 + 2e2
The following figure sketches {e1 , e2 } and {v1 , v2 } as vectors in the Cartesian plane.

Any vector v ∈ R2 can be expressed relatively to B1 or B2 :

v =x1 v1 + x2 v2
=x1 (e1 + e2 ) + x2 (e1 + 2e2 )
=(x1 + x2 )e1 + (x1 + 2x2 )e2 . (55)

Denote by    
y x
[v]B1 = 1 , [v]B2 = 1 . (56)
y2 x2
the coordinates of v relative to B1 and B2 , respectively. Then equation (55) implies that
 
B1 B1 1 1
[v]B1 = MB2 [v]B2 , where MB2 = . (57)
1 2

MBB21 is the matrix associated with the change of basis transformation B2 → B1 . Clearly, MBB21 is
invertible with inverse  
B2 B1 −1

2 −1
MB1 = MB2 = (58)
−1 1
MBB12 is the matrix associated with the change of basis transformation B1 → B2 . Let us see if this is
indeed true. To this end, we consider the vector v = e1 and compute the coordinates of this vector
relative to B2 . We have
      
1 2 −1 1 2
[v]B1 = ⇒ [v]B2 = = (59)
0 −1 1 0 −1
| {z }
B
MB 2
1

The following figure illustrates that the components of v = e1 relative to B2 are indeed x1 = 2 and
x2 = −1.

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Example 13: (Rotations in R2 ) Consider the linear transformation F : R2 → R2 defined as follows

(counterclockwise rotation of an angle θ)
(
F (e1 ) = cos(θ)e1 + sin(θ)e2
, (60)
F (e2 ) = − sin(θ)e1 + sin(θ)e2
where    
1 0
e1 = , e2 = (61)
0 1
is the canonical basis of R2 .

The matrix associated with the transformation F relative to the basis BV = {e1 , e2 } is
 
BV cos(θ) − sin(θ)
ABV (F ) = (2D rotation matrix). (62)
sin(θ) cos(θ)
Any vector with components [v]BV is rotated to a vector F (v) with components
[F (v)]BV = ABBVV [v]BV . (63)
   
2 2
For example, the vector v = has components relative to the canonical basis BV , and it is
1 1
transformed to a vector F (v) with components
    
cos(θ) − sin(θ) 2 2 cos(θ) − sin(θ)
[F (v)]BV = = . (64)
sin(θ) cos(θ) 1 2 sin(θ) + cos(θ)

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In particular, if θ = π/2 (90 degrees counterclockwise rotation) then

 
−1
[F (v)]BV = . (65)
2

The inverse transformation (inverse rotation) is obtained by replacing θ with −θ in (62), i.e.,
 
 BV −1 cos(θ) sin(θ)
ABV (F ) = . (66)
− sin(θ) cos(θ)
It is straightforward to verify that for all θ ∈ [0, 2π] we have
 BV −1 BV
ABV (F ) ABV (F ) = I2 . (67)
In fact,
cos θ + sin2 θ
    2   
cos θ − sin θ cos θ sin θ 0 1 0
= = .
sin θ cos θ − sin θ cos θ 0 cos2 θ + sin2 θ 0 1

The rotation matrix is an orthogonal matrix5 .

Example 14: (Rotations in R3 ) We can define rotations along each of the three axes of a 3D Cartesian
coordinate system, i.e.,

 
cos θ3 − sin θ3 0
R3 =  sin θ3 cos θ3 0
0 0 1

5
In general, we say that A ∈ Mn×n (R) is orthogonal if
AT = A−1 . (68)
This is equivalent to the statement that orthogonal matrices satisfy
AAT = In . (69)

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AM 10 Prof. Daniele Venturi

 
1 0 0
R1 = 0 cos θ1 − sin θ1 
0 sin θ1 cos θ1

 
cos θ2 0 − sin θ2
R2 =  0 1 0 
sin θ2 0 cos θ2

Note that the composition of two rotation in R3 do not commute. For example, R1 R3 6= R3 R1 .

Example 15: (Orthogonal projection) Consider

F : R3 → R3

and the canonical bases of R3

     
 1 0 0 
B3 = {e1 , e2 , e3 } = 0 , 1 , 0
   
0 0 1
 

We define F by mapping the basis B3 as follows

F (e1 ) = e1 , F (e2 ) = e2 , F (e3 ) = 0R3 .

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AM 10 Prof. Daniele Venturi

The associated matrix defines an orthogonal projection onto the (x1 , x2 )

 
1 0 0
P = 0 1 0 . (70)
0 0 0

Note that P 2 = P . The orthogonal projection transformation basically project any vector v ∈ R3
onto the plane spanned by e1 and e2 . If we are interested in a projection onto different plane, we
can use e.g., the 3D rotation matrices Ri and rotate the plane before applying the projection. Note
that with just R1 and R3 we can orient the (x1 , x2 ) plane in all possible directions. Hence,

P (θ1 , θ3 ) = R1 (θ1 )R3 (θ3 )P R3T (θ3 )R1T (θ1 ). (71)

To explain this formula suppose for simplicity that we just rotate the plane (x1 , x2 ) counterclockwise
of an angle θ1 around the x1 axis.

The projection of any object onto such plane is obtained by rotating the object clockwise of an
angle θ1 around x1 (matrix R1T (θ3 ) projecting onto the (x1 , x2 ) plane and then rotating the result
back (matrix R1 (θ3 )). Clearly, (71) satisfies the condition for orthogonal projections,

P 2 (θ1 , θ3 ) = P (θ1 , θ3 ). (72)

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AM 10 Prof. Daniele Venturi

 
v1
Example 16: (Oblique projection) Let v = v2  be a vector of R3 representing the direction of a
v3  
x1
light beam. A light beam passing through an arbitrary point x = x2  has the form

x3
     
y1 v1 x1
y2  = c v2  + x2  where c ∈ R (73)
y3 v3 x3

If we set y3 = 0 we obtain c = −x3 /v3 . With such a value for c, the light beam passing through the
point x intersects the horizontal plane. The linear transformation defined by
 v1
 y1 = − x3 + x1


 v3
v2

y2 = − x3 + x2 (74)


 v3

y3 = 0


defines an oblique projection onto the horizontal plane. The matrix associated with such oblique
projection transformation (relative to the canonical basis of R3 ) is
 
1 0 −v1 /v3
P = 0 1 −v2 /v3  (75)
0 0 0

The oblique projection can be used to compute the shadow of any object in 3D. The following figure
shows the shadow projected by a horse for various angles of the light beam.

Note that for v1 = v2 = 0 the oblique projection reduces to the projection we studied in the previous
example.

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AM 10 Prof. Daniele Venturi

Example 17: Let P4 = span{1, x, x2 , x3 , x4 } be the space of polynomials of degree at most 4. Define
the linear transformation

F : P4 → P3
dp(x)
p(x) → .
dx

The canonical bases of P4 and P3 are

B4 = {1, x, x2 , x3 , x4 },
B3 = {1, x, x2 , x3 }.

We define the derivative transformation by mapping each element of P4 and representing the result
in terms of P3 . This yields

F (1) = 0, F (x) = 1, F (x2 ) = 2x, F (x3 ) = 3x2 , F (x4 ) = 4x3 .

The matrix associated with F (derivative operator) relative to the bases B4 and B3 is
 
0 1 0 0 0
0 0 2 0 0
ABB34 =
0
.
0 0 3 0
0 0 0 0 4
For example, let us compute the derivative of the polynomial

p(x) = 1 − 3x + 6x3 . (76)

The coordinates of p(x) relative to B4 are

 T
[p(x)]B4 = 1 −3 0 6 0 .

This implies that

 
  1  
0 1 0 0 0   −3
  −3  
dp(x) 0 0 2 0 0    0 
= ABB34 [p(x)]B4
 
⇒ =
0 0  =  .
dx B3
0 0 3 0 6 18
0 0 0 0 4 0
0

Therefore we obtained
dp(x)
= −3 + 0x + 18x2 + 0x3 = −3 + 18x2 , (77)
dx
which is indeed the derivative of the polynomial (76).

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