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Chapter 4 Chapter Content
1. Real Vector Spaces
2. Subspaces
3. Linear Independence
4. Basis
5. Dimension
6. Row Space, Column Space, and Nullspace
8. Rank and Nullity
9. Matrix Transformations for Rn to Rm
Definition (Vector Space)
Let V be an arbitrary nonempty set of objects on which two operations are
defined: addition, and multiplication by scalars.
If the following axioms are satisfied by all objects u, v, w in V and all scalars
k and m, then we call V a vector space and we call the objects in V vectors
1. If u and v are objects in V, then u + v is in V.
2. u + v = v + u
3. u + (v + w) = (u + v) + w
4. There is an object 0 in V, called a zero vector for V, such that 0 + u= u + 0 = u
for all u in V.
5. For each u in V, there is an object -u in V, called a negative of u, such that
u + (-u) = (-u) + u = 0.
6. If k is any scalar and u is any object in V, then ku is in V.
7. k (u + v) = ku + kv
8. (k + m) u = ku + mu
9. k (mu) = (km) (u)
10. 1u = u
To Show that a Set with Two Operations is a Vector Space
1.
Identify the set V of objects that will become vectors.
2.
Identify the addition and scalar multiplication operations on V.
3.
Verify Axioms 1(closure under addition) and 6 (closure under
scalar multiplication) ;
that is, adding two vectors in V produces a vector in V, and
multiplying a vector in V by a scalar also produces a vector in V.
4. Confirm that Axioms 2,3,4,5,7,8,9 and 10 hold.
Remarks
• Depending on the application, scalars may be real numbers or complex
numbers.
•Vector spaces in which the scalars are complex numbers are called complex
vector spaces, and those in which the scalars must be real are called real vector
spaces.
• The definition of a vector space specifies neither the nature of the vectors nor
the operations.
•Any kind of object can be a vector, and the operations of addition and scalar
multiplication may not have any relationship or similarity to the standard vector
operations on
.
• The only requirement is that the ten vector space axioms be satisfied.
The Zero Vector Space
Let V consist of a single object, which we denote by 0, and define
0 + 0 = 0 and k 0 = 0 for all scalars k.
It’s easy to check that all the vector space axioms are satisfied.
We called this the zero vector space.
Example ( R n Is a Vector Space)
The set V = R n with the standard operations of addition and scalar multiplication
is a vector space.
(Axioms 1 and 6 follow from the definitions of the
standard operations on R n ; the remaining axioms follow from Theorem 4.1.1.)
The three most important special cases of R n are R (the real numbers), R 2
(the vectors in the plane), and R 3 (the vectors in 3-space).
Example (2×2 Matrices)
Show that the set V of all 2×2 matrices with real entries is a vector space if
vector addition is defined to be matrix addition and vector scalar
multiplication is defined to be matrix scalar multiplication.
 u11 u12 
 v11 v12 
and
v


v

u21 u22 
 21 v22 
Solution: Let u  
(1) we must show that u + v is a 2×2 matrix.
 u11  v11 u12  v12 
uv  

u

v
u

v
 21 21 22 22 
(2) Want to show that u + v = v + u
 u11  v11 u12  v12 
uv  
  vu
u

v
u

v
 21 21 22 22 
(3) Similarly we can show that u + ( v + w ) = ( u + v )+ w.
(4) Define 0 to be
0 0 such that 0  u   u11 u12   u  0  u
0
u


u
0
0

21
22



 u11 u12 

u

(5) Define the negative of u to be
u
 such that

u
 21
22 
0 0
u  u  u  u  

0
0


 ku11 ku12 
ku

(6) If k is any scalar and u is a 2X2 matrix, then
ku
 is 2X2 matrix.
ku
 21
22 
(7)-(9) will be obtained by similar approach.
1u11 1u12   u11 u12 
  u
  u.
1
u
1
u
u
 21
22 
 21 22 
(10) 1u  
Thus, the set V of all 2×2 matrices with real entries is a vector space.
Example: Given the set of all triples of real numbers ( x, y, z ) with the operations
( x, y, z )  ( x ', y ', z ')  ( x  x ', y  y ', z  z ') and
k ( x, y, z )  (kx, y, z )
Determine if it’s a vector space under the given operation.
Solution: We must check all ten properties:
(1) If (x, y, z) and (x’, y’, z’) are triples of real numbers, so is
(x, y, z) + (x’, y’, z’) = (x + x’, y +y’, z + z’).
(2) (x, y, z) + (x’, y’, z’) = (x + x’, y + y’, z + z’)= (x’, y’, z’) + (x, y, z).
(3) (x, y, z) + [(x’, y’, z’) + (x’’, y’’, z’’)] = (x, y, z) + [(x’, y’, z’) + (x’’, y’’, z’’)].
(4) There is an object 0, (0, 0, 0), such that
(0, 0, 0) + (x, y, z) = (x, y, z) + (0, 0, 0)= (x, y, z).
(5) For each positive real x, (-x, -y, -z) acts as the negative:
(x, y, z) + (-x, -y, -z) = (-x, -y, -z) + (x, y, z) =(x, y, z)
(6) If k is a real and (x, y, z) is a triple of real numbers, then k (x, y, z) = (kx,
y, z) is again a triple of real numbers.
(7) k[(x, y, z) + (x’, y’, z’)] = (k(x+x’), y+y’, z+z’) = k(x, y, z) + k(x’, y’, z’)
(8) (k + m) (x, y, z) = ((k + m)x, y, z)

k (x, y, z) + m(x, y, z)
Axiom (8) fails
Thus, the set of all triples of real numbers ( x, y, z ) with the operations is
NOT a vector space under the given operation.
Example.
Let V = R2 and define addition and scalar multiplication operations as follows:
If u = (u1, u2) and v = (v1, v2), then define
u + v = (u1 + v1, u2 + v2)
and if k is any real number, then define
k u = (k u1, 0)
There are values of u for which Axiom 10 fails to hold. For example, if u = (u1,
u2) is such that u2 ≠ 0,then
1u = 1 (u1, u2) = (1 u1, 0) = (u1, 0) ≠ u
Thus, V is not a vector space with the stated operations
Theorem 5.1.1
Let V be a vector space, u be a vector in V, and k a scalar; then:
(a) 0 u = 0
(b) k 0 = 0
(c) (-1) u = -u
(d) If k u = 0 , then k = 0 or u = 0.
4.2 Subspaces
Definition
A subset W of a vector space V is called a subspace of V if W is itself a
vector space under the addition and scalar multiplication defined on V.
Theorem 5.2.1
If W is a set of one or more vectors from a vector space V, then W is a
subspace of V if and only if the following conditions hold:
a) If u and v are vectors in W, then u + v is in W.
b) If k is any scalar and u is any vector in W , then ku is in W.
Remark
Theorem 5.2.1 states that W is a subspace of V if and only if W is a closed under
addition (condition (a)) and closed under scalar multiplication (condition (b)).
Example
All vectors of the form (a, 0, 0) is a subspace of R3.
• The set is closed under vector addition because
(a, 0, 0) + (b, 0, 0) = (a + b, 0, 0)
• It is closed under scalar multiplication because
k(a, 0, 0) = (ka, 0, 0)
Therefore it is a subspace of R3.
Example (Not a Subspace)
Let W be the set of all points (x, y) in R2 such that x ≥ 0 and y ≥ 0. These are the
points in the first quadrant.
The set W is not a subspace of R2 since it is not closed under scalar multiplication.
For example, v = (1, 1) lines in W, but its negative (-1)v = -v = (-1, -1) does not.
Subspaces of Mnn
The set of n×n diagonal matrices forms subspaces of Mnn, since each of these
sets is closed under addition and scalar multiplication.
The set of n×n matrices with integer entries is NOT a subspace of the vector space
Mnn of n×n matrices.
This set is closed under vector addition since the sum of two integers is again an
integer.
However, it is not closed under scalar multiplication since the product ku
where k is real and a is an integer need not be an integer.
Thus, the set is not a subspace.
Linear Combination
Definition in 3.1
A vector w is a linear combination of the vectors v1, v2,…, vr if it can be
expressed in the form
w = k1v1 + k2v2 + · · · + kr vr
where k1, k2, …, kr are scalars.
Example:Vectors in R3 are linear combinations of i, j, and k
Every vector v = (a, b, c) in R3 is expressible as a linear combination of the
standard basis vectors
i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)
Since
v= a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = a i + b j + c k
Example
Consider the vectors u = (1, 2, -1) and v = (6, 4, 2) in R3. Show that w = (9, 2,
7) is a linear combination of u and v and that w′ = (4, -1, 8) is not a linear
combination of u and v.
Solution.
In order for w to be a linear combination of u and v, there must be scalars k1
and k2 such that w = k1u + k2v;
(9, 2, 7) = (k1 + 6k2, 2k1 + 4k2, -k1 + 2k2)
Equating corresponding components gives
k1 + 6k2 = 9
2k1+ 4k2 = 2
-k1 + 2k2 = 7
Solving this system yields k1 = -3, k2 = 2, so
w = -3u + 2v
Similarly, for w‘ to be a linear combination of u and v, there must be scalars k1
and k2 such that w'= k1u + k2v;
(4, -1, 8) = k1(1, 2, -1) + k2(6, 4, 2)
or
(4, -1, 8) = (k1 + 6k2, 2k1 + 4k2, -k1 + 2k2)
Equating corresponding components gives
k1 + 6k2 = 4
2 k1+ 4k2 = -1
- k1 + 2k2 = 8
This system of equation is inconsistent, so no such scalars k1 and k2 exist.
Consequently, w' is not a linear combination of u and v.
Linear Combination and Spanning
Theorem 5.2.3
If v1, v2, …, vr are vectors in a vector space V, then:
(a) The set W of all linear combinations of v1, v2, …, vr is a subspace of V.
(b) W is the smallest subspace of V that contain v1, v2, …, vr in the sense that
every other subspace of V that contain v1, v2, …, vr must contain W.
Example
If v1 and v2 are non-collinear vectors in R3 with their initial points at
the origin, then span{v1, v2}, which consists of all linear combinations
k1v1 + k2v2 is the plane determined by v1 and v2.
Similarly, if v is a nonzero vector in R2 and R3, then span {v}, which
is the set of all scalar multiples kv, is the line determined by v.
Example
Determine whether v1 = (1, 1, 2), v2 = (1, 0, 1), and v3 = (2, 1, 3)
span the vector space R3.
Solution
Is it possible that an arbitrary vector b = (b1, b2, b3) in R3 can be expressed as
a linear combination b = k1v1 + k2v2 + k3v3 ?
b = (b1, b2, b3) = k1(1, 1, 3) + k2(1, 0, 1) + k3(2, 1, 3)
= (k1+k2+2k3, k1+k3, 2k1+k2+3k3)
Or
k1 + k2 + 2k3 = b1
k1 + k3 = b2
2k1 + k2 + 3 k3 = b3
This system is consistent for all values of b1, b2, and b3 if and only if the
coefficient matrix has a nonzero determinant.
However, det(A) = 0, so that v1, v2, and v3, do not span R3.
Solution Space
Solution Space of Homogeneous Systems
If Ax = b is a system of the linear equations, then each vector x that satisfies this
equation is called a solution vector of the system.
Theorem 5.2.2
If Ax = 0 is a homogeneous linear system of m equations in n unknowns, then
the set of solution vectors is a subspace of Rn.
Remark: Theorem 5.2.2 shows that the solution vectors of a homogeneous
linear system form a vector space, which we shall call the solution space of the
system.
Theorem 4.2.5
If S = {v1, v2, …, vr} and S′ = {w1, w2, …, wr} are two sets of vector in a vector
space V, then
span{v1, v2, …, vr} = span{w1, w2, …, wr}
if and only if
each vector in S is a linear combination of these in S′ and each vector in S′ is
a linear combination of these in S.
4. 3 Linearly Independence
Definition
If S = {v1, v2, …, vr} is a nonempty set of vector, then the vector equation
k1v1 + k2v2 + … + krvr= 0
has at least one solution, namely
k1 = 0, k2 = 0, … , kr = 0.
If this the only solution, then S is called a linearly independent set. If there are other
solutions, then S is called a linearly dependent set.
Examples
Given v1 = (2, -1, 0, 3), v2 = (1, 2, 5, -1), and v3 = (7, -1, 5, 8).
Then the set of vectors S = {v1, v2, v3} is linearly dependent, since 3v1 + v2 – v3 = 0.
Example
Let i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1) in R3. Determine it it’s a linear
independent set
Solution: Consider the equation
k1i + k2j + k3k = 0
⇒ k1(1, 0, 0) + k2(0, 1, 0) + k3(0, 0, 1) = (0, 0, 0)
⇒ (k1, k2, k3) = (0, 0, 0)
⇒ The set S = {i, j, k} is linearly independent.
Similarly the vectors
e1 = (1, 0, 0, …,0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, 0, …, 1)
form a linearly independent set in Rn.
Remark:
To check whether a set of vectors is linear independent or not, write down the
linear combination of the vectors and see if their coefficients all equal zero.
Example
Determine whether the vectors v1 = (1, -2, 3), v2 = (5, 6, -1), v3 = (3, 2, 1) form a
linearly dependent set or a linearly independent set.
Solution
Let the vector equation
k1v1 + k2v2 + k3v3 = 0
⇒ k1(1, -2, 3) + k2(5, 6, -1) + k3(3, 2, 1) = (0, 0, 0)
⇒ k1 + 5k2 + 3k3 = 0
-2k1 + 6k2 + 2k3 = 0
3k1 – k2 + k3 = 0
⇒ det(A) = 0
⇒ The system has nontrivial solutions
⇒ v1,v2, and v3 form a linearly dependent set
Theorems
Theorem 4.3.1
A set with two or more vectors is:
(a) Linearly dependent if and only if at least one of the vectors in S is expressible as a
linear combination of the other vectors in S.
(b) Linearly independent if and only if no vector in S is expressible as a linear
combination of the other vectors in S.
Theorem 4.3.2
(a) A finite set of vectors that contains the zero vector is linearly dependent.
(b) A set with exactly one vector is linearly independent if and only if that vector is not
the zero vector.
(c) A set with exactly two vectors is linearly independent if and only if neither vector is
a scalar multiple of the other.
Theorem 4. 3.3
Let S = {v1, v2, …, vr} be a set of vectors in Rn. If r > n, then S is linearly dependent.
Geometric Interpretation of Linear Independence
In R2 and R3, a set of two vectors is linearly independent if and only if the vectors do
not lie on the same line when they are placed with their initial points at the origin.
In R3, a set of three vectors is linearly independent if and only if the vectors do not lie
in the same plane when they are placed with their initial points at the origin.
Section 4.4 Coordinates and Basis
Definition
If V is any vector space and S = {v1, v2, …,vn} is a set of vectors in V, then S is
called a basis for V if the following two conditions hold:
(a) S is linearly independent.
(b) S spans V.
Theorem 5.4.1 (Uniqueness of Basis Representation)
If S = {v1, v2, …,vn} is a basis for a vector space V, then every vector v in V can
be expressed in the form v = c1v1 + c2v2 + … + cnvn in exactly one way.
Coordinates Relative to a Basis
If S = {v1, v2, …, vn} is a basis for a vector space V, and
v = c1v1 + c2v2 + ··· + cnvn
is the expression for a vector v in terms of the basis S,
then the scalars c1, c2, …, cn, are called the coordinates of v relative to the
basis S.
The vector (c1, c2, …, cn) in Rn constructed from these coordinates is called the
coordinate vector of v relative to S; it is denoted by
(v)S = (c1, c2, …, cn)
Remark: Coordinate vectors depend not only on the basis S but also on the
order in which the basis vectors are written. A change in the order of the
basis vectors results in a corresponding change of order for the entries in
the coordinate vector.
Standard Basis for R3
Suppose that i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1), then S = {i, j, k} is a
linearly independent set in R3. This set also spans R3 since any vector v =
(a, b, c) in R3 can be written as
v = (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = ai + bj + ck
Thus, S is a basis for R3; it is called the standard basis for R3.
Looking at the coefficients of i, j, and k, it follows that the coordinates of v
relative to the standard basis are a, b, and c, so
(v)S = (a, b, c)
Comparing this result to v = (a, b, c), we have
v = (v)S
Standard Basis for Rn
If e1 = (1, 0, 0, …, 0), e2 = (0, 1, 0, …, 0), …, en = (0, 0, 0, …, 1), then
S = {e1, e2, …, en} is a linearly independent set in Rn. This set also spans Rn
since any vector v = (v1, v2, …, vn) in Rn can be written as
v = v1e1 + v2e2 + … + vnen
Thus, S is a basis for Rn; it is called the standard basis for Rn.
The coordinates of v = (v1, v2, …, vn) relative to the standard basis are v1 ,v2, …,
vn, thus
(v)S = (v1, v2, …, vn)
As the previous example, we have v = (v)s, so a vector v and its coordinate
vector relative to the standard basis for Rn are the same.
Example
Let v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4). Show that the set S = {v1, v2,
v3} is a basis for R3.
Solution: To show that the set S spans R3, we must show that an arbitrary
vector b = (b1, b2, b3) can be expressed as a linear combination
b = c1v1 + c2v2 + c3v3
of the vectors in S.
Let (b1, b2, b3) = c1(1, 2, 1) + c2(2, 9, 0) + c3(3, 3, 4)
c1 +2c2 +3c3 = b1
2c1+9c2 +3c3 = b2
c1
+4c3 = b3
1 2 3 
Let A be the coefficient matrix  2 9 3 , then det(A) = -1 ≠ 0


So S spans R3.
1 0 4
Example
To show that the set S is linear independent, we must show that the only
solution of c1v1 + c2v2 + c3v3 =0 is a trivial solution.
c1 +2c2 +3c3 = 0
2c1+9c2 +3c3 = 0
c1
+4c3 = 0
Note that det(A) = -1 ≠ 0, so S is linear independent.
So S is a basis for R3.
Example
Let v1 = (1, 2, 1), v2 = (2, 9, 0), and v3 = (3, 3, 4), and S = {v1, v2, v3} be the
basis for R3 in the preceding example.
(a) Find the coordinate vector of v = (5, -1, 9) with respect to S.
(b) Find the vector v in R3 whose coordinate vector with respect to the basis
S is (v)s = (-1, 3, 2).
Solution (a)
We must find scalars c1, c2, c3 such that v = c1v1 + c2v2 + c3v3, or, in
terms of components, (5, -1, 9) = c1(1, 2, 1) + c2(2, 9, 0) + c3(3, 3, 4)
c1 +2c2 +3c3 = 5
2c1+9c2 +3c3 = -1
c1
+4c3 = 9
Solving this, we obtaining c1 = 1, c2 = -1, c3 = 2.
Therefore, (v)s = (1, -1, 2).
Solution
Solution (b)
Using the definition of the coordinate vector (v)s, we obtain
v = (-1)v1 + 3v2 + 2v3 = (11, 31, 7).
Finite-Dimensional
Definition
A nonzero vector space V is called finite-dimensional if it contains a finite set of
vector {v1, v2, …,vn} that forms a basis.
If no such set exists, V is called infinite-dimensional. In addition, we shall
regard the zero vector space to be finite-dimensional.
Example
The vector space Rn is finite-dimensional.
4.5 Dimension
Theorem 4. 5.2
Let V be a finite-dimensional vector space and {v1, v2, …,vn} any basis.
(a) If a set has more than n vector, then it is linearly dependent.
(b) If a set has fewer than n vector, then it does not span V.
Which can be used to prove the following theorem.
Theorem 4.5.1
All bases for a finite-dimensional vector space have the same number of
vectors.
Dimension
Definition
The dimension of a finite-dimensional vector space V, denoted by dim(V), is
defined to be the number of vectors in a basis for V.
In addition, we define the zero vector space to have dimension zero.
Example:
dim(Rn) = n [The standard basis has n vectors]
dim(Mmn) = mn [The standard basis has mn vectors]
Example
Determine a basis for and the dimension of the solution space of the
homogeneous system
2x1 + 2x2 – x3 + x5 = 0
-x1 - x2 + 2x3 – 3x4 + x5 = 0
x1 + x2 – 2x3 – x5 = 0
x3+ x4 + x5 = 0
Solution:
By Gauss-Jordan Elimination method, we have
 2 2 1
 1 1 2

 1 1 2

0 0 1
0 1
3 1
0 1
1 1
0
0 
rref
0

0
1
0

0

0
1
0
0
0
0
1
0
0
0
0
1
0
1
1
0
0
0
0 
0

0
Thus, x1+x2+x5=0, x3+x5=0, x4=0. Solving for the leading variables yields
the general solution of the given system: x1 = -s-t, x2 = s, x3 = -t, x4 = 0, x5 = t
Solution
Therefore, the solution vectors can be written as
 x1    s  t    s   t 
 1  1
x   s   s   0 
1 0
2
  
    
   
 x3    t    0    t   s  0   t  1
  
    
   
 x4   0   0   0 
0 0
 x5   t   0   t 
 0   1 
Which shows that the vectors
 1
 1
1
0
 
 
v1   0  and v2   1
 
 
0
 
0
 0 
 1 
span the solution space.
Since they are also linearly independent, {v1, v2} is a basis, and the solution
space is two-dimensional.
Some Fundamental Theorems
Theorem 4.5.3 (Plus/Minus Theorem)
Let S be a nonempty set of vectors in a vector space V.
(a) If S is a linearly independent set, and if v is a vector in V that is outside of
span(S), then the set S ∪ {v} that results by inserting v into S is still linearly
independent.
(b) If v is a vector in S that is expressible as a linear combination of other
vectors in S, and if S – {v} denotes the set obtained by removing v from S,
then S and S – {v} span the same space; that is, span(S) = span(S – {v})
Theorem 4.5.4
If V is an n-dimensional vector space, and if S is a set in V with exactly
n vectors, then S is a basis for V if either S spans V or S is linearly
independent.
Theorems
Theorem 4.5.5
Let S be a finite set of vectors in a finite-dimensional vector space V.
(a) If S spans V but is not a basis for V, then S can be reduced to a basis for V
by removing appropriate vectors from S.
(b) If S is a linearly independent set that is not already a basis for V, then S can
be enlarged to a basis for V by inserting appropriate vectors into S.
Theorem 4.5.6
If W is a subspace of a finite-dimensional vector space V, then
(a) W is finite-dimensional.
(b) dim(W) ≤ dim(V);
(c) W = V if and only if dim(W) = dim(V).
Section 4.7 Row Space, Column Space, and Nullsapce
Definition. For an mxn matrix
The vectors
 a11
a
 21
 .
A
 .
 .

 am1
a12 ... a1n 
a22 ... a2 n 
.
.
. 

.
.
. 
.
.
. 

am 2 ... amn 
r1   a11 a12 ... a1n 
r2   a21 a22 ... a2 n 
.
.
.
rm   am1 am 2 ... amn 
in Rn formed from the rows of A are called the row vectors of A,
Row Vectors and Column Vectors
And the vectors
 a11 
 a12 
 a1n 
a 
a 
a 
21
22
 
 
 2n 
 . 
 . 
 . 
c1    , c2    ,..., cn   
 . 
 . 
 . 
 . 
 . 
 . 
 
 
 
 am1 
 am 2 
 amn 
In Rn formed from the columns of A are called the column vectors of A.
Nullspace
Theorem
Elementary row operations do not change the nullspace of a matrix.
Example
Find a basis for the nullspace of
 2 2 1 0 1 
 1 1 2 3 1 

A
 1 1 2 0 1


0 0 1 1 1
The nullspace of A is the solution space of the homogeneous system Ax=0.
2x1 + 2x2 – x3 + x5 = 0
-x1 - x2 + 2x3 – 3x4 + x5 = 0
x1 + x2 – 2x3 – x5 = 0
x3+ x4 + x5 = 0.
Nullspace Cont.
Then by the previous example, we know
Form a basis for this space.
 1
 1
1
0
 
 
v1   0  and v2   1
 
 
0
 
0
 0 
 1 
Theorems
Theorem
Elementary row operations do not change the row space of a matrix.
Note: Elementary row operations DO change the column space of a matrix.
However, we have the following theorem
Theorem
If A and B are row equivalent matrices, then
(a)
A given set of column vectors of A is linearly independent if and only if
the corresponding column vectors of B are linearly independent.
(b)
A given set of column vectors of A forms a basis for the column space of
A if and only if the corresponding column vectors of B form a basis for the
column space of B.
Theorems Cont.
Theorem
If a matrix R is in row-echelon form, then the row vectors with the leading 1’s
(the nonzero row vectors) form a basis for the row space of R, and the
column vectors with the leading 1’s of the row vectors form a basis for the
column space of R.
Example
Example
Find bases for the row and column spaces of
 1 3 4 2 5 4 
 2 6 9 1 8 2 

A
 2 6 9 1 9 7 


 1 3 4 2 5 4 
Solution. Since elementary row operations do not change the row space of a
matrix, we can find a basis for the row space of A by finding a basis for the
row space of any row-echelon form of A.
1 3
0 0
R
0 0

0 0
0 14
1 3
0 0
0 0
0 37 
0 4 
1 5 

0 0 
Example
By Theorem, the nonzero row vectors of R form a basis for the row space of R
and hence form a basis for the row space of A.
These basis vectors are
r1  1 3 0 14 0 37 
r2   0 0 1 3 0 4
r3   0 0 0 0 1 5
Note that A ad R may have different column spaces, but from Theorem that if
we can find a set of column vectors of R that forms a basis for the column
space of R, then the corresponding column vectors of A will form a basis for
the column space of A.
Example
Note
1 
0 
0 
0 
1 
0 
c1    , c2    , c5   
0 
0 
1 
 
 
 
0
0
 
 
0 
Form a basis for the column space of R; thus the corresponding column vectors
of A,
1
4
5
2
9
8
c1    , c2    , c5   
2
9
9
 
 
 

1

4
 
 
 5
Form a basis for the column space of A.
Section 4.8 Rank and Nullity
Theorem
If A is any matrix, then the row space and column space of A have the same
dimension.
Definition
The common dimension of the row space and column space of a matrix A is
called the rank of A and is denoted by rank(A);
the dimension of the nullspace of A is called the nullity of A and is denoted by
nullity(A).
Example
Example
Find the rank and nullity of the matrix
 1 2 0 4 5 3
 3 7 2 0 1 4 

A
 2 5 2 4 6 1 


4

9
2

4

4
7


Solution. The reduced row-echelon form of A is
1
0

0

0
0 4 28 37 13
1 2 12 16 5 
0 0
0
0
0

0 0
0
0
0
Example
Since there are two nonzero rows (or, equivalently, two leading 1’s), the row
space and column space are both two-dimensional, so rank(A)=2.
To find the nullity of A, we must find the dimension of the solution space of the
linear system Ax=0. This system can be solved by reducing the augmented
matrix to reduced row-echelon form.
The corresponding system of equations will be
X1-4x3-28x4-37x5+13x6=0
X2-2x3-12x4-16x5+5x6=0
Solving for the leading variables, we have
x1=4x3+28x4+37x5-13x6
X2=2x3+12x4 +16x5-5x6
It follows that the general solution of the system is
Example Cont.
x1=4r+28s+37t-13u
X2=2r+12s +16t-5u
X3=r
X4=s
X5=t
X6=u
 x1 
Equivalently,
 4   28 37 
 13
x 
 2  12  16 
 5 
2
 
     


 x3 
1   0   0 
 0 

r

s

t

u
 
     


x
0
1
0
0
 4
     


 x5 
0  0   1 
 0 
 
     


x
0
0
0
1
     


 6 
Because the four vectors on the right side of the equation form a basis for the
solution space, nullity(A)=4.
Theorems
Theorem
If A is any matrix, then rank(A)=rank(AT).
Theorem (Dimension Theorem for Matrices)
If A is a matrix with n columns, then
Rank(A)+nullity(A)=n
Theorem
If A is an mxn matrix, then
(a) rank(A)= the number of leading variables in the solution of Ax=0.
(b) nullity(A)= the number of parameters in the general solution of Ax=0.
Theorems
Theorem (Equivalent Statements)
If A is an nxn matrix, and if TA: Rn  Rn is multiplication by A, then the following
are equivalent.
(a) A is invertible.
(b) Ax=0 has only the trivial solution.
(c)
The reduced row-echelon form of A is In.
(d) A is expressed as a product of elementary matrices.
(e) Ax=b is consistent for every nx1 matrix b.
(f)
Ax=b has exactly one solution for every nx1 matrix b.
(g) Det(A)0.
(h) The range of TA is Rn.
(i)
TA is one-to-one.
(j)
The column vectors of A are linearly independent.
Theorem Cont.
(k)
(l)
(m)
(n)
(o)
(p)
(q)
The row vectors of A are linearly independent.
The column vectors of A span Rn.
The row vectors of A span Rn.
The column vectors of A form a basis for Rn.
The row vectors of A form a basis for Rn.
A has rank n.
A has nullity 0.
4.9 Transformations from Rn to R m
Functions from R n to R
A function is a rule f that associates with each element in a set A one and only
one element in a set B.
If f associates the element b with the element a, then we write b = f(a) and say
that b is the image of a under f or that f(a) is the value of f at a.
The set A is called the domain of f and the set B is called the codomain of f.
The subset of B consisting of all possible values for f as a varies over A is
called the range of f.
Function from R n to R m
Here, we will be concerned exclusively with transformations from Rn to Rm.
Suppose f1, f2, …, fm are real-valued functions of n real variables, say
w1 = f1(x1,x2,…,xn)
w2 = f2(x1,x2,…,xn)
…
wm = fm(x1,x2,…,xn)
These m equations assign a unique point (w1,w2,…,wm) in Rm to each point
(x1,x2,…,xn) in Rn and thus define a transformation from Rn to Rm. If we
denote this transformation by T: Rn → Rm, then
T (x1,x2,…,xn) = (w1,w2,…,wm)
Example: The equations
w1  x1  x2
w2  3 x1 x2
w3  x12  x22
Defines a transformation T : R 2  R 3 .
With this transformation, the image of the point (x1, x2) is
T ( x1 , x2 )  ( x1  x2 ,3x1 x2 , x12  x22 )
Thus, for example, T(1, -2)=(-1, -6, -3)
Linear Transformations from
R n to R m
A linear transformation (or a linear operator if m = n) T:
defined by equations of the form
w1  a11 x1  a12 x2  ...  a1n xn
w2  a21 x1  a22 x2  ...  a2 n xn
...
wm  am1 x1  am 2 x2  ...  amn xn
or
R n→ R m
 w1   a11 a12
w  a
 2    21 a22
  
  
 wm   am1 am 2
is
a1n   x1 
a2 n   x2 
 
 
amn   xm 
or
w = Ax
The matrix A = [aij] is called the standard matrix for the linear transformation T,
and T is called multiplication by A.
Example: If the linear transformation T : R 4  R 3 is defined by the equations
w1  2 x1  3x2  x3  5 x4
w2  4 x1  x2  2 x3  x4
w3  5 x1  x2  4 x3
Find the standard matrix for T, and calculate T (1, 3, 0, 2)
Solution: T can be expressed as
 x1 
 w1   2 3 1 5  
 w    4 1 2 1   x2 
 2 
 x 
 w3   5 1 4 0   3 
 x4 
So the standard matrix for T is
 2 3 1 5
A   4 1 2 1 
5 1 4 0 
Furthermore, if ( x1 , x2 , x3 , x4 )  (1, 3,0, 2)
w1  2 x1  3x2  x3  5 x4  1
w2  4 x1  x2  2 x3  x4  3
w3  5 x1  x2  4 x3  8
Thus, T (1, 3, 0, 2)  (1,3,8)
Or
1 
 w1   2 3 1 5   1 
 w    4 1 2 1   3  3
 2 
 0   
 w3   5 1 4 0    8
2 
Remarks
Notations:
If it is important to emphasize that A is the standard matrix for T. We denote the
m
n
R m. Thus, TA(x) = Ax
linear transformation T: R n → R by TA: R→
We can also denote the standard matrix for T by the symbol [T], or T(x) = [T]x
Remark:
We have establish a correspondence between m×n matrices and linear
transformations from R n to R m :
To each matrix A there corresponds a linear transformation TA (multiplication by
A), and to each linear transformation T:
→R n R ,m there corresponds an
m×n matrix [T] (the standard matrix for T).
Properties of Matrix Transformations
The following theorem lists four basic properties of matrix transformations that
follow from the properties of matrix multiplication.
Theorem 4.9.2
If TA: Rn Rm and TB: Rn Rm are matrix multiplications, and if TA(x)=TB(x) for
every vector x in Rn, then A=B.
Examples
m
n
Zero Transformation from R to R
If 0 is the m×n zero matrix and 0 is the zero vector in R n, then for every
vector x in R n
T0(x) = 0x = 0
n
So multiplication by zero maps every vector in R into the zero vector in
.R m . We call T0 the zero transformation from R n to R m .
Identity operator on R n
If I is the n×n identity, then for every vector in R n
TI(x) = Ix = x
So multiplication by I maps every vector in R n into itself. We call TI the
identity operator on R n .
A Procedure for Finding Standard Matrices
Reflection Operators
2
In general, operators on R and R 3 that map each vector into its symmetric
image about some line or plane are called reflection operators.
Such operators are linear.
Projection Operators
In general, a projection operator (or more precisely an orthogonal
projection operator) on R 2 or R 3 is any operator that maps each
vector into its orthogonal projection on a line or plane through the origin.
The projection operators are linear.
Rotation Operators
2
An operator that rotate each vector in R through a fixed angle θ is called a
rotation operator on R 2 .
A Rotation of Vectors in R3
• A rotation of vectors in R3 is usually described
in relation to a ray emanating from the origin, called
the axis of rotation.
• As a vector revolves around the axis of rotation
it sweeps out some portion of a cone.
• The angle of rotation is described as “clockwise”
or “counterclockwise” in relation to a viewpoint that is
along the axis of rotation looking toward the origin.
•
The counterclockwise direction for a rotation about its axis can be
determined by a “right hand rule”.
Example: Use matrix multiplication to find the image of the vector (1, 1) when it is
rotated through an angle of 30 degree (  / 6 )
x 
x

Solution: the image of the vector
 y  is
 
cos  / 6  sin  / 6   x   3 / 2
w
  y  
sin

/
6
cos

/
6

    1/ 2
 3
1 
x  y
1/ 2   x   2
2 
   
3 / 2   y   1
3 
x

y

2 
2
So
 3/2
w
 1/ 2
 3 1
1/ 2  1  2 

   
3 / 2  1 1  3 


 2 
Dilation and Contraction Operators
2
If k is a nonnegative scalar, the operator on R or R 3 is called a contraction
with factor k if 0 ≤ k ≤ 1 and a dilation with factor k if k ≥ 1 .
Expansion and Compressions
In a dilation or contraction of R2 or R3, all coordinates are multiplied by a factor
k. If only one of the coordinates is multiplied by k, then the resulting
operator is called an expansion or compression with factor k.
Shears
A matrix operator of the form T(x, y)=(x+ky, y) is called the shear in the xdirection with factor k.
Similarly, a matrix operator of the form T(x, y)=(x, y+kx) is called the shear in
the y-direction with factor k.
4.10 Properties of Matrix Transformations
Compositions of Linear Transformations
n
k
k
m
If TA : R → R and TB : R → R are linear transformations, then for
n
k
each x in R one can first compute TA(x), which is a vector in R , and TB
then one can compute TB(TA(x)), which is a vector in R m .
Thus, the application of TA followed by TB produces a transformation from
to R m . This transformation is called the composition of TB with TA and is
denoted by T T . Thus
B
A
The composition TB TA
(TB TA )(x)  (TB (TA (x))
is linear since
(TB TA )(x)  (TB (TA (x))  B( Ax)  ( BA)x
The standard matrix for TB TA is BA. That is,
TB TA  TBA
Rn
Remark:
TB TA  TBA captures an important idea: Multiplying matrices is equivalent to
composing the corresponding linear transformations in the right-to-left order of
the factors.
n
k
k
m
Alternatively, If T1 : R  R and T2 : R  R are linear transformations, then because
the standard matrix for the composition T2 T1 is the product of the standard matrices
of T2 and T1, we have
T2
T1   T2  T1 
Example: Find the standard matrix for the linear operator T : R 2  R 2 that first reflects
A vector about the y-axis, then reflects the resulting vector about the x-axis.
Solution: The linear transformation T can be expressed as the composition
T  T2 T1
Where T1 is the reflection about the y-axis, and T2 is the reflection about
The x-axis.
 1 0
,
1


T1    0
Sine the standard matrix for T is
1
T2   0

T   T2
0
1
T1   T2  T1 
1 0   1 0  1 0 
T

  0 1  0 1   0 1


 

Which is called the reflection about the origin.
Note: the composition is NOT commutative.
Example: Let T1 : R 2  R 2 be the reflection operator about the line y=x, and let
T2 : R 2  R 2 be the orthogonal projection on the y-axis. Then
T1
0 1  0 0 0 1 
T2   T1  T2   
 0 1   0 0
1
0


 

T2
0 0 0 1  0 0
T1   T2  T1   
 1 0   1 0 
0
1


 

T2
T1   T1 T2 
Thus, T2 T1 and T2 T1 have different effects on a vector x.
One–to-One Matrix Transformations
Linearity Properties
Section 5.1 Eigenvalue and Eigenvector
In general, the image of a vector x under multiplication by a square matrix A
differs from x in both magnitude and direction. However, in the special case
where x is an eigenvector of A, multiplication by A leaves the direction
unchanged.
Depending on the sign and magnitude of the eigenvalue λ corresponding to x,
the operation Ax= λx compresses or stretches x by a factor of λ, with a
reversal of direction in the case where λ is negative.
Computing Eigenvalues and Eigenvectors
3 0 
Example. Find the eigenvalues of the matrix A  

8 1
 3 0
Solution.
det( A  I ) 
 (  3)(  1)  0
8   1
This shows that the eigenvalues of A are λ=3 and λ=-1.
Finding Eigenvectors and Bases for Eigenspaces
Since the eigenvectors corresponding to an eigenvalue λ of a matrix A