Linear Algebra. Session 8 - Texas A&M Universityroquesol/Math_304_Fall_2017...Session 8 Abstract...

Abstract Linear Algebra

Linear Algebra. Session 8

Dr. Marco A Roque Sol

08/01/2017

Dr. Marco A Roque Sol Linear Algebra. Session 8

Abstract Linear AlgebraRange and kernel.Matrix transformations. Matrix of a linear transformation. Similar matrices.

Range and kernel.

Range and kernel

Let V,W be vector spaces and L : V→W, be a linear mapping.

Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)

The kernel of L, denoted by ker(L) is the set of all vectors v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel

Let

V,W be vector spaces and L : V→W, be a linear mapping.

Definition.





Range and kernel.

Range and kernel

Let V,W

be vector spaces and L : V→W, be a linear mapping.

Definition.





Range and kernel.

Range and kernel

Let V,W be vector spaces and

L : V→W, be a linear mapping.

Definition.





Range and kernel.

Range and kernel

Let V,W be vector spaces and L : V→W,

be a linear mapping.

Definition.





Range and kernel.

Range and kernel


Definition.





Range and kernel.

Range and kernel


Definition.

The range

(or image) of L is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image)

of L is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L

is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set

of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors

w ∈W suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W

suchthat w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat

w = L(v) for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat w = L(v)

for some v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat w = L(v) for some

v ∈ V. The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V.

The range of L is denoted by L(V)




Range and kernel.

Range and kernel


Definition.

The range (or image) of L is the set of all vectors w ∈W suchthat w = L(v) for some v ∈ V. The range of L

is denoted by L(V)




Range and kernel.

Range and kernel


Definition.





Range and kernel.

Range and kernel


Definition.


The kernel

of L, denoted by ker(L) is the set of all vectors v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L,

denoted by ker(L) is the set of all vectors v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L, denoted by ker(L)

is the set of all vectors v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L, denoted by ker(L) is the set

of all vectors v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L, denoted by ker(L) is the set of all vectors

v ∈ Vsuch that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L, denoted by ker(L) is the set of all vectors v ∈ V

such that L(v) = 0.



Range and kernel.

Range and kernel


Definition.


The kernel of L, denoted by ker(L) is the set of all vectors v ∈ Vsuch that

L(v) = 0.



Range and kernel.

Range and kernel


Definition.





Range and kernel.

Example 8.1

Consider the linear tranformation, L : R3 → R3, given by

L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z

= AXFind ker(L) and range of L

Solution

The kernel, ker(L), is the nullspace of the matrix A. To findker(L), we apply row reduction to the matrix A: 1 0 −11 2 −1

1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1

Consider

the linear tranformation, L : R3 → R3, given by

L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1

Consider the linear tranformation,

L : R3 → R3, given by

L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1

Consider the linear tranformation, L : R3 → R3,

given by

L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

=

1 0 −11 2 −11 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z

= AX

Find ker(L) and range of L

Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z

= AXFind

ker(L) and range of L

Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z

= AXFind ker(L) and

range of L

Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z

= AXFind ker(L) and range of

L

Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel,

ker(L), is the nullspace of the matrix A. To findker(L), we apply row reduction to the matrix A: 1 0 −11 2 −1

1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L),

is the nullspace of the matrix A. To findker(L), we apply row reduction to the matrix A: 1 0 −11 2 −1

1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace

of the matrix A. To findker(L), we apply row reduction to the matrix A: 1 0 −11 2 −1

1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace of the matrix A.

To findker(L), we apply row reduction to the matrix A: 1 0 −11 2 −1

1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace of the matrix A. To find

ker(L), we apply row reduction to the matrix A: 1 0 −11 2 −11 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace of the matrix A. To findker(L),

we apply row reduction to the matrix A: 1 0 −11 2 −11 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace of the matrix A. To findker(L), we apply row reduction

to the matrix A: 1 0 −11 2 −11 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution

The kernel, ker(L), is the nullspace of the matrix A. To findker(L), we apply row reduction to the matrix A:

1 0 −11 2 −11 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒

1 0 −10 2 00 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒

1 0 −10 1 00 0 0

⇒



Range and kernel.

Example 8.1


L

xyz

= 1 0 −11 2 −1

1 0 −1

= xy

z


Solution


1 0 −1

⇒ 1 0 −10 2 0

0 0 0

⇒ 1 0 −10 1 0

0 0 0

⇒Dr. Marco A Roque Sol Linear Algebra. Session 8


Range and kernel.

Hence (x , y , z) belongs to ker(L) if x − z = y = 0. It follows thatker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.

Hence

(x , y , z) belongs to ker(L) if x − z = y = 0. It follows thatker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z)

belongs to ker(L) if x − z = y = 0. It follows thatker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z) belongs to ker(L)

if x − z = y = 0. It follows thatker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z) belongs to ker(L) if x − z = y = 0.

It follows thatker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z) belongs to ker(L) if x − z = y = 0. It follows that

ker(L) is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z) belongs to ker(L) if x − z = y = 0. It follows thatker(L)

is the line spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Hence (x , y , z) belongs to ker(L) if x − z = y = 0. It follows thatker(L) is the line

spanned by (1, 0, 1)

Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since

L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L

is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

=

x

111

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+

y

020

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+

z

−1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then,

The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range

of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L,

is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned

by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1),

(0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and

(−1,−1,−1). It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1).

It follows that L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that

L(R3) is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3)

is the plane spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1

then, The range of L, is spanned by vectors (1, 1, 1), (0, 2, 0), and(−1,−1,−1). It follows that L(R3) is the plane

spanned by(1, 1, 1), (0, 2, 0).



Range and kernel.


Since L is given by

L

xyz

= x 11

1

+ y 02

0

+ z −1−1−1




Range and kernel.

Example 8.2

Consider the linear tranformation, L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′. Find range and ker(L) of L

Solution

According to the theory of differential equations, the initial valueproblem

u(x)′′−2u(x)′+u(x)′ = g(x); u(x0) = u0; u′(x0) = u′0; u′′(x0) = u′′0

has a unique solution for any g ∈ C (R) and any u0, u′0, u′′0 . ItFollows that L(C (R)3) = C (R)



Range and kernel.

Example 8.2

Consider

the linear tranformation, L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′. Find range and ker(L) of L

Solution






Range and kernel.

Example 8.2

Consider the linear tranformation,

L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′. Find range and ker(L) of L

Solution






Range and kernel.

Example 8.2

Consider the linear tranformation, L : C (R)3 → C (R),

given byL(u) = u′′ − 2u′ + u′. Find range and ker(L) of L

Solution






Range and kernel.

Example 8.2

Consider the linear tranformation, L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′.

Find range and ker(L) of L

Solution






Range and kernel.

Example 8.2

Consider the linear tranformation, L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′. Find

range and ker(L) of L

Solution






Range and kernel.

Example 8.2

Consider the linear tranformation, L : C (R)3 → C (R), given byL(u) = u′′ − 2u′ + u′. Find range and ker(L)

of L

Solution






Range and kernel.

Example 8.2


Solution






Range and kernel.

Example 8.2


Solution

According

to the theory of differential equations, the initial valueproblem





Range and kernel.

Example 8.2


Solution

According to the theory

of differential equations, the initial valueproblem





Range and kernel.

Example 8.2


Solution

According to the theory of differential equations,

the initial valueproblem





Range and kernel.

Example 8.2


Solution






Range and kernel.

Example 8.2


Solution


u(x)′′−2u(x)′+u(x)′ = g(x);

u(x0) = u0; u′(x0) = u

′0; u′′(x0) = u

′′0




Range and kernel.

Example 8.2


Solution






Range and kernel.

Example 8.2


Solution



has a unique solution

for any g ∈ C (R) and any u0, u′0, u′′0 . ItFollows that L(C (R)3) = C (R)



Range and kernel.

Example 8.2


Solution



has a unique solution for any g ∈ C (R) and

any u0, u′0, u′′0 . It

Follows that L(C (R)3) = C (R)



Range and kernel.

Example 8.2


Solution



has a unique solution for any g ∈ C (R) and any u0, u′0, u′′0 .

ItFollows that L(C (R)3) = C (R)



Range and kernel.

Example 8.2


Solution



has a unique solution for any g ∈ C (R) and any u0, u′0, u′′0 . ItFollows that

L(C (R)3) = C (R)



Range and kernel.

Example 8.2


Solution






Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3

Now, the functions xex , ex , 1 satisfy L(xex) = L(ex) = L(1) = 0and W (xex , ex , 1) 6= 0. Therefore,ker(L) = Span(xex , ex , 1)

General linear equations

Definition.A linear equation is an equation of the form

L(x) = b

where L : V→W, is a linear mapping, b is a given vector from W,and x is an unknown vector from V.



Range and kernel.

Also,

the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation

l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 )

which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is

alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping

l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3,

becomes invertible whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible

whenrestricted to ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to

ker(L). Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L).

Hence dim (ker(L)) = 3




L(x) = b




Range and kernel.

Also, the initial data evaluation l(u) = u → (u0, u′0, u′′0 ) which is alinear mapping l(u) : C (R3)→ R3, becomes invertible whenrestricted to ker(L). Hence

dim (ker(L)) = 3




L(x) = b




Range and kernel.





L(x) = b




Range and kernel.


Now,

the functions xex , ex , 1 satisfy L(xex) = L(ex) = L(1) = 0and W (xex , ex , 1) 6= 0. Therefore,ker(L) = Span(xex , ex , 1)



L(x) = b




Range and kernel.


Now, the functions xex , ex , 1

satisfy L(xex) = L(ex) = L(1) = 0and W (xex , ex , 1) 6= 0. Therefore,ker(L) = Span(xex , ex , 1)



L(x) = b




Range and kernel.


Now, the functions xex , ex , 1 satisfy L(xex) = L(ex) = L(1) = 0and

W (xex , ex , 1) 6= 0. Therefore,ker(L) = Span(xex , ex , 1)



L(x) = b




Range and kernel.


Now, the functions xex , ex , 1 satisfy L(xex) = L(ex) = L(1) = 0and W (xex , ex , 1) 6= 0.

Therefore,ker(L) = Span(xex , ex , 1)



L(x) = b




Range and kernel.


Now, the functions xex , ex , 1 satisfy L(xex) = L(ex) = L(1) = 0and W (xex , ex , 1) 6= 0. Therefore,

ker(L) = Span(xex , ex , 1)



L(x) = b




Range and kernel.





L(x) = b




Range and kernel.




Definition.A linear equation

is an equation of the form

L(x) = b




Range and kernel.




Definition.A linear equation is an equation

of the form

L(x) = b




Range and kernel.





L(x) = b




Range and kernel.





L(x) = b

where

L : V→W, is a linear mapping, b is a given vector from W,and x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W,

is a linear mapping, b is a given vector from W,and x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping,

b is a given vector from W,and x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping, b

is a given vector from W,and x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping, b is a given vector

from W,and x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping, b is a given vector from W,and

x is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping, b is a given vector from W,and x

is an unknown vector from V.



Range and kernel.





L(x) = b

where L : V→W, is a linear mapping, b is a given vector from W,and x is an unknown vector

from V.



Range and kernel.





L(x) = b




Range and kernel.

The range of L is the set of all vectors b ∈W such that theequation L(x) = b has a solution.

The kernel of L is the solution set of the homogeneous linearequation L(x) = 0 .

Theorem

If the linear equation L(x) = b is solvable and dim (ker(L))


Range and kernel.

The range

of L is the set of all vectors b ∈W such that theequation L(x) = b has a solution.


Theorem



Range and kernel.

The range of L

is the set of all vectors b ∈W such that theequation L(x) = b has a solution.


Theorem



Range and kernel.

The range of L is the set

of all vectors b ∈W such that theequation L(x) = b has a solution.


Theorem



Range and kernel.

The range of L is the set of all vectors

b ∈W such that theequation L(x) = b has a solution.


Theorem



Range and kernel.

The range of L is the set of all vectors b ∈W

such that theequation L(x) = b has a solution.


Theorem

If

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Linear Algebra. Session 8 - Texas A&M Universityroquesol/Math_304_Fall_2017...Session 8 Abstract...

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