1. Lecture 5
2 Linear algebra (continued)
4 Linear maps
5 Linear systems of equations
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2. Linear maps (4)
Example
Let U = Kn, V = Km and A ∈ Km×n. Consider the map f : Kn → Km
defined by
f (u) = Au
given by multiplication of the column vector v with the matrix A. By the
properties of multiplication
A(u + v) = Au + Av for all u, v ∈ Kn
A(λu) = λAu for all λ ∈ K, u ∈ Kn.
Therefore f is a linear map.
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3. Linear maps (5)
Let V and W be vector spaces of dimensions n and m respectively and let
f : V −→ W be a linear map. Choose bases {v1, . . . , vn} and
{w1, . . . , wm} for V and W . The images of the basis vectors of V are
vectors in W which can be written as a unique linear combination of the
basis vectors. Hence we can define coefficients aij ∈ K by the following
equations
f (vj ) =
m
X
i=1
aij wi .
Now let x be an arbitrary vector in V , with coordinates xj with respect to
the chosen basis of V , i.e. x =
Pn
j=1 xj vj . By linearity of f we have
f (x) =
n
X
j=1
xj f (vj ) =
n
X
j=1
m
X
i=1
aij xj wi .
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4. Linear maps (6)
Let yi denote the coordinates of f (x) with respect to the chosen basis of
W . Then by the previous equation
yi =
n
X
j=1
aij xj .
Define the coordinate vectors
e
x =
x1
x2
.
.
.
xn
∈ Kn
, e
y =
y1
y2
.
.
.
ym
∈ Km
and the representing matrix of f by A = (aij ) ∈ Km×n. Then we have
e
y = Ae
x.
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5. Linear maps (7)
Hence after a choice of bases we can express any linear map
y = f (x)
in the form
e
y = Ae
x.
Note that the coordinate vectors as well as the representing matrix all
depend on the choice of bases for the vector spaces V and W . In
general, if we choose different bases then the coordinate vectors and the
representing matrix will change.
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6. Linear maps (8)
Let U, V , W be vector spaces of dimension k, m and n, respectively, and
f : U → V , g : V → W linear maps. Choose bases u1, . . . , uk, v1, . . . , vm
and w1, . . . , wn for the respective vector spaces. Denote the representing
matrices of f and g by A ∈ Km×k and B ∈ Kn×m.
Theorem
The representing matrix of the composition g ◦ f with respect to the given
bases is equal to the matrix product BA.
Proof.
Let C = (cij ) denote the representing matrix of g ◦ f . Then
n
X
i=1
cij wi = (g ◦ f )(uj ) = g
m
X
l=1
alj vl
!
=
n
X
i=1
m
X
l=1
alj bil wi .
Comparing coefficients we get cij =
Pm
l=1 bil alj , hence C = BA.
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7. Linear maps (9)
Let V and W be finite dimensional vector spaces.
Theorem
Let f : V → W be an isomorphism. Then the dimensions of V and W are
the same and the representing matrix with respect to any bases is
invertible.
Corollary
Let f : V → W be a linear map between vector spaces of the same
dimension and A a representing matrix of f . Then f is an isomorphism if
and only if det A 6= 0.
Theorem
Let f : V → W be a linear map between vector spaces of the same
dimension. Then f is an isomorphism if and only if it is injective,
i.e. ker f = {0}, if and only if it is surjective.
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8. Linear systems of equations (1)
A linear system of equations is a set of m linear equations in n
unknowns of the form
a11x1 + a12x2 + . . . + a1nxn = b1
a21x1 + a22x2 + . . . + a2nxn = b2
.
.
.
am1x1 + am2x2 + . . . + amnxn = bm
where the aij and bk are given numbers in K and the x1, . . . , xn are
unknowns that are to be determined. Define a matrix A = (aij ) and
column vectors x = (x1, . . . , xn)T and b = (b1, . . . , bm)T ; then these
equations can be written in the short form
Ax = b.
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9. Linear systems of equations (2)
Let Ax = b be a linear system of equations, where A and b are given and
x is the unknown. The following can happen:
The system has no solution at all.
The system has a unique solution.
The system has a whole set of (infinitely many) solutions.1
A system of the form Ax = 0 with b = 0 is called homogeneous. If b 6= 0
the system Ax = b is called inhomogeneous. If the system is in the
so-called echelon form, it is easy to determine all solutions.
1
Again because we are working over K = R or C.
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10. Linear systems of equations (3)
Example
The following system has no solution at all:
5 −2 −3
0 7 1
0 0 0
x1
x2
x3
=
0
2
9
.
The last equation reads
0x1 + 0x2 + 0x3 = 9.
This equation has no solution, hence the whole system has no solution.
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11. Linear systems of equations (4)
Example
The following system has a unique solution:
5 −2 −3
0 7 1
0 0 1
x1
x2
x3
=
0
2
9
.
The last equation reads x3 = 9. Then the second equation reads
7x2 + x3 = 7x2 + 9 = 2.
This implies x2 = −1. Now the first equation reads
5x1 − 2x2 − 3x3 = 5x1 + 2 − 27 = 0.
This implies x1 = 5. Therefore we have found the unique solution
x = (5, −1, 9)T .
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12. Linear systems of equations (5)
Example
The following system has infinitely many solutions:
3 −1 7
0 1 2
0 0 0
x1
x2
x3
=
12
3
0
.
The last equation is trivially satisfied for all numbers x1, x2, x3. The second
equation reads
x2 + 2x3 = 3.
We can choose one of the variables freely, say x3 = λ. This implies
x2 = 3 − 2λ. Now the first equation reads
3x1 − x2 + 7x3 = 3x1 − 3 + 2λ + 7λ = 12.
This implies x1 = 5 − 3λ. Therefore we have found the set of solutions
{(5 − 3λ, 3 − 2λ, λ)T | λ ∈ R}. 12 / 16
13. Linear systems of equations (6)
A linear system of equations Ax = b is in echelon form if it looks like
a11 a12 a13 . . . a1n
0 a22 a23 . . . a2n
0 0 a33 . . . a3n
· · · . . . ·
0 0 0 . . . 0
· · · . . . ·
0 0 0 . . . 0
x1
x2
x3
.
.
.
xn
=
b1
b2
b3
.
.
.
.
.
.
bm
(some of the aii can be zero as well!). It is then possible, as in the
examples, to determine the set of all solutions by back substitution.
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14. Linear systems of equations (7)
The following operations on a linear system of equations Ax = b do not
change the solution set:
Interchange of two equations.
Multiplication of an equation by a non-zero constant c.
Addition of a constant multiple of one equation to another equation.
Instead of doing these operations on the equations, one can do them on
the augmented matrix (A | b):
(A | b) =
a11 a12 . . . a1n
a12 a22 . . . a2n
· · . . . ·
am1 am2 . . . amn
b1
b2
·
bm
It is always possible to transform any given linear system of equations to
echelon form using only the above operations. Having determined the
solutions one can check them by plugging them into the equations we
started with.
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15. Linear systems of equations (8)
Example
Let us bring the following system into echelon form:
0 0 3 −2 3
2 −1 4 1 6
0 0 6 −3 10
0 0 0 4 20
5
1
12
0
Interchange the first and second row:
2 −1 4 1 6
0 0 3 −2 3
0 0 6 −3 10
0 0 0 4 20
1
5
12
0
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16. Linear systems of equations (9)
Example
Add (−2)-times the second row to the third row:
2 −1 4 1 6
0 0 3 −2 3
0 0 0 1 4
0 0 0 4 20
1
5
2
0
Add (−4)-times the third row to the fourth row:
2 −1 4 1 6
0 0 3 −2 3
0 0 0 1 4
0 0 0 0 4
1
5
2
−8
The system is now in echelon form.
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