2Linear maps
IB Linear Algebra
2.4 Change of basis
Suppose we have a linear map
α
:
U → V
. Given a basis
{e
i
}
for
U
, and a basis
{f
i
} for V , we can obtain a matrix A.
U V
F
m
F
n
α
A
(e
i
) (f
i
)
We now want to consider what happens when we have two different basis
{u
i
}
and
{e
i
}
of
U
. These will then give rise to two different maps from
F
m
to our
space
U
, and the two basis can be related by a change-of-basis map
P
. We can
put them in the following diagram:
U U
F
m
F
m
ι
U
P
(u
i
) (e
i
)
where
ι
U
is the identity map. If we perform a change of basis for both
U
and
V
,
we can stitch the diagrams together as
U U V V
F
m
F
m
F
n
F
n
ι
U α
ι
V
P
(u
i
)
B
A
(e
i
) (f
i
)
Q
(v
i
)
Then if we want a matrix representing the map
U → V
with respect to bases
(u
i
) and (v
i
), we can write it as the composition
B = Q
−1
AP.
We can write this as a theorem:
Theorem.
Suppose that (
e
1
, ··· , e
m
) and (
u
1
, ··· , u
m
) are basis for a finite-
dimensional vector space
U
over
F
, and (
f
1
, ··· , f
n
) and (
v
1
, ··· , v
n
) are basis
of a finite-dimensional vector space V over F.
Let
α
:
U → V
be a linear map represented by a matrix
A
with respect to
(e
i
) and (f
i
) and by B with respect to (u
i
) and (v
i
). Then
B = Q
−1
AP,
where P and Q are given by
u
i
=
m
X
k=1
P
ki
e
k
, v
i
=
n
X
k=1
Q
ki
f
k
.
Note that one can view
P
as the matrix representing the identity map
i
U
from
U
with basis (
u
i
) to
U
with basis (
e
i
), and similarly for
Q
. So both are
invertible.
Proof. On the one hand, we have
α(u
i
) =
n
X
j=1
B
ji
v
j
=
X
j
X
`
B
ji
Q
`j
f
`
=
X
`
[QB]
`i
f
`
.
On the other hand, we can write
α(u
i
) = α
m
X
k=1
P
ki
e
k
!
=
m
X
k=1
P
ki
X
`
A
`k
f
`
=
X
`
[AP ]
`i
f
`
.
Since the f
`
are linearly independent, we conclude that
QB = AP.
Since Q is invertible, we get B = Q
−1
AP .
Definition
(Equivalent matrices)
.
We say
A, B ∈ Mat
n,m
(
F
) are equivalent if
there are invertible matrices
P ∈ Mat
m
(
F
),
Q ∈ Mat
n
(
F
) such that
B
=
Q
−1
AP
.
Since
GL
K
(
F
) =
{A ∈ Mat
k
(
F
) :
A is invertible}
is a group, for each
k ≥
1,
this is indeed an equivalence relation. The equivalence classes are orbits under
the action of GL
m
(F) ×GL
n
(F), given by
GL
m
(F) ×GL
n
(F) ×Mat
n,m
(F) → Mat(F)
(P, Q, A) 7→ QAP
−1
.
Two matrices are equivalent if and only if they represent the same linear map
with respect to different basis.
Corollary.
If
A ∈ Mat
n,m
(
F
), then there exists invertible matrices
P ∈
GL
m
(F), Q ∈ GL
n
(F) so that
Q
−1
AP =
I
r
0
0 0
for some 0 ≤ r ≤ min(m, n).
This is just a rephrasing of the proposition we had last time. But this tells
us there are min(m, n) + 1 orbits of the action above parametrized by r.
Definition (Column and row rank). If A ∈ Mat
n,m
(F), then
–
The column rank of
A
, written
r
(
A
), is the dimension of the subspace of
F
n
spanned by the columns of A.
–
The row rank of
A
, written
r
(
A
), is the dimension of the subspace of
F
m
spanned by the rows of A. Alternatively, it is the column rank of A
T
.
There is no a priori reason why these should be equal to each other. However,
it turns out they are always equal.
Note that if
α
:
F
m
→ F
n
is the linear map represented by
A
(with respect to
the standard basis), then
r
(
A
) =
r
(
α
), i.e. the column rank is the rank. Moreover,
since the rank of a map is independent of the basis, equivalent matrices have
the same column rank.
Theorem.
If
A ∈ Mat
n,m
(
F
), then
r
(
A
) =
r
(
A
T
), i.e. the row rank is equivalent
to the column rank.
Proof. We know that there are some invertible P, Q such that
Q
−1
AP =
I
r
0
0 0
,
where r = r(A). We can transpose this whole equation to obtain
(Q
−1
AP )
T
= P
T
A
T
(Q
T
)
−1
=
I
r
0
0 0
So r(A
T
) = r.