Lecture 29
Andrei Antonenko
April 23, 2003
1
Change of the matrix of an operator
Last time we studied what happens with the coordinates of the vector if we change the basis
from the “old” to the “new” one. We figured out that there exists a changeofbasis matrix,
and if vector
x
had coordinates
(
x
1
, x
2
, . . . , x
n
) in the “old” basis
{
e
1
, e
2
, . . . , e
n
}
and coordinates
(
x
0
1
, x
0
2
, . . . , x
0
n
) in the “new” basis
{
e
0
1
, e
0
2
, . . . , e
0
n
}
and
C
is a changeofbasis matrix such that
(
e
0
1
, e
0
2
, . . . , e
0
n
) = (
e
1
, e
2
, . . . , e
n
)
C
then
x
1
x
2
.
.
.
x
n
=
C
x
0
1
x
0
2
.
.
.
x
0
n
On this lecture we will study what happens with the matrix of a linear operator if we take
another basis in the space.
Let
V
be a vector space, and let
{
e
1
, e
2
, . . . , e
n
}
be the “old” basis and
{
e
0
1
, e
0
2
, . . . , e
0
n
}
be
the “new” basis, and the changeofbasis matrix from the “old” basis to the “new” one is
C
,
such that
(
e
0
1
, e
0
2
, . . . , e
0
n
) = (
e
1
, e
2
, . . . , e
n
)
C.
(1)
Multiplying this equation by
C

1
we have
(
e
0
1
, e
0
2
, . . . , e
0
n
)
C

1
= (
e
1
, e
2
, . . . , e
n
)
.
(2)
Let
A
be a linear operator in the vector space
V
, and its matrix with respect to the “old” basis
is
A
. Then by definition of the matrix of the operator,
(
A
(
e
1
)
,
A
(
e
2
)
, . . . ,
A
(
e
n
)) = (
e
1
, e
2
, . . . , e
n
)
A.
(3)
1
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Now we can get the expression for the matrix of
A
is “new” basis. Applying
A
to both sides
of the equality (1), we get
(
A
(
e
0
1
)
,
A
(
e
0
2
)
, . . . ,
A
(
e
0
n
)) = (
A
(
e
1
)
,
A
(
e
2
)
, . . . ,
A
(
e
n
))
C
by (1)
= (
e
1
, e
2
, . . . , e
n
)
AC
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 Spring '03
 Andant
 Linear Algebra, ........., ann en

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