MAS 213: Linear Algebra II.
Problem list for Week #7.
Solutions.
Problem 1:
(Problem 2.5.2, 2.5.3, various parts, from [FIS].)
In the parts below you are given a vector space
V
and two bases for it,
β
and
β
′
. Find the change-of-coordinates matrix which takes
β
′
-coordinates
to
β
-coordinates.
(i)
V
=
R
2
,
β
=
{
(1
,
0)
,
(0
,
1)
}
, and
β
′
=
{
(
a
1
, a
2
)
,
(
b
1
, b
2
)
}
.
(ii)
V
=
P
2
(
R
),
β
=
{
2
x
2
−
x,
3
x
2
+ 1
, x
2
}
and
β
′
=
{
1
, x, x
2
}
.
(iii)
V
=
P
2
(
R
),
β
=
{
x
2
−
x, x
2
+ 1
, x
−
1
}
,
β
′
=
{
5
x
2
−
2
x
−
3
,
−
2
x
2
+ 5
x
+ 5
,
2
x
2
−
x
−
3
}
.
Solution to part (i).
This is the standard procedure.
We go through the basis we are moving
“from” (in this case it is
β
′
), expressing each vector as a linear combination
of the vectors of the basis we are moving “to”. The coeﬃcients we find then
form the columns of the change-of-coordinates matrix.
•
(
a
1
, a
2
) =
a
1
∗
(1
,
0) +
a
2
∗
(0
,
1).
•
(
b
1
, b
2
) =
b
1
∗
(1
,
0) +
b
2
∗
(0
,
1).
The coeﬃcients are assembled into the “change-of-coeﬃcients” matrix:
Q
= [
I
]
β
β
′
=
[
a
1
b
1
a
2
b
2
]
.
Solution to part (ii).
We do the same thing here.
This part is going to be different, though,
because now we are moving
from
the standard ordered basis, not
to
it.
That makes the calculations a bit harder.
We need to express the three vectors in
β
′
as linear combinations of the
vectors in
β
. The equations we get can be solved simultaneously by solving
the following augmented matrix:
0
1
0
1
0
0
−
1
0
0
0
1
0
2
3
1
0
0
1
→
1
0
0
0
−
1
0
0
1
0
1
0
0
2
3
1
0
0
1
→
1
0
0
0
−
1
0
0
1
0
1
0
0
0
0
1
−
3
2
1
.
Let’s decode exactly what we have just calculated. We have just calculated
that:
•
1 = (0)
∗
(2
x
2
−
x
) + (1)
∗
(3
x
2
+ 1) + (
−
3)
∗
(
x
2
).
•
x
= (
−
1)
∗
(2
x
2
−
x
) + (0)
∗
(3
x
2
+ 1) + (2)
∗
(
x
2
).
•
x
2
= (0)
∗
(2
x
2
−
x
) + (0)
∗
(3
x
2
+ 1) + (1)
∗
(
x
2
).
1