Math 311 second practice exam answers
1.
(
D
2

I
)
u
(
x
) =
u
(
x
)

u
(
x
) =

4 cos 2
x

cos 2
x
=

5 cos 2
x.
2.
To show that these polynomials span the whole vector space, we note that for any
a
0
, a
1
, a
2
,
a
0
+
a
1
x
+
a
2
x
2
=
a
0
1 +
a
1
(
x
+
x
2
) + (
a
2

a
1
)
x
2
.
To show that these polynomials are linearly independent, suppose that for some
a, b, c
,
a
1 +
b
(
x
+
x
2
) +
cx
2
= 0
.
Then
a
+
bx
+ (
b
+
c
)
x
2
= 0
,
from which is follows that
a
= 0
,
b
= 0
,
b
+
c
= 0
, so also
c
= 0
.
3.
Row reduction brings the matrix
1
1

1
2
4
2
3
5
1
to the form
1
1

1
0
1
2
0
0
0
.
The first two columns are lead columns, so the first two of the original vectors
Span
1
2
3
,
1
4
5
form a basis for this subspace. In particular, that subspace has dimension
2
.
4.
2

2
1
0
1
3
0
0
1
.
Since the matrix is uppertriangular, we see right away that the eigenvalues are
λ
= 1
,
2
. For
λ
= 1
, the
eigenvectors are elements of the null space
Null
1

2
1
0
0
3
0
0
0
=
Null
1

2
0
0
0
1
0
0
0
= Span
2
1
0
.
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 Spring '08
 Anshelvich
 Math, Linear Algebra, Polynomials, Vector Space, 4 Span, 311 second

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