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Quiz VII
1. The matrix
A
=
1
0
1
0
2
0
1
0
1
is symmetric. Find a basis for
R
3
consisting of
eigenvectors for
A
.
Solution:
Since
A
is symmetric,
there will be
an
orthonormal
basis consisting of
eigenvectors of
A
, but you do not need to use that to solve the problem. First ﬁnd
the eigenvalues
det
1

λ
0
1
0
2

λ
0
1
0
1

λ
= (1

λ
)
2
(2

λ
)

(2

λ
) = (2

λ
)(

2
λ
+
λ
2
)
.
So the eigenvalues are 2 and 0. Solving for the eigenspaces,
⃗v
1
=
1
0

1
is an
eigenvector for the eigenvalue 0. For the eigenvalue 2,
A

2
I
=

1
0
1
0
0
0
1
0

1
,
and the reduced row echelon form of that is just
1
0

1
0
0
0
0
0
0
So a basis for the
eigenspace
E
2
is
{
⃗v
2
=
0
1
0
,⃗v
3
=
1
0
1
}
. The basis
B
=
{
⃗v
1
,⃗v
2
,⃗v
3
}
is a basis of
eigenvectors. Note that, if you divide
⃗v
1
and
⃗v
3
by their lengths (both lengths are
√
2), this becomes an orthonormal basis.
2. Find a basis for
R
2
such that the matrix of
T
(
⃗x
) =
(
3
2

1
1
)
⃗x
with respect to
this basis has the form
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 Fall '08
 lee
 Eigenvectors, Vectors

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