CS205 Homework #3 Solutions
Problem 1
Consider an
n
×
n
matrix
A
.
1. Show that if
A
has distinct eigenvalues all the corresponding eigenvectors are linearly
independent.
2. Show that if
A
has a full set of eigenvectors (i.e. any eigenvalue
λ
with multiplicity
k
has
k
corresponding linearly independent eigenvectors), it can be written as
A
=
QΛQ

1
where
Λ
is a diagonal matrix of
A
’s eigenvalues and
Q
’s columns are
A
’s eigenvectors.
Hint: show that
AQ
=
QΛ
and that
Q
is invertible.
3. If
A
is symmetric show that any two eigenvectors corresponding to diﬀerent eigenvalues
are orthogonal.
4. If
A
is symmetric show that it has a full set of eigenvectors. Hint: If (
λ
,
q
) is an
eigenvalue, eigenvector (
q
normalized) pair and
λ
is of multiplicity
k >
1, show that
A

λ
qq
T
has an eigenvalue of
λ
with multiplicity
k

1. To show that consider the
Householder matrix
H
such that
Hq
=
e
1
and note that
HAH

1
=
HAH
and
A
are
similar.
5. If
A
is symmetric show that it can be written as
A
=
QΛQ
T
for an orthogonal matrix
Q
. (You may use the result of (4) even if you didn’t prove it)
Solution
1. Assume that the eigenvectors
q
1
,
q
2
, . . . ,
q
n
are
not
linearly independent. Therefore,
among these eigenvectors there is one (say
q
1
) which can be written as a linear combi
nation of some of the others (say
q
2
, . . . ,
q
k
) where
k
≤
n
. Without loss of generality,
we can assume that
q
2
, . . . ,
q
k
are linearly independent (we can keep removing vectors
from a linearly dependent set until it becomes independent), therefore the decomposi
tion of
q
1
into a linear combination
q
1
=
∑
k
i
=2
α
k
q
k
is unique. However
q
1
=
k
X
i
=2
α
k
q
k
⇒
Aq
1
=
A
±
k
X
i
=2
α
k
q
k
!
⇒
λ
1
q
1
=
k
X
i
=2
λ
k
α
k
q
k
⇒
q
1
=
k
X
i
=2
λ
k
λ
1
α
k
q
k
In the third equation above, we assumed that
λ
1
6
= 0. This is valid, since otherwise
the same equation would indicate that there is a linear combination of
q
2
, . . . ,
q
k
that
is equal to zero (which contradicts their linear independence). From the last equation
above and the uniqueness of the decomposition of
q
1
into a linear combination of
q
2
, . . . ,
q
k
we have
λ
k
λ
1
α
k
=
α
k
. There must be an
α
k
0
6
= 0
,
2
≤
k
0
≤
k
, otherwise we
would have
q
1
=
0
. Thus
λ
1
=
λ
k
0
which is a contradiction.
1
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View Full Document2. Since
A
has a full set of eigenvectors, we can write
Aq
i
=
λ
i
q
i
, i
= 1
, . . . , n
.
We can collect all these equations into the matrix equation
AQ
=
QΛ
, where
Q
has the
n
eigenvectors as columns, and
Λ
=
diag
(
λ
1
, . . . , λ
n
). From (1) we have
that the columns of
Q
are linearly independent, therefore it is invertible.
Thus
AQ
=
QΛ
⇒
A
=
QΛQ

1
.
3. If
λ
1
, λ
2
are two distinct eigenvalues and
q
1
,
q
2
are their corresponding eigenvectors
(there could be several eigenvectors per eigenvalue, if they have multiplicity higher
than 1), then
q
T
1
Aq
2
=
q
T
1
(
Aq
2
) =
λ
2
(
q
T
1
q
2
)
q
T
1
Aq
2
=
q
T
1
A
T
q
2
= (
Aq
1
)
T
q
2
=
λ
1
(
q
T
1
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 Fall '07
 Fedkiw
 Linear Algebra, linearly independent eigenvectors, ann − λ

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