Math 250 quad Fall 2006
Selected Solutions
HH 3.7.10: Computing Operator Norm
Let
A
be an
`
×
k
matrix. Let
A
T
denote the transpose of
A
.
(a) Show that
A
T
A
is symmetric.
(b) Show that all eigenvalues of
A
T
A
are nonnegative, and that they are all positive if and only if the
kernel of
A
is zero (i.e.,
A
if and only if
A
is onetoone).
(c) Show that

A

= max
λ
√
λ
, where
λ
ranges over all eigenvalues of
A
T
A
.
Response. For part (a), we appeal to the easily veriﬁed fact that the transpose map
A
→
A
T
is
antimultiplicative: (
AB
)
T
=
B
T
A
T
. Given this, we can compute that (
A
T
A
)
T
= (
A
T
)(
A
T
)
T
=
A
T
A
. This
means that
A
T
A
equals its own transpose, i.e., it is symmetric.
(b) Suppose that
λ
is an eigenvalue of
A
T
A
, with corresponding eigenvector
v
. (We assume that
v
6
= 0. A number is not called an eigenvalue unless there is a nonzero eigenvector attached to it.) Then
(
A
T
A
)
v
=
λ
v
. Take the inner product of this equation with
v
, and use the fact that
v
•
u
=
u
T
v
. Then,
taking into account also that matrix multiplication is associative, we see that
(
A
T
A
)
v
•
v
=
v
T
(
A
T
A
v
) = (
v
T
A
T
)(
A
v
) = (
A
v
)
T
(
A
v
) =
A
(
v
)
•
A
(
v
)
.
On the other hand if
A
T
A
(
v
) =
λ
v
, then (
A
T
A
)(
v
)
•
v
=
λ
v
•
v
, so we can write
λ
=
A
v
•
A
v
v
•
v
, and this is
clearly nonnegative. It is positive unless
A
v
= 0, that is, unless
A
annihilates the nonzero vector
v
.
(c) We can rewrite the expression
A
v
•
A
v
v
•
v
=

A
v

2

v

2
=
±

A
v


v

²
2
.
In the ﬁnal form, we recognize this as the square of the stretching factor for
v
by the linear transformation
A
. The operator norm

A

is by deﬁnition the largest possible value over
v
of this stretching factor. As
is discussed in the note on “Operator Norm”, this may also be taken as the maximum over the unit sphere
S
1
=
{
v
:

v

= 1
}
of the function

A
v

. Clearly this is the same as the square root of the maximum value over
S
1
of the function

A
v

2
= (
A
v
)
•
(
A
v
) = (
A
T
A
)
v
•
v
=
Q
A
T
A
(
v
). We can think of ﬁnding this maximum
value as a constrained maximization problem: maximize
Q
A
T
A
subject to the condition
Q
I
(
v
) =
v
•
v
= 1.
We can apply the doctrine of Lagrange multipliers to this problem. This says that at the maximum
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 Fall '06
 RogerHowe
 Math, triangle, maximum value

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