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Homework Solutions
18.335 - Fall 2004
1.1
jdet(A)j = 1. Show
det(B), then A + B is
Let A be an orthogonal matrix. Prove that
that if B is also orthogonal and
det(A) =
singular.
(det A)2 = det A det A = det A det AT = det AAT = det I = 1
A + B is singular i¤ AT (A + B) = I + AT B is. AT B is orthogonal so all its
eigenvalues are 1 or -1. Since their product is equal to det AT B = 1 then at
least one of the eigenvalues of AT B must be -1. Let the corresponding vector
be x. Then (I + AT B)x = x x = 0, so I + AT B is singular and so is A + B.
Second proof: det(A + B) = det(AT ) det(A + B) det(B T ) =
det(AT AB T + AT BB T ) = det(AT + B T ) = det(A + B), so det(A +
B) = 0.
1.2
Trefethen 2.5
(a) Let be an eigenvalue of S and v its corresponding eigenvector so that
2
Sv = v ) v Sv = v v = kvk : We also have v Sv = v S v = v Sv:
This implies that =
) is imaginary.
(b) If (I S) v = 0 for v 6= 0 then Sv = v and this means that 1 is an eigenvalue
of S, a contradiction to (a).
(c) We have:
Q Q =
h
(I
1
S)
= (I + S ) (I
= (I
i
(I + S) (I
S )
S) (I + S)
= (I + S)
1
(I
1
1
(I
S) (I
(I
1
S)
S)
1
(I + S)
(I + S)
S)
1
(I + S)
S)
1
(I + S) = I
where we have used that if AB = BA and B is invertible that AB
B 1A
1.3
1
=
Trefethen 3.2
kAxk
: Choose an eigenvalue of A and let x 6= 0
x6=0 kxk
kAx k
k x k
j j kx k
such that Ax = x : Then
=
=
= : Thus we have
kx k
kx k
kx k
kAxk
kAk = sup
j j : So kAk j j and since this is true for any eigenvalue
x6=0 kxk
of A we get kAk sup fj j ; eigenvalue of Ag = (A) :
We know that kAk = sup
1
1.4
Trefethen 3.3
s
(a) By de…nition kxk1 = max jxi j
1 i m
m
P
j=1
x2j = kxk2 : Equality is achieved
when we have a vector with only one non-zero component.
s
m
q
P
p
(b) Again, using the de…nition kxk2 =
x2j
m max jxi j = m kxk1 :
1 i m
j=1
We have equality for a vector whose components are equal to each other.
(c) Denoting by rj the j-th row of A we have kAk1 = max krj k1 : For some
1 j m
p
vector v 2 Cn , v = (1; :::; 1) = n andsusing the 2-norm de…nition we
m
P
1
2
get kAk2 = sup kAxk2 kAvk2 = p
krj k1 : These yield kAk1 =
n
j=1
jxj=1
s
m
P
p
2
max krj k1
krj k1
n kAk2 : Equality is achieved for a matrix
1 j m
j=1
which is zero everywhere except along a row of ones.
s
s
m
m
P
P
1
2
2
(d) Using the notation from part (c), kAk2 = p
krj k1
krj k1
n j=1
j=1
p
p
m max krj k1 = m kAk1 : We get equality for a square matrix which
1 j m
is zero everywhere except along a column of ones.
1.5
Prove that
kxy kF =
s
n P
n
P
j=1 i=1
kxy kF = kxy k2 = kxk2 kyk2
2
jxi yj j =
s
n
P
i=1
2
jxi j
s
n
P
j=1
for any
x and y 2 Cn :
2
jyj j = kxk2 kyk2
kxk2 jy zj
kxy zk2
kxy k2 = sup
= sup
: This ratio is maximized if z==y; so
n
n
kzk
kzk2
z2 C
z2 C
2
2
that jy zj = kyk2 ; thus completing the proof.
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