Problem EE221A Set 3 Solutions - Fall 2011
Problem 1.
a) A w.r.t. the standard basis is, by inspection,
00
0 4 .
02
2
AE = 1
0
b) Now consider the diagram from LN3, p.8. We are dealing with exactly this situation; we have one matrix
representation, and two bases, but we are using them in both the domain and the codomain so we have all the
ingredients. So the matrices P and Q for the similarity transform in this case are,
P=
e1
e2
e3
=
b1
b2
b3
1
b1
b2
b3
,
since the matrix formed from the E basis vectors is just the identity; and
b1
Q=
b2
b3
1
e1
e2
e3
=
b1
b2
b3
1
= P 1 .
Let AB be the matrix representation of A w.r.t. B . From the diagram, we have
AB = QAE P
= P 1 AE P
b1
=
b2
12
= 0 0
21
16
1
7
=
15
21
1
b3
AE b1 b2 b3
1
0
2 00
1
5 1 0 4 0
1
0 02
2
4 12
32 6
6
12
0
5
1
2
0
1
Problem 2. Representation of a linear map. This is straightforward from the denition of matrix representation,
0
0
.
..
.
1
.
.
..
A= 0 1
.0
.
..
.
. 0
.
0
1
Problem 3. Norms.
n
2
n
n
2
2
Proof. 1st inequality: Consider the Cauchy-Schwarz inequality, ( i=1 xi yi )
i=1 xi
i=1 yi . Now, let
2
2
y = 1 (vector of all ones). Then we have x 1 n x 2 which is equivalent to the rst inequality.
2
2
2nd inequality: Note that x 2 x 1 x 2 x 1 . Consider that
x
2
2
2
2
= |x1 | + + |xn | ,
while
x
2
1
2
= (|x1 | + + |xn |)
2
2
2
= |x1 | + |x1 | |x2 | + + |x1 | |xn | + |x2 | + |x2 | |x1 | + + |xn | |xn1 | + |xn |
=x
2
2
+ (cross terms),
showing the second inequality.
2
Problem 4.
Proof. First note that the problem implies that A Fmn . By denition,
A
Consider Au
1
n
j =1
=
respectively. Then Au
1
Aj uj
1
n
j =1
1,i
= sup
uU
Au 1
.
u1
, where Aj and uj represent the j -th column of A and the j -th component of u
Aj
1
|uj |. Let Amax be the column of A with the maximum 1-norm; that is,
m
Amax =
Then Au
1
n
j =1
Amax |uj | = Amax
n
j =1
|aij | .
max
j {1,...,n}
i=1
|uj | = Amax u 1 . So we have that
Au 1
Amax .
u1
T
Now, it remains to nd a u such that equality holds. Chose u = (0, . . . , 1, . . . 0) , where the 1 is in the k -th
component that such Au pulls out a column of A having the maximum 1-norm. Note that u 1 = 1, and we see
then that
Au 1
= Amax .
u1
Thus in this case the supremum is achieved and we have the desired result.
Problem 5.
Proof. Straightforward; we simply use properties of the inner product at each step:
x+y
2
+ xy
2
= x + y, x + y + x y, x y
= x + y, x + x + y, y + x y, x + x y, y
= ( x, x + y + y , x + y + x, x y + y, x y )
= ( x, x + x, y + y , x + y , y + x, x + x, y + y, x + y, y )
=2 x
2
+2 y
2
+
x, y + y , x x, y + x, y
=2 x
2
+2 y
2
+
x, y + x, y
=2 x
2
+2 y
2
+ ( x, y + x, y )
=2 x
2
+2 y
2
+ x, y y
=2 x
2
+2 y
2
Problem 6. We will show that the adjoint map A : Cn Cn is identical to matrix multiplication by the complex
conjugate transpose of A. Initially we will use the notation Aa for the matrix representation of the adjoint of A and
reserve the notation v for the complex conjugate transpose of v . First, we know that we can represent A (w.r.t.
the standard basis of Cn ) by a matrix in Cnn ; call this matrix A. Then we can use the dening property of the
adjoint to write,
Au, v = u, Aa v
u A v = u Aa v
Now, this must hold for all u, v Cn . Choose u = ei , v = ej (where ek is a vector that is all zeros except for 1 in
the k -th entry). This will give,
a = aa ,
ij
ij
for all i, j {1, . . . , n}. Thus Aa = A ; it is no accident that we use the notation for both adjoints and complex
conjugate transpose.
3
Problem 7. Continuity and Linearity.
Proof. Let A : (U, F ) (V, F ) with dim U = n and dim V = m be a linear map. Let x, y U , x = y , and
z = x y . Since A is a linear map between nite dimensional vector spaces we can represent it by a matrix A.
Now, the induced norm,
Az
A i := sup
z
z U,z =0
=
Given some
Az A
i
z.
> 0, let
=
A
i
So
x y = z < =
Az < A i = A
i
A
=
Az <
i
and we have continuity.
Alternatively, we can also use the induced matrix norm to show Lipschitz continuity,
x, y U, Ax Ay < K x y ,
where K > A i , which shows that the map is Lipschitz continuous, and thus is continuous (LC = C , note
that the reverse implication is not true!).