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TA: Yankun Wang
Economics 617
Problem Set 3 Solution Key
1. [Matrix Multiplication and Transpose]
Suppose
A
and
B
are symmetric
n
×
n
matrices.
(a) Under what conditions on
A
and
B
is the matrix
AB
also symmetric?
Explain.
So
lu
t
ion
: Wha
tw
eknowi
s
A
=
A
0
,B
=
B
0
,
andwewan
tasu
ﬃ
cient
condition such that
(
AB
)
0
=
AB.
Notice that
(
AB
)
0
=
B
0
A
0
=
BA,
thus as
long as we have
BA
=
AB, AB
is symmetric.
Notice that this problem is just asking for a su
ﬃ
cient condition, so you
might come up with di
f
erent conditions: that’s OK.
(b) Under the conditions identi
f
ed by you in (i), is the matrix
BA
also
symmetric? Explain.
Solution: Yes. Because
(
BA
)
0
=
A
0
B
0
=
AB
=
BA,
where the last step
follows from our condition.
2. [NonSingular and Invertible Matrices]
(a) If
A
is an
n
×
n
nonsingular matrix, we know that there exists an
n
×
n
matrix
B
such that
BA
=
AB
=
I.
Show that this
B
is unique.
Proof: Suppose there exists a matrix
C
such that
CA
=
AC
=
I,
and
C
is not the same as
B.
Then:
B
=
BI
=
B
(
AC
)=(
BA
)
C
=
IC
=
C.
Thus we get contradiction.
(b) Suppose
A
is an
n
×
n
nonsingular matrix, and
B
is an
n
×
n
matrix
satisfying
AB
=
I.
Does it follow that
BA
=
I
?Exp
la
in
.
Proof: Yes. Prof. Mitra actually proved this in class. Or you can look
at Page 15 and 16 in the Lectures Notes. Notice that why we can
f
nd a
X
such at
AX
=
I,
(
X
is like
B
in our problem), and why we can
f
nd a
Y
such
at
YA
=
I,
and
f
nally, how we prove that
X
=
Y.
(c) Suppose
A
is an
n
×
n
nonsingular matrix, and
B
is an
n
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View Full DocumentSolution: Yes. The reason is:
AB
=
A
×
£
B
1
B
2
...B
n
¤
=[00
...
0]
,
so we must have
AB
i
=0
,i
=1
, ..., n.
Since all the column of
A
are
linearly independent,
B
i
=0
,i
=1
, ..., n.
Thus
B
is the null matrix.
3. [Proof of the Rank Theorem]
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 Fall '08
 STAFF
 Economics

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