=
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
x
y
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
3
3
/
/
/
/
/
/
/
/
/
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
x
x
x
1
2
3
0
0
0
0
0
0
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
1
1 2
1 2
0
1 2
1 2
1
1
0
−
−
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
/
/
/
/
1
0
0
0
1
0
0
0
1
1
2
3
/
/
/
.
y
y
y
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
The determinant of the product matrix is the product of the three determinants.
The determinant of the center
matrix is 1/2.
The determinants of the diagonal matrices are the products of the diagonal elements.
Therefore, the Jacobian is
J
=
abs(
∂
x
/
∂
y
′
)= ½(
x
1
x
2
x
3
)/(
y
1
y
2
y
3
) = 2(
y
1
/
y
2
) (after making the substitutions for
x
i
).
20.
Prove that exchanging two columns of a square matrix reverses the sign of its determinant.
(
Hint:
use a
permutation matrix. See Exercise 6.)
Exchanging the first two columns of a matrix is equivalent to postmultiplying it by a permutation
matrix
B
= [
e
2
,
e
1
,
e
3
,
e
4
,...] where
e
i
is the
i
th column of an identity matrix.
Thus, the determinant of the matrix
is 
AB
 = 
A
 
B
.
The question turns on the determinant of
B
. Assume that
A
and
B
have
n
columns. To obtain
the determinant of
B
, merely expand it along the first row. The only nonzero term in the determinant is (1)
I
n

1
 = 1, where
I
n
1
is the (
n
1)
(
n
1) identity matrix.
This completes the proof.
×
21.
Suppose
x
=
x
(
z
) where
z
is a scalar.
What is
∂
[(
x
′
Ax
)/(
x
′
Bx
)]/
z
?
The required derivatives are given in Exercise 16.
Let
g
=
∂
x
/
∂
z
and let the numerator and
denominator be
a
and
b
, respectively. Then,
∂
(
a
/
b
)/
∂
z
=
[
b
(
∂
a
/
∂
z
) 
a
(
∂
b
/
∂
z
)]/
b
2
=
[
x
′
Bx
(2
x
′
Ag
) 
x
′
Ax
(2
x
′
Bg
)] / (
x
′
Bx
)
2
=
2[
x
′
Ax
/
x
′
Bx
][
x
′
Ag
/
x
′
Ax

x
′
Bg
/
x
′
Bx
].
22.
Suppose
y
is an
n
×
1 vector and
X
is an
n
×
K
matrix.
The projection of
y
into the column space of
X
is
defined in the text after equation (255),
=
Xb
.
Now, consider the projection of
y
*
=
c
y
into the column
space of
X
*
=
XP
where
c
is a scalar and
P
is a nonsingular
K
ˆ
y
×
K
matrix.
Find the projection of
y
*
into the
column space of
X
*
.
Prove that the cosine of the angle between
y
*
and its projection into the column space of
X
*
is the same as that between
y
and its projection into the column space of
X
.
How do you interpret this
result?
The projection of
y
*
into the column space of
X
*
is
X
*
b
*
where
b
*
is the solution to the set of
equations
X
*
′
y
*
=
X
*
′
X
*
b
*
or
P
′
X
′
(
c
y
)
=
P
′
X
′
XPb
*
.
Since
P
is nonsingular,
P
′
has an inverse.
Premultiplying the equation by (
P
′
)
1
, we have
c
X
′
y
=
X
′
X
(
Pb
*
)
or
X
′
y
=
X
′
X
[(1/
c
)
Pb
*
].
Therefore, in
terms of the original
y
and
X
, we see that
b
=
(1/
c
)
Pb
*
which implies
b
*
=
c
P
1
b
.
The projection is
X
*
b
*
=
(
XP
)(
c
P
1
b
)
=
c
Xb
.
We conclude, therefore, that the projection of
y
*
into the column space of
X
*
is a
multiple
c
of the projection of
y
into the space of
X
.
This makes some sense, since, if
P
is a nonsingular
matrix, the column space of
X
*
is exactly the same as the same as that of
X
.
The cosine of the angle between
y
*
and its projection is that between
c
y
and
c
Xb
.
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 Spring '10
 Dr.Fang
 Linear Algebra, Determinant, Xq

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