# Problem sheet 5 The Hessian matrix by f34q4h6

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```									Problem sheet 5: The Hessian matrix.

Trond Stølen Gustavsen
September, 21, 2009

Problem 1.

Solution. See answers in FMEA.

Problem 2.

Solution. (a) We have
f (x, y) = ax2 + 2bxy + cy 2 + px + qy + r.
The Hessian matrix is
f11   f12                 2a   2b
f =                       =                  .
f21   f22                 2b   2c
The leading principal minors are
2a       2b
D1 = 2a and D2 =                        = 4(ac − b2 ).
2b       2c
So if a > 0 and ac − b2 > 0 the Hessian is positive deﬁnite, so the function is strictly convex.
If a < 0 and ac − b2 > 0 we have (−1)1 D1 > 0 and (−1)2 D2 > 0, so the Hessian is negative
deﬁnite. This means that the function is strictly concave.
(b) We know from a theorem in Lecture 6 that a function is convex if and only if the Hessian
is positive semideﬁnite. We also know that a symmetric matrix is positive semideﬁnite if
all principal minors are positive or zero. The principal minors are
2a 2b
, 2a and 2c.
2b 2c
Thus the Hessian is positive semideﬁnite if and only if ac − b2 ≥ 0, a ≥ 0 and c ≥ 0. Thus
this is a suﬃcient and a necessary condition for the function f to be convex. Similarly we
get that the Hessian is negative deﬁnite if and only if ac − b2 ≥ 0 and a ≤ 0 and c ≤ 0, so
this is a necessary and suﬃcient condition for the function f to be concave.

Problem 3.

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Solution. One approach is to use the result of the previous problem. We will however solve
the problem directly.
The function
f (x, y) = −6x2 + (2a + 4)xy − y 2 + 4ay
has the following partial derivatives:
f1 = −12x + (2a + 4)y
f2 = (2a + 4)x − 2y + 4a.
The Hessian matrix becomes
f11   f12             −12     2a + 4
f =                   =                              .
f21   f22            2a + 4    −2
The principal minors are
−12 2a + 4
−12, −2 and D2 =                              = 8 − 4a2 − 16a
2a + 4 −2
Since principal minors of order 1 are negative, the function is never convex. The function
is concave if and only if
√                √
8 − 4a2 − 16a ≥ 0 ⇐⇒ −2 − 6 ≤ a ≤ −2 + 6.

Problem 4. Consider the function
f (x, y) = x4 + 16y 4 + 32xy 3 + 8x3 y + 24x2 y 2 .
Find the Hessian matrix. Show that f is convex.

Solution. The Hessian matrix is
48xy + 12x2 + 48y 2           96xy + 24x2 + 96y 2
f =
96xy + 24x2 + 96y 2          192xy + 48x2 + 192y 2
Completing the squares we see that
2              2
12 (x + 2y)        24 (x + 2y)
f =                2                   2     .
24 (x + 2y)        48 (x + 2y)
2                        2
The principal minors of order one are 12 (x + 2y) ≥ 0 and 48 (x + 2y) ≥ 0. The principal
minor of order two is
2             2
12 (x + 2y) 24 (x + 2y)
2             2  = 0.
24 (x + 2y) 48 (x + 2y)
Thus all principal minors are non-negative, thus the function is convex (but not strictly
convex).

Problem 5.

Solution. See answers in FMEA.

Problem 6.

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Solution. (a) π(x, y) = 13x + 8y − C(x, y) = 9x + 6y − 0.04x2 + 0.01xy − 0.01y 2 − 500.
(b) π1 = 9 − 0.08x + 0.01y and π2 = 6 + 0.01x − 0.02y. The Hessian matrix is
−0.08 0.01
.
0.01 −0.02
The leading principal minors are
−0.08 0.01
D1 = −0.08 and D2 =                                  = 0.0015 > 0
0.01 −0.02
Thus the function concave.
Solving π1 = 9 − 0.08x + 0.01y = 0 and π2 = 6 + 0.01x − 0.02y = 0, we get that (160, 380)
as the only stationary point which has to be a maximum.

Problem 7. Consider the function f deﬁned on the subset S = {(x, y, z) : z > 0} of R3 by
f (x, y, z) = 2xy + x2 + y 2 + z 3 .
Show that S is convex. Find the stationary points of f . Find the Hessian matrix. Is f
concave or convex? Does f have a global extreme point?

Solution. If (x1 , y1 , z1 ) and (x2 , y2 , z2 ) are two points in S. We must show that the line
segment
[(x1 , y1 , z1 ), (x2 , y2 , z2 )] = {(x, y, z)|(x, y, z) = s(x1 , y1 , z1 ) + (1 − s)(x2 , y2 , z2 ), s ∈ [0, 1]}
is in S. For a point on the line segment we have z = sz1 + (1 − s)z2 . Since either s, z1 , (1 − s)
and z2 are non negative, z is non negative. Since either s or (1 − s) is positive, z must be
positive. This shows that the line segment is contained in S.
The ﬁrst order conditions are
f1 = 2x + 2y = 0
f2 = 2x + 2y = 0
f3 = 3z 2 = 0
From this we see that
x = −y
z=0
We conclude that there are no stationary points in S.
The Hessian matrix is                             
2 2 0
 2 2 0 .
0 0 6z
Solving
2−λ        2        0
2    2−λ         0    = −λ (λ − 4) (λ − 6z)
0       0     6z − λ
The eigenvalues are thus λ = 0, λ = 4 and λ = 6z. Since all eigenvalues are ≥ 0, we
conclude that the Hessian is positive semideﬁnite, hence the function is convex. Since there
are no stationary points in S (and since S is open), f does not have a global minimum or
maximum.

Problem 8.

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Solution. The function
f (x, y, z) = x4 + y 4 + z 4 + x2 + y 2 − xy + zy + z 2
has the Hessian matrix,
12x2 + 2
                                      
−1                0
   −1     12y 2 + 2           1       .
0         1           12z 2 + 2
The leading principal minors are
D1 = 12x2 + 2 > 0
12x2 + 2   −1
D2 =
−1     12y 2 + 2
= 24x2 + 24y 2 + 144x2 y 2 + 3 > 0
12x2 + 2   −1               0
D3 =      −1     12y 2 + 2          1
0         1          12z 2 + 2
= 36x2 + 48y 2 + 36z 2 + 288x2 y 2 + 288x2 z 2 + 288y 2 z 2 + 1728x2 y 2 z 2 + 4 > 0
So the Hessian is positive deﬁnite, and hence f is convex.

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