Conrail Case Study (A) Memo
(due in class 11/28)
1. Why does CSX want to buy Conrail? Why can CSX justify paying a premium
to acquire Conrail?
2. Why would the Surface Transportation Board (STB) likely approve the merger
(i.e., why might the STB not be to
INTERCO Case Study Memo (due start of class 9/12)
1. Assess Intercos financial performance. Why is Interco a target of a hostile
takeover attempt?
2. As a member of Intercos board, you are presented a historical Premiums
Paid Analysis in Exhibit 10. This
Bethlehem Case Study Memo (due start of class 9/26)
1. Many U.S. corporations have offered defined-benefit (DB) pension plans. Who are all
the stakeholders in this U.S. corporate defined-benefit system?
2. For what type of firms and institutions does offe
Dividend Policy at Linear Technology Case Study (due 10/26)
QUESTIONS
1. As an investor yourself, would you rather a firm pays you a lot of dividends or would you
rather simply earn capital gains? In other words, would you rather your return for investing
AT&T Case Study Memo
(due in class 11/9)
1. Review AT&Ts past financial policies and financing choices. Were these
appropriate for the nature of the business?
2. In what fundamental ways will AT&Ts business change in the near future?
3. In view of AT&Ts c
determinant of A is the product of the eigenvalues. And if
condition I holds, we already know that these
eigenvalues are positive. But we also have to deal with
every upper left submatrix Ak . The trick is to look at all
nonzero vectors whose last nk comp
TAx > 0 all > 0 orthogonal Similar matrix: B = M1AM
(B) = (A) x(B) = M1 x(A) Projection: P = P 2 = P T = 1;0
column space; nullspace Reflection: I 2uuT = 1;1,.,1
u;u Rank-1 matrix: uvT = v Tu;0,.,0 u; v Inverse: A
1 1/(A) eigenvectors of A Shift: A+cI (A)
see why. If you multiply any R by a matrix Q with
orthonormal columns, then (QR) T (QR) = R TQ TQR = R T
IR = A. Therefore QR is another choice. Applications of
positive definite matrices are developed in my earlier
book Introduction to Applied Mathematic
diagonal) is between 1 and n. You can see this in Figure
6.6, where the horizontal distance to the ellipse (where
a11x 2 = 1) is between the shortest distance and the
longest distance: 1 n 1 a11 1 1 which is 1
a11 n. The diagonal entries of any symmetric
has a spectral decomposition into A = 1P1+kPk ,
where Pi is the projection onto the eigenspace for i .
Since there is a full set of eigenvectors, the projections
add up to the identity. And since the eigenspace are
orthogonal, two projections produce zero
is also practical, because the pivots can locate the
eigenvalues: A has positive pivots A2I has a negative
pivot A = 3 3 0 3 10 7 0 7 8 A2I = 1 3 0 3
8 7 0 7 6 . A has positive eigenvalues, by our test. But
we know that min is smaller than 2, because subt
can work with the nonsingular A+ I and AI, and at the
end let 0.) Proof. We want to borrow a trick from
topology. Suppose C is linked to an orthogonal matrix Q
by a continuous chain of nonsingular matrices C(t). At t =
0 and t = 1, C(0) = C and C(1) = Q.
space. Then A + has the explicit formula U T (U U T ) 1 (L
TL) 1L T . Why is A +b in the row space with U T at the
front? Why does A TAA+b = A Tb, so that x + = A +b
satisfies the normal equation as it should? 22. Explain
why AA+ and A +A are projection m
lower dimension. The major axis Of this cross section
cannot be longer than the major axis of the whole
ellipsoid: 1(B) 1(A). But the major axis of the cross
section is at least as long as the second axis of the
original ellipsoid: 1(B) 2(A). Similarly th
Singular Value Decomposition 371 that A must have
independent columns.) In the reverse order A = S 0Q, the
matrix S 0 is the symmetric positive definite square root
of AAT . 4. Least Squares For a rectangular system Ax = b.
the least-squares solution come
tet . The top equation is du1/dt = u1 + u2, and its
solution is 1 2 t 2 e t . When has multiplicity m with
only one eigenvector, the extra factor t appears m1
times. These powers and exponentials of J are a part of
the solutions uk and u(t). The other par
1AS = . Then the Jordan form coincides with the
diagonal . This is impossible for a defective
(nondiagonalizable) matrix. For every missing
eigenvector, the Jordan form will have a 1 just above its
main diagonal. The eigenvalues appear on the diagonal
bec
, but it is n by n. Remark 5. Here is the reason that Avj =
juj . Start with A TAvj = 2 j v j : Multiply by A AATAv j =
2 j Avj (2) This says that Avj is an eigenvector of AAT !
We just moved parentheses to (AAT )(Avj). The length of
this eigenvector Avj
combination acb 2 may be negative. This occurred in
both examples, when b dominated a and c. It also occurs
if a and c have opposite signs. Then two directions give
opposite resultsin one direction f increases, in the
other it decreases. It is useful to c
eigenvalues and eigenmatrices for A T = A. 16. (a) Find
an orthogonal Q so that Q 1AQ = if A = 1 1 1 1 1 1
1 1 1 and = 0 0 0 0 0 0 0 0 3 . 5.6
Similarity Transformations 337 Then find a second pair of
orthonormal eigenvectors x1, x2 for = 0. (b) Verify th
Example 2. In our original f the coefficient 2b = 4 was
positive. Does this ensure a minimum? Again the answer
is no; the sign of b is of no importance! Even though its
second derivatives are positive, 2x 2 +4xy+y 2 is not
positive definite. Neither F nor
that: P 1BP = " 0 1 1 0#"1 0 1 1#"0 1 1 0# = " 1 1 0 1# = J.
(A) From A to J, we go first to T as in equation (4). Then
change 2 to 1: U 1AU = " 1 2 0 1# = T and then M1T M
= " 1 1 0 1# = J. Example 5. A = 0 1 2 0 0 1 0 0 0
and B = 0 0 1 0 0 0 0 0 0 . Ze
There are no higher-order terms or lower-order terms
only second-order. The function is zero at x = (0,.,0), and
its first derivatives are zero. The tangent is flat; this is a
stationary point. We have to decide if x = 0 is a minimum
or a maximum or a sad
decomposed x TAx into a sum of squares: Sum of squares
ax2 +2bxy+cy2 = a x+ b a y 2 + acb 2 a y 2 . (1) Those
coefficients a and (ac b 2 )/a are the pivots for a 2 by 2
matrix. For larger matrices the pivots still give a simple
test for positive definiten
surface z = f(x,y) will then be shaped like a bowl, resting
on the origin (Figure 6.1). If the stationary point of F is at
x = , y = , the only change would be to use the second
derivatives at , : Quadratic part of F f(x,y) = x 2 2 2F
x 2 (,) +xy 2F x y
devoted more to theory than to applications. The Jordan
form is independent of this triangular form.) 5R There is a
unitary matrix M = U such that U 1AU = T is triangular.
The eigenvalues of A appear along the diagonal of this
similar matrix T. Proof. Eve
(Q )( Q T ), and (Q Q T )(Q Q T ). 7. If A = QQ
T is symmetric positive definite, then R = Q Q T is its
symmetric positive definite square root. Why does R have
positive eigenvalues? Compute R and verify R 2 = A for A
= " 10 6 6 10# and A = " 10 6 6 10# .
eigenvalues. The signs of the eigenvalues are often
crucial. For stability in differential equations, we needed
negative eigenvalues so that e t would decay. The new
and highly important problem is to recognize a minimum
point. This arises throughout scie
are in trouble. An indefinite equation y 2 1 9y 2 2 = 1
describes a hyperbola and not an ellipse. A hyperbola is a
cross-section through a saddle, and an ellipse is a crosssection through a bowl. The change from x to y = Q T x
rotates the axes of the spac