18.303 Problem Set 1 Solutions
Problem 1: 10+(5+10)+10 points
(a) It has a solution if b is in C(AT A) = N (A)⊥ , and since N (A) is spanned by n this means
it has a solution if b ⊥ n, i.e. nT b = 0 . Given any solution x of AT Ax = b, we can get
another solution if we add anything in N (AT A) = N (A) to x. i.e. all solutions are given by
x + αn for any scalar α. (Notice that, for this problem, the left nullspace of A is irrelevant.)
(b) Suppose that A is a real 21 × 21 matrix with eigenvalues 1, 12 , 212 , 213 ,. . ., 2120 and corresponding
eigenvectors x0 , x1 , x2 , . . . , x20 , respectively (that is, the eigenvalues are λk = 2−k for k =
0, . . . , 20).
(i) Since it has 21 distinct eigenvalues, A must be diagonalizable—defective matrices only
occur when there are repeated roots of the characteristic polynomial. It is not necessarily
symmetric: symmetric matrices have real eigenvalues, but the converse is not necessarily
true. [For example, suppose C is a symmetric matrix; a non-symmetric matrix with
the same real eigenvalues is the similar matrix A = B −1 CB for any non-orthogonal
invertible matrix B.]
(ii) First, let us write the solution of the recurrence in a simpler form: y2 = Ay1 = A2 y0 ,
y3 = Ay2 = A3 y0 , . . ., yn = An y0 . (You can also prove this more formally by induction,
but the pattern is obvious.)
P
Since A is diagonalizable, we can write y0 in the P
basis of the eigenvectors: y0 = k ck xk
for some coefficients ck . Then yn = An y0 = k ck λnk xn , since An simply multiplies
each eigenvector by λn . However, all of the eigenvalues have |λ| < 1 except for λ0 = 1,
which means that λn → 0 for large n. Thus, for large n, everything except for the x0
term goes to zero, and yn ≈ c0 x0 . However, we can be more explicit than this because
A was given to be symmetric in this part, which means that the eigenvectors are orxT y0
x0 xT
xT y
thogonal and we can write ck = xTk xk0 . Hence, yn ≈ T0 x0 = T 0 y0 for large n,
k
x0 x0
x0 x0
where for aesthetic reasons I rewrote it in the form of a projection matrix x0 xT0 /xT0 x0 .
If you want, you could use a more accurate approximation, and write yn as the sum
of thehx0 term plus the
i lowest-order correction by including the next-biggest term x1 :
x xT
x xT
yn ≈ xT0 x00 + 21n x1T x11 y0 .
0
1
Note that you can not assume that the eigenvectors are normalized to length = 1,
i.e. you cannot assume here that xT0 x0 = kx0 k2 = 1. Although one can always choose
the eigenvectors to be scaled in this way, it is not automatic, and their length was not
specified in this problem.
(c) Suppose x is an eigenvector of A, i.e. Ax = λx. Take the dot product of both sides with
x , i.e. multiply both sides with x∗ = xT (noting that we cannot assume that x is real),
obtaining:
x∗ Ax
= λx∗ x = λkxk2
∗
=
(A∗ x) x
=
(−Ax)∗ x = (−λx)∗ x = −λ̄kxk2 ,
where in the third line we have used the fact that A is real-antisymmetric, so that A∗ = AT =
−A. Then, since by definition an eigenvector has x 6= 0 and thus kxk 6= 0, by comparing the
1
first and third lines we obtain λ = −λ̄ , and hence the eigenvalue is purely imaginary.
Problem 2: 10+5+10 points
(a) The eigenfunctions must still be sines, cosines, or exponentials, as noted in class, with the
boundary conditions determining which one. u(0) = 0 means that they must be sines. To get
u0 (L) = 0, they must reach a maximum or a minimum at L (a phase that is an odd multiple
of π/2), i.e. the eigenfunctions must be of the form:
1 πx
un (x) = sin n +
2 L
π 2
] . (Of course,
for integers n = 0, 1, 2, . . ., with corresponding eigenvalues λn = −[(n + 12 ) L
any constant multiple of this works as well, as well as n < 0, but those variations do not
give linearly independent solutions.) We can check that these are orthogonal, using the trig.
identity sin A sin B = [cos(A − B) − cos(A + B)]/2:
Z
L
un (x)um (x)dx
=
0
=
1
2
Z
L
πx πx i
cos (n − m)
− cos (n + m + 1)
dx
L
L
0

L 
πx sin
(n
+
m
+
1)
L  sin (n − m) πx
L
L 
−
2π
n−m
n+m+1
h
0
=
0
for n 6= m (and hence distinct eigenvalues λn 6= λm ), since the sine of any integer multiple of
π is zero.
(b) The are not a vector space, e.g. 2v(x) will not have the same boundary values.
(c) If we simply write v(x) = u(x) + 2 + x, for example, then u(0) = 0, u0 (L) = 0 implies v(0) = 2,
d2
v 0 (L) = 1; that is, q(x) = u(x) − v(x) = −2 − x. Plugging v = u − q into dx
2 v(x) = g(x)
d2
00
00
yields dx2 u(x) = f (x) with f (x) = g(x) + q (x) = g(x). That is, r(x) = q (x) = 0.
This is not the only possible answer, of course. For example, q(x) = −2 −
gives r(x) = q 00 (x) = −1/L.
x2
2L
also works, and
Problem 3: 10+15 points.
In 18.303, we will mostly solve PDEs in convenient arbitrary units where we don’t have any pesky
dimensionful constants flapping around the equations and annoying us. This represents no loss of
generality because, even if someone hands you a a PDE in SI units, you can always rescale things
to get back into dimensionless form.
∂
(a) If we rescale x̃ = x/X and t̃ = t/T , then by a simple 18.02 exercise ∂t
= T1 ∂∂t̃ (dt = T dt̃)
˜ 2 . Therefore, the Schrodinger equation in these new coordinates
and similarly ∇2 = X12 ∇
becomes:
~2 ˜ 2
~ ∂ ψ̃
−
∇ + V (x) ψ̃ = i
.
2
2mX
T ∂ t̃
We then obtain the desired equation
h
i
˜ 2 + V (X x̃) ψ̃ = i ∂ ψ̃
−∇
∂ t̃
2
√
by choosing X = ~/ 2m and T = ~.
This looks a bit strange because our x̃ and t̃ are now in “weird” units (not dimensionless). If we
wanted to “fix” this, we would divide through the whole equation by some typical energy scale
V0 . This will replace
V by the dimensionless Ṽ (x̃) = V (X x̃)/V0 , and in that case we would
√
obtain X = ~/ 2mV0 and T = ~/V0 , which have units of distance and time respectively so
that x̃ and t̃ become dimensionless quantities.
˜ 2 u = ∂ 2 2u . As
(b) As suggested, let’s begin by rewriting the u equation in “tilde” coordinates: ∇
∂ t̃
in the previous part, we want to relate these to the “real” coordinates by some rescaling:
x̃ = x/X and t̃ = t/T for some X and T . Similar to (or rather, the reverse of) the previous
T 2 ∂2u
part, this will give the transformed equation ∇2 u = X
2 ∂t2 in the (x, t) coordinates.
Now, what are X and T ? First, the X rescaling has to transform from x̃ on the unit circle to
x on the circle of radius R, so clearly X = R = 0.2 m. Second, by comparison of the u and h
equations, c = X/T and thus we must have T = X/c = R/c = 0.001333̄ s.1
Finally, in addition to rescaling the coordinates, u can be related to h by some overall scale
factor H, since this just corresponds to multiplying both sides of the wave equation by H:
u(x̃, ỹ, t̃) = h(Rx̃, Rỹ, Rt̃/c)/H = h(x, y, t)/H. To get H, we must compare the initial conditions: h(Rx̃, Rỹ, 0) = (0.1 m) sin[1 − (x̃2 + ỹ 2 )] versus u(x̃, ỹ, 0) = sin[1 − (x̃2 + ỹ 2 )]. Clearly,
H = 0.1 m. Thus:
h(x, y, t) = (0.1 m) u(x/R, y/R, ct/R) .
1 This
corresponds to a typical timescale of this system, which here is the time for waves to propagate from the
center to the edge of the drum.
3