Homework/Programming 7. (Solution)
Math. 417
I will give the solution for the work out problems in this note.
Problem 1. (Workout) A symmetric matrix is always diagonalizable (this
is a spectral theorem). This problem involves the explicit form of the eigenvectors for A3 and the computation of the corresponding eigenvaluess.
(a) Show that the vectors φj ∈ Rn given by
φj = (sin(πj/(n + 1)), sin(2πj/(n + 1)), . . . , sin(nπj/(n + 1)))t ,
for j = 1, 2, . . . , n, are the n eigenvectors of A3 . Hint: compute the l′ th
entry of A3 φj and, using trigonometric identities, show that it equals a
multiple (not depending on l) times (φj )l .
(b) Compute the corresponding eigenvalues (the multiple appearing in in
Part (a) above).
(c) Let G be the matrix corresponding to the Jacobi method applied to the
problem A3 x = b with x, b ∈ Rn . Show that ρ(G) is less than one. (Hint:
Write G as a linear combination of the matrix A3 and the identity matrix
so that the eigenvalues of G can be trivially calculated from those of A3 .)
Solution:
Part(a): Note that it is important to keep the indices straight in this problem. There is the index j which enumerates the eigenvectors, j = 1, . . . , n.
There are also indices which refer to components in the vectors.
The first thing to do is to apply the matrix A3 to the vector φj . After studing
the structure of A3 and φj , you should obtain

2 sin(πj/(n + 1)) − sin(2πj/(n + 1))





when l = 1,






 − sin((l − 1)πj/(n + 1)) + 2 sin(lπj/(n + 1))
− sin((l + 1)πj/(n + 1))
(1.1)
(A3 φj )l =



when l = 2, . . . , n − 1,





− sin((n − 1)πj/(n + 1)) + 2 sin(nπj/(n + 1))




when l = n.
Here the subscript l on the right denotes the l’th entry of the vector A3 φj .
We need to show that
(1.2)
(A3 φj )l = λj (φj )l
1
2
for some constant λj not depending on l.
We consider the intermediate case, i.e., l = 2, . . . , n − 1. Using the trigonometric identity:
sin(a + b) = sin(a) cos(b) + cos(a) sin(b)
we find
sin((l−1)πj/(n+1)) = sin(lπj/(n+1)) cos(πj/(n+1))−cos(lπj/(n+1)) sin(πj/(n+1))
(here we took a = lπj/(n + 1) and b = −πj/(n + 1) and used the identities
cos(−x) = cos(x) and sin(−x) = − sin(x)). Similarly,
sin((l+1)πj/(n+1)) = sin(lπj/(n+1)) cos(πj/(n+1))+cos(lπj/(n+1)) sin(πj/(n+1))
(here we took a = lπj/(n + 1) and b = πj/(n + 1) and used the identities
cos(−x) = cos(x) and sin(−x) = − sin(x)). Adding the above two equations
and multiplying by minus one shows that
(A3 φj )l = − sin((l − 1)πj/(n + 1)) + 2 sin(lπj/(n + 1)) − sin((l + 1)πj/(n + 1))
= (2 − 2 cos(πj/(n + 1))) sin(lπj/(n + 1))
so that (1.2) holds with λj = 2 − 2 cos(πj/(n + 1)) when l = 2, . . . , n − 1.
We need only show that (1.2) holds for l = 1 and l = n (with the same value
of λj ). We note that
2 sin(πj/(n + 1)) − sin(2πj/(n + 1))
= − sin(0πj/(n + 1)) + 2 sin(1πj/(n + 1)) − sin(2πj/(n + 1))
as the first term on the right hand side is zero. Now that right hand side
has the same form as the intermediate value case (it is just l = 1) so the
argument for the intermediate value case implies that (1.2) holds for l = 1.
Similarly,
− sin((n − 1)πj/(n + 1)) + 2 sin(nπj/(n + 1))
= − sin((n − 1)πj/(n + 1)) + 2 sin(nπj/(n + 1)) − sin((n + 1)πj/(n + 1))
since the last term on the right is sin(πj) = 0. This again reduces to the
intermediate case with l = n. This completes part (a).
Part(b): There is really nothing to do here except observe that the constants of part (a) are the eigenvalues, i.e., the eigenvalues of A3 are {2 −
2 cos(πj/(n + 1)) : j = 1, . . . , n}.
Part(c): This was done in class.
GJ = −D−1 (L + U ) = −D−1 (L + D + U − D) = I − D−1 A3 .
3
As D = 2I, GJ = (I − 1/2A3 ). It follows that GJ and A3 share the same
eigenvectors and λj (GJ ) = 1 − 1/2λj = cos(πj/(n + 1)). Now,
n
ρ(GJ ) = max | cos(πj/(n + 1))|.
j=1
As cos(x) is a decreasing function on [0, π], it absolute value takes on the
maximum at one of the two endpoints, i.e.,
ρ(GJ ) = max{cos(π/(n + 1)), | cos(πn/(n + 1))|}.
Now, using the trigonometic identity for cos(a + b) gives
cos(πn/(n + 1)) = cos(π[(n + 1) − 1]/(n + 1))
= cos(π) cos(π/(n + 1)) + sin(π) sin(π/(n + 1)) = − cos(π/(n + 1)),
i.e, ρ(Gj ) = cos(π/(n + 1)).
Problem 2. (Workout) Write down the arrays VAL, CIND and RIND corresponding to the matrix A3 when n = 6.
Solution: I will write down the matrices in modified CRS structure.
(i) VAL={2,-1,2,-1,-1,2,-1,-1,2,-1,-1,2,-1,-1,2,-1}
(ii) CIND={1,2,2,1,3,3,2,4,4,3,5,5,4,6,6,5}
(iii) RIND={1,3,6,9,12,15,17}