Math 191 Wilkening
Spring 2023
Homework 6
due Sat, Mar 11, 2:00 PM
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1. (2 points) (I.6.12, page 42, Strang.) The following matrix is singular with rank one: 1 2 1 2
A = 2 2 1 2 = 4 2 4 .
Find three linearly independent eigenvectors and the corresponding eigenvalues.
2. (3 points) (I.6.14, page 42, Strang.) Suppose A has eigenvalues 0,3,5 with corresponding eigenvectors u, v, w.
(a) Give a basis for the nullspace and a basis for the column space (range) of A. (b) Find a particular solution to Ax = v + w. Find the general solution.
(c) Ax = u has no solution. If it did, then ——— would be in the range of A.
3. (3 points) The Fibonacci sequence {xn}∞n=0 = (0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . . ) satisfies the
3-term recurrence
This can be turned into a one-step iteration by combining consecutive components of x into a vector:
xn+1 =xn−1 +xn, n=1,2,3,….
(un)0 xn un = (u ) = x
n=0,1,2,3,….
Note that un satisfies
u0=e1= 1 , un+1=Aun, (n≥0), A= 1 1 .
It follows that
un = Anu0, (n ≥ 0). So if we diagonalize A = UΛU−1, we have
un = UΛnU−1u0, xn = eT0 UΛnU−1e1.
Diagonalize A and work out the resulting formula for xn. What is the first n for which xn ≥ 10100?
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4. (2 points) (I.7.6, page 52, Strang.) The following matrix M is antisymmetric and also ———. So all of its eigenvalues are purely imaginary and they also satisfy |λ| = 1. (∥Mx∥ = ∥x∥ for every x implies that ∥λx∥ = ∥x∥ if x is an eigenvector with eigenvalue λ.) Find all four eigenvalues from the trace of M:
M = √1 −1 0 −1 1 can only have eigenvalues i or −i.
3 − 1 1 0 − 1 −1 −1 1 0
5.(2points)(I.7.9,page53,Strang.)IfSissymmetricandorthogonal,thenST =SandST =S−1.
(a) Show how symmetry and orthogonality lead to S2 = I.
(b) What are the possible eigenvalues of S? Describe all possible Λ that can result from diago- nalizing S.
6. (4 points) (Variant of I.7.10, page 53, Strang.) If A is a symmetric n × n matrix and X is square and invertible, XT AX is said to be congruent to A. In this case, show that XT AX and A have the same number of positive, negative and zero eigenvalues. This is known as the “law of inertia.” (Hint: let V+, V0 and V− denote the subspaces of Rn spanned by the eigenvectors of A with positive, zero and negative eigenvalues, respectively. Let ν+, ν0 and ν− denote their dimensions. Similarly, let W+, W0 and W− denote the subspaces of Rn spanned by the eigenvectors of XT AX with positive, zero and negative eigenvalues, respectively, and let μ+, μ0 and μ− denote their dimensions. Note that ν+ +ν0 +ν− = n = μ+ +μ0 +μ−. Show that ν0 = μ0. Then show that assuming ν+ > μ+ leads to a contradiction by producing a nonzero vector in V+ ∩ X(W0 + W−) and explaining why this is impossible. Reach a similar contradiction by assuming μ+ > ν+.)
7. (3 points) (I.7.16, page 54, Strang.) For what numbers c and d are S and T positive definite? Test their 3 leading determinants:
c 1 1 1 2 3 S=1 c 1 and T=2 d 4
8. (1 point) (I.7.18, page 54, Strang.) A positive definite matrix cannot have a zero (or even worse, a negative number) on its main diagonal. Show that the following matrix fails to have xT Sx > 0:
4 1 1x 1
x1 x2 x3 1 0 2x2 isnotpositivewhenx=? 125 x3
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9. (3 points) (I.7.21, page 54, Strang.) Draw the tilted ellipse x2 + xy + y2 = 1 and find the half- lengths of its axes from the eigenvalues of the corresponding matrix S for which x21 + x1x2 + x2 = xTSx.
10. (2 points) (Slight modification of I.7.22, page 54, Strang.) In the Cholesky factorization A = RT R, with R = √DLT , the square roots of the pivots are on the diagonal of R. Find R (upper triangular) for
9 0 0 1 1 1 A1=0 1 2 A2=1 2 2.
11. (2 points) (I.7.27, page 55, Strang.) For which a and c is the following matrix positive definite? For which a and c is it positive semidefinite (which includes positive definite)?
a a a S=a a+c a−c
Note: all 5 tests are possible. The energy xT Sx equals a(x1 + x2 + x3)2 + c(x2 − x3)2. Derive this
if you decide to solve it this way.
12. (3 points) (Variant of I.8.1-2, page 68, Strang.) Suppose S = ST is a real symmetric matrix with orthonormal eigenvectors v1, . . . , vn. Any vector x ∈ Rn can be written as a linear combination x = c1v1 + · · · + cnvn. Explain these two formulas:
xTx=c21 +c2 +···+c2n, xTSx=λ1c21 +λ2c2 +···+λnc2n. These combine to give a formula for the Rayleigh quotient,
ρ(x)= xTSx = λ1c21 +···+λnc2n. x T x c 21 + · · · + c 2n
Explain why the largest eigenvalue is the maximum value of ρ(x) over x ̸= 0 in Rn.