Å A Š × ç ü ç î ¬tpç5t =Ú (!€bç t æ) [u 3Ü/ 1Ü]

Å A Š × ç
ü ç   ^î ¬tpç5t ^{= Ú} (!€bç t æ) [u 3Ü/  1Ü]

Part I. LINEAR ALGEBRA

1. (8 points) Let A be a 3 × 4 matrix over the real number field R, and let {(2, 3, 1, 0)} be a basis for the nullspace of A.

(a) What is the rank of A and the complete solution to Ax = 0?

Solution: The dimension of the nullspace is 1, so the rank of A is 4 − 1 = 3. The complete solution to Ax = 0 is x = c · (2, 3, 1, 0) for any constant c.

(b) Find a basis for the column space of A^T.

Solution: The column space of A^T is the row space of A. The nonzero rows of the row reduced echelon form



1 0 −2 0 0 1 −3 0

0 0 0 1



 form a basis.

2. (a) (4 points) The linear transformation T : R²→ R²reflects a vector about the line y = −x and then projects that vector orthogonally onto the x-axis. Find the standard matrix for T .

Solution: T

·1 0

¸

=

·0 0

¸ and T

·0 1

¸

=

·−1 0

¸

so the matrix representation for T is

·0 −1

0 0

¸ .

(b) (4 points) Suppose T : R⁴→ R² is a linear transformation with T (1, 0, 0, 1) = (2, 3) and T (0, 1, 1, 0) = (1, 5). Find T (6, 2, 2, 6).

Solution: Let v1= (1, 0, 0, 1) and v2= (0, 1, 1, 0). Then v = (6, 2, 2, 6) = 6v1+ 2v2. By linearity, T (v) = T (6v1+ 2v2) = 6 T (v1) + 2 T (v2) = 6 (2, 3) + 2 (1, 5) = (14, 28)

3. (8 points) Suppose the 3 × 3 matrix A over R has eigenvalues 0, 1, and 2 with eigenvectors v0, v1, v2, respectively.

(a) What is the trace of A − 2I?

Solution: A − 2I has eigenvalues −2, −1, 0 so its trace is −3.

(b) Solve the equation Ax = v1+ v2for x.

Solution: x = av0+ v1+¹₂v2.

4. (16 points) Let V be the vector space R³over R. The following matrix is a projection matrix on V : P =₂₁¹



 1 2 −4

2 4 −8

−4 −8 16



.

(a) What subspace W of V does P project onto?

Solution: The projection matrix P projects onto the column space of P which is the line c · (1, 2, −4).

(b) What is the distance from that subspace W to b = (5, 4, −2)?

Solution: The vector from b to the subspace is

e = b − Pb =



 5 4

−2



 −²¹₂₁



 1 2

−4



 =



4 2 2





and the distance is

kek =p

4²+ 2²+ 2²= 2√ 6.

(c) What are the three eigenvalues of P?

Solution: Since P projects onto a line, its three eigenvalues are 0, 0, 1. The eigenvectors for 0 are vectors orthogonal to (1, 2, −4). The eigenvector for 1 is (1, 2, −4).

(d) If you solve^du_dt = −Pu (notice minus sign) starting from u(0), the solution u(t) approaches a steady state as t → ∞. Describe that limit vector u(∞)?

Å A Š × ç
ü ç   ^î ¬tpç5t ^{= Ú} (!€bç t æ) [u 3Ü/  2Ü]

Solution: The solution u(t) to the differential equation has the form u(t) = v₁e^−t+ v₂where v₁is in W and v₂is in the orthogonal complement of W . Then u(∞) = v₂, which is the projection of u(0) onto the orthogonal complement of W . That is, u(∞) = u(0) − P(u(0)).

5. (10 points) Let n denote a positive integer, V denote an n-dimensional vector space, and T denote a linear operator on V . Suppose v ∈ V is a nonzero vector such that T^kv = 0 for some positive integer k. Show that Tⁿv = 0.

Solution: Suppose k is the smallest positive integer such that T^kv = 0. The vectors v, T v, T²v, . . . , T^k−1v are linearly indepen- dent so k ≤ n:

Suppose c0v + c1T v + c2T²v + · · · + ck−1T^k−1v = 0. Applying T^k−1 to both sides we get c0T^k−1v = 0 and so c0= 0. Now applying T^k−2to both sides we get c₁T^k−1v = 0 and so c₁= 0. Continuing in this fashion, we see that c_j= 0 for all j.

Part II. ADVANCED CALCULUS

6. (10 points) Let x₁=^π₂ and suppose that x_nare defined inductively for n = 2, 3, . . . by cos(x_n+1) = ^sin(x_xnⁿ⁾, 0 < x_n+1< ^π₂, for n = 1, 2, . . ..

(a) Prove that x_n+1< x_nfor n = 1, 2, . . .. Hint: You may find cos(x) =_dx^d sin(x) and the Mean Value Theorem useful.

Solution: By the Mean Value Theorem, there exists a 0 < c < xnsuch that sin xn= sin xn−sin 0 = cos c (xn−0) = cos c xn. This implies that xn+1= c < xn.

(b) Show that the sequence {x_n}^∞_n=1is convergent.

Solution: Since the set {xn|n ∈ N} is nonempty and bounded from below by 0, l = inf{xn|n ∈ N} exists. The definition of l implies that for eachε> 0 there exists an xkfor some k ∈ N such that l +ε> xk. The result of part (a) says that {xn}^∞_n=1is decreasing which implies that l +ε> xk≥ xn> 0 for all n ≥ k. Therefore, we get |xn− l| <εfor all n ≥ k which means that lim

n→∞xn= l.

(c) Find explicitly the limit x of the sequence {xn}^∞_n=1.

Solution: Let f (x) = x cos x − sin x for x ∈ [0,^π₂]. To find all possible limits is equivalent to find zeros of f over [0,^π₂].

Note that f (0) = 0 and f⁰(x) = −x sin x < 0 for all x ∈ (0,^π₂], we can conclude that x = 0 is the only zero of f . Hence,

n→∞limx_n= 0.

7. (a) (5 points) State what it means for a sequence { f_n(x)}^∞_n=1of real valued functions on a set X ⊂ R^pto converge uniformly to a function f on X.

Solution: { fn}^∞_n=1is said to converge uniformly on X to f if for eachε> 0 there exists a K(ε) ∈ N such that for all n ≥ K(ε) and x ∈ X then | fn(x) − f (x)| <ε.

(b) (5 points) Let F : [0, 1] × [0, 1] → R be a continuous function on the unit square. Let fn(t) = F(¹_n,t) and f (t) = F(0,t).

Use the definition in part (a) to show that { fn(x)}^∞_n=1converges uniformly to f on the interval [0, 1].

Solution: Since F is continuous on [0, 1] × [0, 1], F is uniformly continuous there. This implies that for anyε> 0 and any t ∈ [0, 1], there exists aδ =δ(ε) > 0 such that if |x − 0| <δ then |F(x,t) − F(0,t)| <ε. Letting K(ε) = [_δ¹] + 1, where [_δ¹] is defined to be the greatest integer less than or equal to _δ¹, we note that n ≥ K(ε) > _δ¹ implies that ¹_n <δ. Thus, | f_n(t) − f (t)| = |F(¹_n,t) − F(0,t)| <εfor all n ≥ K(ε). Thus { f_n}^∞_n=1converges uniformly to f on [0, 1]

8. (10 points) For each integer k ≥ 1, let fk: R → R be differentiable satisfying | f_k⁰(x)| ≤ 1, for all x ∈ R, and fk(0) = 0.

(a) For each x ∈ R, prove that the set { fk(x)}^∞_k=1is bounded.

Solution: For each x ∈ R, since | f_k(x)| = | f_k(x) − f_k(0)| ≤ | f_k⁰(c_k) (x − 0)| ≤ |x|, where c_klies between x and 0, the set { f_k(x)}^∞_k=1is bounded.

(b) Use Cantor’s diagonal method to show that there is an increasing sequence n1< n2< n3< · · · of positive integers such that, for every x ∈ Q, we have { f_nk(x)} is a convergent sequence of real numbers.

Å A Š × ç
ü ç   ^î ¬tpç5t ^{= Ú} (!€bç t æ) [u 3Ü/  3Ü]

Solution: Let Q = {x₁, x₂, · · · }. Since { f_k(x₁)} is bounded, we can extract a convergent subsequence, denoted { f_k¹(x₁)}, out of { f_k(x₁)}. Next, the boundedness of { f_k¹(x₂)} implies that we can extract a convergent subsequence, denoted { f_k²(x₂)}, out of { f_k¹(x₂)}. Continuing this way, the boundedness of { f_k^j(x_j+1)} implies that we can extract a convergent subsequence { f_k^j+1(x_j+1)}, out of { f_k^j(x_j+1)} for each j ≥ 1. Let f_nk = f_k^kfor each k ≥ 1. Then { f_nk} is a subsequence of { f_n} and { f_nk} converges at each x_j∈ Q.

9. (10 points) A function f : R → R is said to be convex if for all x, y ∈ R,λ ∈ [0, 1], f (λx + (1 −λ)y) ≤λf (x) + (1 −λ) f (y).

Suppose that f : R → R is convex and that f⁰⁰(x) exists for all x ∈ R.

(a) For any x < y and 0 < h < y − x, prove that

f (x + h) ≤y − x − h

y − x · f (x) + h

y − x· f (y), and f (y − h) ≤ h

y − x· f (x) +y − x − h y − x · f (y).

Solution: Setting x + h =λx + (1 −λ)y, we getλ =^y−x−h_y−x and 1 −λ=_y−x^h . Thus f (x + h) = f (^y−x−h_y−x x +_y−x^h y) ≤^y−x−h_y−x f (x) +_y−x^h f (y).

Setting y − h =βx + (1 −β)y, we getβ=_y−x^h and 1 −β =^y−x−h_y−x . Thus

f (y − h) = f (_y−x^h x +^y−x−h_y−x y) ≤_y−x^h f (x) +^y−x−h_y−x f (y).

(b) Prove that f⁰(x) ≤ f⁰(y) whenever x ≤ y.

Solution: Since

f⁰(x) − f⁰(y) = lim

h→0⁺

f (x+h)− f (x)

h − f (y−h)− f (y)

−h = lim

h→0⁺

f (x+h)+ f (y−h)− f (x)− f (y)

h ≤ lim

h→0⁺

f (x)+ f (y)− f (x)− f (y)

h = 0,

where we have used the result of part (a) in the last inequality. We have shown that f⁰(x) ≤ f⁰(y) whenever x ≤ y.

(c) Prove that f⁰⁰(x) ≥ 0 for all x ∈ R.

Solution: The result of part (b) implies that f⁰⁰(x) = lim

h→0⁺

f⁰(x+h)− f⁰(x)

h ≥ 0, we have f⁰⁰(x) ≥ 0 for all x ∈ R.

10. (10 points) Let g : R^p→ R^pbelong to class C¹(R^p), i.e. Dg(x) exists for all x ∈ R^pand the mapping x → Dg(x) is continuous.

Assume that there is a constant a such that kDg(x)k ≤ a < 1 for each x ∈ R^p.

(a) Show that the function f (x) = x + g(x) for x ∈ R^psatisfies k f (x₁) − f (x₂) − (x₁− x₂)k ≤ akx₁− x₂k for all x₁, x₂∈ R^p. Solution: The Mean Value Theorem implies that there exists a z =λx₁+ (1 −λ)x₂∈ R^pfor someλ ∈ [0, 1] such that

k f (x1) − f (x2) − (x1− x2)k = kDg(z) · (x1− x2)k ≤ akx1− x2k.

(b) Show that f in part (a) is a bijection of R^pinto R^p.

Solution: Since D f = I + Dg and the eigenvalues of Dg are bounded by a < 1, D f is invertible everywhere. The Inverse Function Theorem implies that f is a local bijection. Since the result of part (a) says that f is a global one-to-one function of R^pinto R^p, f is a bijection of R^pinto R^p.