1. a)Let A = (\begin{matrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ) Show that for any 3 × 3 matrix B such

that AB = BA we can find c_{0}, c_{1}, c_{2} ∈ ℝ such that

B = c_{0}Id + c_{1}A + c_{2}A^{2}

Proof. If you write the matrix elements of matricies AB and BA you find that AB = BA iff there exists a sequence c_{k}, 0 ≤ k ≤ 2 such that b_{ij} = 0 for i > j and b_{ij} = c_{j−i} for i ≤ j. Then it is easy to see that

B = c_{0}Id + c_{1}A + c_{2}A^{2}

b)Generalize the statement [and IF possible the proof] to the case of n × n matricies.

Let A = (a_{ij}) be an n × n matrix such that a_{ij} = 1 if j − i = 1 and a_{ij} = 0 otherwise. Let B be an n × n matrix such that AB = BA. Then there exists a sequence c_{k}, 0 ≤ k ≤ n − 1 such that

B = ∑_{0≤i≤n−1} c_{i}A^{i}

Proof. If you write the matrix elements of matricies AB and BA you find that AB = BA iff there exists a sequence c_{k}, 0 ≤ k ≤ n − 1 such that b_{ij} = 0 for i > j and b_{ij} = c_{j−i} for i ≤ j. Then it is easy to see that

B = ∑_{0≤i≤n−1} c_{i}A^{i}

2.Let V be the set of sequences v̄ = (v_{n}), 1 ≤ n < ∞ such that v_{n+2} = v_{n+1} + v_{n}, 1 ≤ n < ∞. We define the operations of addition and the scalar multiplication on V by

v̄′ + v̄′′ := v̄, v̄ := (v′_{n} + v′′_{n}), 1 ≤ n < ∞ for v̄′ = (v′_{n}), v̄′′ = (v′′_{n})

cv̄ = (cv_{n}), 1 ≤ n < ∞ for v̄ = (v_{n})

a) Show that V is a finite-dimensional vector space and find it’s dimension.

I’ll leave for you to check the axioms.

Here are two ways to find the dimension of V.

Consider the map R : V → ℝ^{2} given by R(v̄) := (v_{1}, v_{2}). One can easily see that R is an isomorphism of vector spaces. Therefore dim(V) = 2.

Another proof. Let v̄′ = (v′_{n}) ∈ V the sequence such that v′_{1} = 1, v′_{2} = 0, [v′_{3} = v′_{1} + v′_{2} = 1, ...] and v̄′′ = (v′′_{n}) ∈ V the sequence such that v′′_{1} = 0, v′′_{2} = 1. It is easy to see that B := (v̄′, v̄′′) is a basis of V. So dim(V) = 2.

b) For any sequence v̄ = (v_{n}) ∈ V we denote by T v̄ the sequence T v̄ := w̄ where w̄ = (w_{n}), 1 ≤ n < ∞, w_{n} := v_{n+1}.

Show that for any v ∈ V we have T v̄ ∈ V,

that the operator T : V → V is linear and

write a matrix A_{T}^{B} for T in a basis B of V [ choose one].

I’ll leave for you to check that T is linear. It is clear that T v̄′ = v̄′′, T v̄′′ = v̄′ + v̄′′. So the operator T in the basis B is given by the matrix

(\begin{matrix}0 & 1 \\ 1 & 1 \end{matrix} )

3.Let V be a vector space of dimension 5, V′, V′′ ⊂ V subspaces such that dimV′ = dimV′ = 3. Show that V′ ∩ V′′ ≠ {¯0}.

Proof. We assume that V′∩V′′ = {¯0} and show that this assumption leads to a contrudiction.

Let (e′_{1}, e′_{2}, e′_{3}) be a basis of V′ and (e′′_{1}, e′′_{2}, e′′_{3}) be a basis of V′′.

Claim. If V′ ∩ V′′ = {¯0} then 6 vectors (e′_{1}, e′_{2}, e′_{3}, e′′_{1}, e′′_{2}, e′′_{3}) are linearly independent.

Proof of the claim. Let (c′_{1}, c′_{2}, c′_{3}, c′′_{1}, c′′_{2}, c′′_{3}) ∈ ℝ be numbers such that c′_{1}e′_{1} + c′_{2}e′_{2} + c′_{3}e′_{3} + c′′_{1}e′′_{1} + ce′′_{2} + ce′′_{3} = ¯0. Then

c′_{1}e′_{1} + c′_{2}e′_{2} + c′_{3}e′_{3} = −[c′′_{1}e′′_{1} + ce′′_{2} + c_{3}e′′_{3}]

. But the right side belongs to V′ and the left side side belongs to V′′. Since V′ ∩ V′′ = {¯0} we see that c′_{1}e′_{1} + c′_{2}e′_{2} + c′_{3}e′_{3} = −[c′′_{1}e′′_{1} + ce′′_{2}e′′_{2} + ce′′_{3}e′′_{3}] = ¯0. Since (e′_{1}, e′_{2}, e′_{3}) is a basis of V′ and (e′′_{1}, e′′_{2}, e′′_{3}) is a basis of V′′ we see that (c′_{1}, c′_{2}, c′_{3}, c′′_{1}, c′′_{2}, c′′_{3}) = (0, 0, 0, 0, 0, 0). We see that (e′_{1}, e′_{2}, e′_{3}, e′′_{1}, e′′_{2}, e′′_{3}) are linearly independent.

Since dimV = 5 we know that one can not find 6 linearly independent vectors in V. The contrudiction. So our assumption that V′∩V′′ = {¯0} was wrong and V′ ∩ V′′ ≠ {¯0}.

4. Let S^{1} := {(x, y) ∈ ℝ^{2}|x^{2} + y^{2} = 1}.

a) Show that the set S^{1} ⊂ ℝ^{2} is compact. Proof one easily checks that S^{1} is bounded and that the complement ℝ^{2} − S^{1} is open. So S^{1} is bounded and closed and therefore compact.

b) Show that S^{1} is connected [ that is show that for any closed subset X of S^{1} such that the set S^{1} − X is closed we either have X = ∅ or X = S^{1}.

Consider the map f : ℝ → S^{1} given by

θ → (cos(θ), sin(θ)), θ ∈ ℝ

It is easy to see that f is continuous. Therefore for any closed subset Y ⊂ S^{1} the preimage f^{−1}(Y) ⊂ ℝ is also closed.

Let X be a closed subset X of S^{1} such that the set S^{1} − X is also closed, Z := f^{−1}(X) ⊂ ℝ. Then ℝ − Z = f^{−1}(S^{1} − X). Since X and S^{1} − X are closed we see that Z and ℝ − Z are also closed. But we know [ a mid-term exam] that this is possible only if either Z = ∅ or Z = ℝ. But in this case either have X = ∅ or X = S^{1}.

5. Let f be a function on ℝ, f(x) := \frac{sin(x^{2})}{1+x^{2}}.

Find whether f is a uniformely continuous function on ℝ [and give a proof to justify the conclusion].

Here are two ways to prove that f is a uniformely continuous.

The first proof. You check that the derivative f′(x) is bounded. That is there exists C > 0 such that |f′(z)| < C for all z ∈ ℝ.

Now we prove that f is a uniformely continuous. For this we have to show that for any ϵ > there esists δ > 0 such that for all x, y ∈ ℝ, |x − y| < δ we have |f(x) − f(y)| ≤ ϵ.

Take δ = ϵ/C. By the MVT for any x, y ∈ ℝ there exists z ∈ ℝ such that f(x)−f(y) = f′(z)(x−y). Since |f′(z)| < C for all z ∈ ℝ we have |f(x)−f(y)| ≤ C|x−y|. So if |x−y| < δ we have |f(x)−f(y)| ≤ Cδ = ϵ.

The second proof. We check that f(z) → 0 for |z| → ∞.

Now we prove that f is a uniformely continuous. For this we have to show that for any ϵ > there esists δ > 0 such that for all x, y ∈ ℝ, |x − y| < δ we have |f(x) − f(y)| ≤ ϵ.

Since f(z) → 0 for |z| → ∞ we can find N > 1 such that |f(z)| < ϵ/2 for |z| > N − 1. Now the interval [−N, N] is compact. Since f is continuous it’s restriction to the interval [−N, N] is uniformely continuous. Therefore we can find δ > 0 such that for all x, y ∈ [−N, N], |x−y| < δ we have |f(x) − f(y)| ≤ ϵ. We can assume that δ < 1.

Now I claim that for all x, y ∈ ℝ, |x−y| < δ we have |f(x)−f(y)| ≤ ϵ. Really, since |x − y| < δ < 1 either x, y ∈ [−N, N] or |x| > N − 1 and |y| > N − 1.

In the first case [ by the definition of δ] we have |f(x) − f(y)| ≤ ϵ. In the second case we have |f(x)| < ϵ/2 and |f(y)| < ϵ/2 and there-

fore |f(x) − f(y)| ≤ ϵ.

6. Let V_{n} be the space of polynomials p(x) such that degree p ≤ n. Consider a function F : V_{10} → V_{20} given by F (p) := p^{2}.

a) Show that the function F is differentiable at all p ∈ V where V := V_{10}.

and

b) Find D_{F} (p) : V_{10} → V_{20} for p(x) ≡ 1.

Proof. Let W := V_{20}. For any p ∈ V_{n} we have F (p + h) − F (p) = 2hp + h^{2}. Define L_{P} : V → W by L_{P} (h) := 2ph. It is clear that L_{P} is a linear map and |F (p + h) − F (p) − L_{P} (h)|/|h| = |h|^{2}/|h| → 0 for |h| → 0. So the function F is differentiable at all p ∈ V and the linear map D_{F} (p) : V_{10} → V_{20} is given by D_{F} (p)(h) = 2ph.

c)Show that all partial derivatives of F of order 3 are equal to 0.

If you write F in coordinates F = (F_{i}, 0 ≤ i ≤ 20), F_{i} = F_{i}(x_{j}), 0 ≤ j ≤ 10 you will see that all the functions F − i are quandratic polynomials. Therefore all partial derivatives of F of order 3 are equal to 0.

7. Let f be a function on ℝ − {−1} given by f(x) := ln(1 + x) and let p_{f}^{n}(1)(x) := ∑_{0≤i≤n} \frac{f^{(i)}(1)}{i!}(x − 1)^{i} be the Taylor polynomial for f at the point 1.

Show that |p_{f}^{n}(1)(0)| < 1/2^{n}.

Proof. We first show that the Taylor series p_{f}^{n}(1)(x) := ∑_{0≤i<∞} \frac{f^{(i)}(1)}{i!}(x− 1)^{i} is convergent for x = 0. For this we have to show that R_{n}(0) → 0 for n → ∞ where R_{n}(x) is the remainder. By the formula for the remainder [see p.182] we have R_{n}(0) = \frac{f^{(n+1)}(c)}{(n+1)!}(c − 1)^{(n+1)} for some c, 0 ≤ c ≤ 1. By induction you can show that

\frac{f^{(}n + 1)(c)}{(n + 1)!} = (−1)^{n}(1 + c)^{−(n+1)/(n+1)}, n ≥ 0

. So R_{n}(0) → 0 for n → ∞.

We have f(0) = ln(1) = 0 and therefore

R_{n}(0) = f(0)−p_{f}^{n}(1)(x) = ∑_{n+1≤i<∞} \frac{f^{(i)}(1)}{i!}(x−1)^{i} = ∑_{n≤i<∞}(−1)^{n}2^{−(n+1)}/(n+1)

Now it is easy to show that |R_{n}(0)| < 1/2^{n}

8. Let u(x, y) be a differentiable function on ℝ^{2} such that u(x, x^{2}) = 1 for all x ∈ ℝ. Suppose that ∂u/∂x(2, 4) = 1. Find ∂u/∂y(2, 4).

Let f(x) := u(x, x^{2}). Compute ∂f/∂x in two ways. First of all f(x) ≡ 1. So ∂f/∂x ≡ 0. On the other hand we can apply the chain rule. Then we obtain

∂f/∂x(a) = ∂u/∂x(a, a^{2}) + 2a∂u/∂y(a, a^{2})

In particular ∂f/∂x(2) = ∂u/∂x(2, 4) + 4∂u/∂y(2, 4). So 0 = 1 + 4∂u/∂y(2, 4). Or ∂u/∂y(2, 4) = −1/4

9. a)Let B : ℝ^{2} × ℝ^{2} → ℝ be an antisymmetric bilinear form B such that B is not identically zero. [ That is there exists a pair (v̄, w̄) of vectors in ℝ^{2} such that B(v̄, w̄) ≠ 0]. Show that there exists a basis f_{1}, f_{2} such that B(f_{1}, f_{2}) = 1.

Let (v̄, w̄) be vectors in ℝ^{2} such that B(v̄, w̄) = r ≠ 0. Take f_{1} := v̄, f_{2} := w̄/r. Then B(f_{1}, f_{2}) = 1.

Claim. (f_{1}, f_{2}) are linearly independent.

Proof. Suppose that there exist a_{1}, a_{2} ∈ ℝ such that a_{1}f_{1} +a_{2}f_{2} = ¯0. Then we have 0 = B(¯0, f_{1}) = B(a_{1}f_{1} + a_{2}f_{2}, f_{1}) = −a_{2}. So a_{2} = 0. Analogously one shows that a_{1} = 0. Claim is proved.

remark. Since dimℝ^{2} = 2 the linear independence of (f_{1}, f_{2}) implies that (f_{1}, f_{2}) is a basis of ℝ^{2}.

b)Let B : ℝ^{3} × ℝ^{3} → ℝ be an antisymmetric bilinear form B such that B is not identically zero. Show that there exists a basis f_{1}, f_{2}, f_{3} in ℝ^{3} such that B(f_{1}, f_{2}) = 1, B(f, f_{3}) = 0 for all f ∈ ℝ^{3}

As before we can find f_{1}, f_{2} ∈ ℝ^{3} such that B(f_{1}, f_{2}) = 1. Consider now a pair of linear equations on v ∈ ℝ^{3} given by B(f_{1}, v) = 0, B(f_{2}, v) = 0. Since dimℝ^{3} = 3 > 2 this system of equations have a nonzero solution v ≠ ¯0.

Claim. v is not a linear combination of f_{1}, f_{2}.

Proof of Claim. Assume that v = a_{1}f_{1} + a_{2}f_{2}. Then B((f_{1}, v) = a_{1} and B((f_{2}, v) = a_{2} So a_{1} = a_{2} = 0. But by the construction v ≠ ¯0.

Since v is not a linear combination of f_{1}, f_{2} and [ as we have seen before] f_{1}, f_{2} are linearly independent three vectors f_{1}, f_{2}, v ∈ ℝ^{3} are linearly independent (?). Since dim(ℝ^{3}) = 3 we see that (f_{1}, f_{2}, v) is a basis of ℝ^{3}.

By the construction B(f_{1}, v) = 0, B(f_{2}, v) = 0. since B is antilinear we have B(v, v) = 0. Therefore (?) B(f, v) = 0 for all f ∈ ℝ^{3} and we can take f_{3} = v.