Chapter 2 Euclidean Space

2.1 Algebra: Vectors

2.1.1.

Write down any three specific nonzero vectors $u$, $v$, $w$ from $\mathbb{R}^3$ and any two specific nonzero scalars $a, b$ from $\mathbb{R}$. Compute $u+v, aw, b(v+w), (a+b)u$, $u+v +w, abw$, and the additive inverse to $u$.

Solution:

Let $$u = (1,0,0) \\ v = (0,1,0) \\ w = (0,0,1) \\ a = 1, b = -1 \\ u + v = (1, 1, 0) \\ aw = w = (0,0,1) \\ b(v + w) = (0,-1,-1) \\ (a+b)u = 0 \\ u+v+w = (1,1,1) \\ ab(w) = (0,0,-1) \\ $$

2.1.2.

Working in $\mathbb{R}^2$, give a geometric proof that if we view the vectors $x$ and $y$ as arrows from $0$ and form the parallelogram $P$ with these arrows as two of its sides, then the diagonal $z$ starting at $0$ is the vector sum $x + y$ viewed as an arrow.

Proof:

See the figure below.

2.1.3.

Verify that $\mathbb{R}^n$ satisfies vector space axioms (A2), (A3), (D1).

Proof:

(A2) $0$ is an additive identity

$$0 + x = (0, \cdots, 0) + (x_1, \cdots, x_n) \\ = (0 + x_1, \cdots, 0 + x_n) \\ = (x_1, \cdots, x_n) \\ = x$$

$\square$

(A3) Existence of additive inverses

Let $x = (x_1, \cdots, x_n)$ and $y = (-x_1, \cdots, -x_n)$

$$y + x = (-x_1 + x_1, \cdots, -x_n + x_n) \\ = (0, \cdots, 0)$$

So $y$ is the additive inverse of $x$.

$\square$

(D1) Scalar multiplication distributes over scalar addition

$$(a+b)x = (a+b) (x_1, \cdots, x_n) \\ = ((a+b)x_1, \cdots, (a+b)x_n) \\ = (ax_1 + bx_1, \cdots, ax_n + bx_n) \\ = (ax_1, \cdots, ax_n) + (bx_1, \cdots, bx_n) \\ = ax + bx$$

$\square$

2.1.4.

Are all the field axioms used in verifying that Euclidean space satisfies the vector space axioms?

Solution: Let's check the vector space axioms one by one.

(A1) Addition is associative uses (Fa1)

(A2) 0 is an additive identity uses (Fa2)

(A3) Existence of additive inverses (Fa3)

(A4) Addition is commutative (Fa4)

(M1) Scalar multiplication is associative (Fm1)

(M2) 1 is a multiplicative identity (Fm2)

(D1) Scalar multiplication distributes over scalar addition (Fd1)

(D2) Scalar multiplication distributes over vector addition (Fd1)

So we can see (Fm3) and (Fm4) are not used.

$\square$

2.1.5.

Show that $0$ is the unique additive identity in $\mathbb{R}^n$. Show that each vector $x ∈ \mathbb{R}^n$ has a unique additive inverse, which can therefore be denoted $−x$. (And it follows that vector subtraction can now be defined,

$$- : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}^n, \quad x - y = x + (-y)$$

for all $x, y \in \mathbb{R}^n$.

Show that $0x = 0$ for all $x ∈ \mathbb{R}^n$.

Proof:

Assume $x = (x_1, \cdots, x_n)$ is a additive identity and $y = (y_1, \cdots, y_n) \in \mathbb{R}^n$. Then $x + y = y$, i.e.

$$x_1 + y_1 = y_1, \cdots , x_n + y_n = y_n$$

So $x_1 = \cdots = x_n = 0$, so $x = 0$. And thus $0$ is unique.

Assume $x = (x_1, \cdots, x_n)$ and $y (y_1, \cdots, y_n) \in \mathbb{R}^n$ is $x$'s additive inverse.

$$x + y = (x_1 + y_1, \cdots, x_n + y_n) = 0$$

So

$$y = (-x_1, \cdots, -x_n)$$

The additive inverse is unique.

$$0x = 0 (x_1, \cdots, x_n) = (0x_1, \cdots, 0x_n) = (0, \cdots, 0)$$

$\square$

2.1.6.

Repeat the previous exercise, but with $\mathbb{R}^n$ replaced by an arbitrary vector space $V$ over a field $F$. (Work with the axioms.)

$$0 = 0' + 0 = 0 + 0' = 0'$$

So $0$ is unique.

$$y = y + 0 = y + (x + y') = (y + x) + y' = 0 + y' = y'$$

So the additive inverse is unique.

$$0x + 1x = (0 + 1) x = 1 x = x$$

Then we can add $-x$ on both side

$$0x + x + (-x) = 0x \\ \Rightarrow \\ x + (-x) = 0 \\ \Rightarrow \\ 0x = 0$$

$\square$

2.1.7.

Show the uniqueness of the additive identity and the additive inverse using only (A1), (A2), (A3). (This is tricky; the opening pages of some books on group theory will help.)

Proof:

I cannot figure it for now.

The uniqueness of the additive identity. Assume $0$ and $0'$ are 2 additive identities. Then

$$0' + 0 = 0 \\ 0 + 0' = 0' \\ 0' + 0' = 0' \\ $$

2.1.8.

Let $x$ and $y$ be noncollinear vectors in $\mathbb{R}^3$. Give a geometric description of the set of all linear combinations of $x$ and $y$.

Solution: All the linear combinations of $x$ and $y$ form a plane that includes $0, x, y$.

$\square$

2.1.9

Which of the following sets are bases of $\mathbb{R}^3$?

$$S_1 = \{(1,0,0),(1,1,0),(1,1,1)\} \\ S_2 = \{(1,0,0),(0,1,0),(0,0,1),(1,1,1)\} \\ S_3 = \{(1,1,0),(0,1,1)\} \\ S_4 = \{(1,1,0),(0,1,1),(1,0,-1)\} \\ $$

How many elements do you think a basis for $\mathbb{R}^n$ must have? Give (without proof) geometric descriptions of all bases of $\mathbb{R}^2$, of $\mathbb{R}^3$.

Solution:

For $S_1$, given $(x,y,z)$, we have

$$\begin{cases} x &= a + b + c\\ y &= b + c \\ z &= c \\ \end{cases}$$

Then we can have

$$\begin{cases} a = x - y\\ b = y - z\\ c = z\\ \end{cases}$$

So $S_1$ is a base.

For $S_2$, it is note a base. For example $(1,1,1)$ can be expressed by multiple ways.

For $S_3$

$$\begin{cases} x &= a \\ y &= a + b \\ z &= b \\ \end{cases}$$

Then we cannot represent $(1,3,1)$ with $S_3$.

For $S_4$,

We have

$$\begin{cases} x &= a + c\\ y &= a + b \\ z &= b - c\\ \end{cases}$$

Then we have to have $z = y - x$, so we cannot represent $(1,1,1)$ with $S_4$.

For $\mathbb{R}^n$, it needs $n$ element for a basis.

For $\mathbb{R}^2$, if $\{f_1, f_2\}$ is a basis, then geometrically, they should not be in one line.

For $\mathbb{R}^3$, if $\{f_1, f_2, f_3\}$ is a basis, then geometrically, they should not be in the same plane.

$\square$

2.1.10.

Recall the field $\mathbb{C}$ of complex numbers. Define complex $n$-space $\mathbb{C}^n$ analogously to $\mathbb{R}^n$:

$$\mathbb{C}^n = \{ (z_1, \cdots, z_n) : z_i \in \mathbb{C} \text{ for } i = 1, \cdots, n \}$$

and endow it with addition and scalar multiplication defined by the same formulas as for $\mathbb{R}^n$. You may take for granted that under these definitions, $\mathbb{C}^n$ satisfies the vector space axioms with scalar multiplication by scalars from $\mathbb{R}$, and also $\mathbb{C}^n$ satisfies the vector space axioms with scalar multiplication by scalars from $\mathbb{C}$. That is, using language that was introduced briefly in this section, $\mathbb{C}^n$ can be viewed as a vector space over $\mathbb{R}$ and also, separately, as a vector space over $\mathbb{C}$. Give a basis for each of these vector spaces.

Solution:

For the first one

$$S_1 = \{ (1,\cdots,0), (i,\cdots,0), \cdots (0,\cdots,1), (0,\cdots,i) \}$$

For the second one

$$S_2 = \{ (1,\cdots,0), \cdots (0,\cdots,1) \}$$

$\square$

2.2 Geometry: Length and Angle

2.2.1

Let

$$x = (\frac{\sqrt[]{3}}{2}, -\frac{1}{2}, 0), y = (\frac{1}{2}, \frac{\sqrt[]{3}}{2}, 1), z = (1, 1, 1)$$

Compute $⟨x,x⟩, ⟨x,y⟩, ⟨y,z⟩, |x|, |y|, |z|, θ_{x,y}, θ_{y,e_1}, θ_{z,e_2}$

Solution:

$$⟨x,x⟩ = \frac{3}{4} + \frac{1}{4} + 0 = 1 \\ ⟨x,y⟩ = \frac{\sqrt[]{3}}{4} - \frac{\sqrt[]{3}}{4} + 0 = 0 \\ ⟨y,z⟩ = \frac{1}{2} + \frac{\sqrt[]{3}}{2} + 1 = \frac{3 + \sqrt[]{3}}{2} \\ |x| = 1 \\ |y| = \sqrt[]{2} \\ |z| = \sqrt[]{3} \\ θ_{x,y} = \frac{⟨x,y⟩}{|x| |y|} = \frac{0}{|x| |y|} = 0 \\ θ_{y,e_1} = \frac{⟨y,e_1⟩}{|y| |e_1|} = \frac{1/2}{|x| |y|} = \frac{1}{2 \sqrt[]{2}} \\ θ_{z,e_2} = \frac{1}{\sqrt[]{3}}$$

$\square$

2.2.2.

Show that the points $x = (2,−1,3,1), y = (4,2,1,4), z = (1,3,6,1)$ form the vertices of a triangle in $\mathbb{R}^4$ with two equal angles.

Solution:

$$d(x,y) = |x - y| = \sqrt[]{⟨x-y,x-y⟩} = \sqrt[]{⟨(-2,-3,2,-3),(-2,-3,2,-3)⟩} = \sqrt[]{4+9+4+9} = \sqrt[]{26} \\ d(y,z) = |y-z| = \sqrt[]{⟨y-z,y-z⟩} = \sqrt[]{⟨(3,-1,-5,3),(3,-1,-5,3)⟩} = \sqrt[]{9+1+25+9} = \sqrt[]{44} \\ d(z,x) = |z-x| = \sqrt[]{⟨z-x,z-x⟩} = \sqrt[]{⟨(-1,4,3,0),(-1,4,3,0)⟩} = \sqrt[]{1+16+9+0} = \sqrt[]{26} \\ $$

$\square$

2.2.3.

Explain why for all $x ∈ \mathbb{R}^n$, $x = \sum_{j=1}^{n} ⟨x,e_j⟩e_j$.

Proof:

Note

$$⟨x,e_j⟩ = ⟨(x_1, \cdots, x_n),(0, \cdots, 1, \cdots, 0)⟩ = x_j \\ x_j e_j = (0, \cdots, x_j, \cdots, 0) \\ \sum_{j=1}^{n} x_j e_j = (x_1, \cdots, x_n) = x$$

$\square$

2.2.4.

Prove the inner product properties.

Proof:

(IP1) The inner product is positive definite: $⟨x,x⟩ ≥ 0$ for all $x ∈ \mathbb{R}^n$, with equality if and only if $x = 0$.

$$⟨x,x⟩ = x_1^2 + \cdots + x_n^2 \geq 0$$

It equals 0 only when $x_1 = \cdots = x_n = 0$.

$\square$

(IP2) The inner product is symmetric: $⟨x,y⟩= ⟨y,x⟩$ for all $x,y ∈ \mathbb{R}^n$

$$⟨x,y⟩ = x_1 y_1 + \cdots + x_n y_n = y_1 x_1 + \cdots + y_n x_n = ⟨y,x⟩$$

$\square$

(IP3) The inner product is bilinear:

$$⟨x+x',y⟩ = (x_1 + x'_1) y_1 + \cdots + (x_n + x'_n) y_n \\ = (x_1 y_1 + \cdots + x_n y_n) + (x'_1 y_1 + \cdots + x'_n y_n) \\ = ⟨x,y⟩ + ⟨x',y⟩$$

$$⟨ax,y⟩ = (ax_1) y_1 + \cdots + (ax_n) y_n \\ = a (x_1 y_1 + \cdots + x_n y_n) \\ = a ⟨x,y⟩$$

$\square$

2.2.5.

Use the inner product properties and the definition of the modulus in terms of the inner product to prove the modulus properties.

Proof:

(Mod1) The modulus is positive: $|x| ≥ 0$ for all $x ∈ \mathbb{R}^n$, with equality if and only if $x = 0$.

$$⟨x,x⟩ \geq 0 \\ \Rightarrow \\ |x| = \sqrt[]{⟨x,x⟩} \geq 0$$

$⟨x,x⟩ = 0$ if and only if $x = 0$.

$\square$

(Mod2) The modulus is absolute-homogeneous: $|ax|= |a||x|$ for all $a ∈ R$ and $x ∈ \mathbb{R}^n$.

$$⟨ax,ax⟩ = a ⟨x,ax⟩ = a^2 ⟨x,x⟩$$

So $|ax| = \sqrt[]{⟨ax,ax⟩} = \sqrt[]{a^2 ⟨x,x⟩} = |a||x|$

$\square$

2.2.6.

In the text, the modulus is defined in terms of the inner product. Prove that this can be turned around by showing that for every $x,y ∈ \mathbb{R}^n$,

$$⟨x,y⟩ = \frac{|x+y|^2 - |x-y|^2}{4}.$$

Proof:

$$|x+y|^2 = ⟨x+y,x+y⟩ = ⟨x,x⟩ + 2 ⟨x,y⟩ + ⟨y,y⟩ \\ |x-y|^2 = ⟨x-y,x-y⟩ = ⟨x,x⟩ - 2 ⟨x,y⟩ + ⟨y,y⟩ \\ $$

So the equation holds.

$\square$

2.2.7.

Prove the full triangle inequality: for every $x,y ∈ \mathbb{R}^n$,

$$||x| - |y|| \leq |x \pm y | \leq |x| + |y|$$

Do not do this by writing three more variants of the proof of the triangle inequality, but by substituting suitably into the basic triangle inequality, which is already proved.

Proof:

1. $$|x - y| \leq |x| + |y|$$

Note

$$\begin{split} |x-y| & = |x + (-y)| \\ & \leq |x| + |-y| \\ & = |x| + |y| \\ \end{split}$$

2. $$||x| - |y|| \leq |x + y|$$

Note

$$\begin{split} |x| &= |(-y) + (x+y)| \\ & \leq |-y| + |x+y| \\ & = |y| + |x+y| \\ \end{split} \\ \Rightarrow \\ |x| - |y| \leq |x+y|$$

And similarly

$$\begin{split} |y| &= |(-x) + (x+y)| \\ & \leq |-x| + |x+y| \\ & = |x| + |x+y| \\ \end{split} \\ \Rightarrow \\ |y| - |x| \leq |x+y|$$

Combine these 2 together, we have

$$||x| - |y|| \leq |x + y|$$

3. $$||x| - |y|| \leq |x - y|$$

This is similar to 2.

$\square$

2.2.8.

Let $x = (x_1,...,x_n) ∈ \mathbb{R}^n$. Prove the size bounds: for every $j ∈ \{1,...,n\}$,

$$|x_j| \leq |x| \leq \sum_{j=1}^{n} |x_j|.$$

(One approach is to start by noting that $x_j= ⟨x,e_j⟩$ and recalling equation (2.1).) When can each $≤$ be an $=$?

Proof:

From the hint:

$$\begin{split} |x_j| &= |⟨x,e_j⟩| \\ & \leq |x||e_j| \\ & \text{Cauchy-Schwarz inequality} \\ & = |x| \cdot 1 \\ & = |x| \end{split}$$

This equality holds when $x = a \cdot e_j$.

Note

$$\begin{split} |x| &= \left| \sum_{j=1}^{n} x_j e_j \right| \\ & \leq \sum_{j=1}^{n} \left| x_j e_j \right| \\ & \text{Generalized triangle inequality}\\ & = \sum_{j=1}^{n} | x_j | | e_j | \\ & \text{The modulus is absolute-homogeneous}\\ & = \sum_{j=1}^{n} | x_j | \\ \end{split}$$

This equality holds when only only one of $x_j$ is not $0$.

$\square$

2.2.9.

Prove the distance properties.

Proof:

(D1) Distance is positive: $d(x,y) ≥ 0$ for all $x,y ∈ \mathbb{R}^n$, and $d(x,y) = 0$ if and only if $x = y$.

Note

$$d(x,y) = |x-y| \geq 0$$

The equality holds only when $x-y = 0$, i.e. $x = y$.

$\square$

(D2) Distance is symmetric: $d(x,y) = d(y,x)$ for all $x,y ∈ \mathbb{R}^n$.

Note:

$$d(x,y) = |x-y| = |y-x| = d(y,x)$$

$\square$

$\square$

(D3) Triangle inequality: $d(x,z) ≤ d(x,y)+d(y,z)$ for all $x,y,z ∈ \mathbb{R}^n$.

Note:

$$\begin{split} d(x,z) &= |x-z| \\ &= |(x-y)+(y-z)| \\ & \leq |x-y| + |y-z| \\ &= d(x,y) + d(y,z) \end{split}$$

$\square$

2.2.10

Working in $\mathbb{R}^2$, depict the nonzero vectors $x$ and $y$ as arrows from the origin and depict $x−y$ as an arrow from the endpoint of $y$ to the endpoint of $x$. Let $θ$ denote the angle (in the usual geometric sense) between $x$ and $y$. Use the law of cosines to show that

$$\cos θ = \frac{⟨x,y⟩}{|x||y|}$$

Proof

Geometric proof:

$$\frac{⟨x,y⟩}{|x||y|} = \frac{x_1 y_1 + x_2 y_2}{|x||y|} = \frac{1}{|x|} \left( \frac{x_1 y_1 + x_2 y_2}{|y|} \right)$$

As shown in the figure below, $XD \perp OY$, $AC \perp OY$ and $XB \perp AC$.

We just need to show

$$\frac{x_1 y_1 + x_2 y_2}{|y|} = OD$$

Note

$$\frac{y_1}{y} = \cos \angle AOC \\ \Rightarrow \\ \frac{x_1 y_1}{|y|} = OC$$

$$\frac{y_2}{y} = \cos \angle AXB \\ \Rightarrow \\ \frac{x_2 y_2}{|y|} = CD$$

Then we finished the proof.

Proof with triangle equalities:

$$x_1 = |x| \cos θ_x, x_2 = |x| \sin θ_x, y_1 = |y| \cos θ_y, y_2 = |y| \sin θ_y,$$

Then

$$⟨x,y⟩ = |x||y| \cos θ_x \cos θ_y + |x||y| \sin θ_x \sin θ_y \\ = |x||y| \cos (θ_x - θ_y) = |x||y| \cos θ$$

This also proved.

$\square$

2.2.11.

Prove that for every nonzero $x ∈ \mathbb{R}^n$, $\sum_{i = 1}^{n} \cos ^2 θ_{x, e_i} = 1$.

Proof:

$$\cos ^2 θ_{x, e_i} = \sum_{i = 1}^{n} \left( \frac{⟨x,e_i⟩}{|x||e_i|} \right)^2 \\ \sum_{i = 1}^{n} \frac{x_i^2}{|x|^2} \\ = \frac{1}{|x|^2} \sum_{i = 1}^{n} x_i^2 \\ = 1$$

$\square$

2.2.12.

Prove that two nonzero vectors $x, y$ are orthogonal if and only if

$$|x+y|^2 = |x|^2 +|y|^2.$$

Proof:

$$\begin{split} |x+y|^2 = ⟨x+y,x+y⟩ = |x|^2 + 2 ⟨x,y⟩ + |y|^2 \end{split}$$

Then $|x+y|^2 = |x|^2 +|y|^2$ means $⟨x,y⟩=0$.

$\square$

2.2.13.

Use vectors in $\mathbb{R}^2$ to show that the diagonals of a parallelogram are perpendicular if and only if the parallelogram is a rhombus.

Proof:

The 2 diagonals of the parallelogram are $x+y, x-y$.

$$⟨x+y,x-y⟩ = |x|^2 - |y|^2$$

If $x+y \perp x-y$, then $|x|^2 - |y|^2=0$, i.e. $|x| = |y|$.

On the other hand, if $|x| = |y|$, then $⟨x+y,x-y⟩$.

$\square$

2.2.14.

Use vectors to show that every angle inscribed in a semicircle is right.

Let $(x,y)$ on the semicircle, then $x^2+y^2=1$.

consider

$(x,y) - (-1,0) = (x+1,y)$ and $(x,y) - (1,0) = (x-1,y)$.

Then

$$⟨(x+1,y), (x-1, y)⟩ = x^2 - 1 + y^2 = 0$$

$\square$

2.2.15.

Let $x$ and $y$ be vectors, with $x$ nonzero.Define the parallel component of $y$ along $x$ and the normal component of $y$ to $x$ to be

$$y_{(\parallel x)} = \frac{⟨x,y⟩}{|x|^2} x \qquad \text{and} \qquad y_{(\perp x)} = y - y_{(\parallel x)}$$

Answer:

(a) Show that $y = y_{(\parallel x)} + y_{(\perp x)}$. show that $y_{(\parallel x)}$ is a scalar multiple of $x$. show that $y_{(\perp x)}$ is orthogonal to $x$. Show that the decomposition of $y$ as a sum of vectors parallel and perpendicular to $x$ is unique. Draw an illustration.

Proof:

$$y_{(\parallel x)} + y_{(\perp x)} = \\ y_{(\parallel x)} + (y - y_{(\parallel x)}) = \\ y$$

$$y_{(\parallel x)} = \frac{⟨x,y⟩}{|x|^2} x$$

Since $\frac{⟨x,y⟩}{|x|^2}$ is a scalar, then $y_{(\parallel x)}$ is a scalar multiple of $x$.

$$⟨y - y_{(\parallel x)},y_{(\parallel x)}) = \\ ⟨|x|^2 y - ⟨x,y⟩x, ⟨x,y⟩x⟩ \\ = |x|^2 ⟨x,y⟩ ⟨y,x⟩ - ⟨x,y⟩^2 ⟨x,x⟩ \\ = 0$$

So $y_{(\perp x)}$ is orthogonal to $y_{(\parallel x)}$, thus is orthogonal to $x$.

In general if $x \perp y$ and $ax + by = 0, x,y \neq 0$, then

$$0 = ⟨0, x⟩ = ⟨ax + by,x⟩ = a ⟨x,x⟩ + b ⟨y,x⟩ = a ⟨x,x⟩ \geq 0 \\ \Rightarrow \\ a = 0$$

For the same reason $b = 0$.

With this, the decomposition of $y$ as a sum of vectors parallel and perpendicular to $x$ is unique.

$\square$

(b) Show that

$$|y|^2 = |y_{(\parallel x)}|^2 + |y_{(\perp x)}|^2$$

What theorem from classical geometry does this encompass?

Proof:

$$|y|^2 = \\ |y_{(\parallel x)} + y_{(\perp x)}|^2 = \\ |y_{(\parallel x)}|^2 + |y_{(\perp x)}|^2 + 2 ⟨y_{(\parallel x)}, y_{(\perp x)}⟩ \\ = |y_{(\parallel x)}|^2 + |y_{(\perp x)}|^2 \\ $$

It's Pythagorean theorem or Gougu theorem.

$\square$

(c) Explain why it follows from (b) that

$$|y_{(\parallel x)}|^2 \leq |y|^2$$

with equality if and only if $y$ is a scalar multiple of $x$. Use this inequality to give another proof of the Cauchy–Schwarz inequality. This argument gives the geometric content of Cauchy–Schwarz: the parallel component of one vector along another is at most as long as the original vector.

Proof:

This is because $|y_{(\perp x)}|^2 \geq 0$.

The equality holds when $|y_{(\perp x)}|^2 = 0$. In this case

$$y = y_{(\parallel x)} = \frac{⟨x,y⟩}{|x|^2} x$$

To prove

$$|⟨x,y⟩| \leq |x||y|$$

Note

$$|⟨x,y⟩| = |⟨x,y_{(\parallel x)}+y_{(\perp x)}⟩| \\ = |⟨x,y_{(\parallel x)}⟩ + 0|\\ = |⟨x,y_{(\parallel x)}⟩|$$

Now in general, if $a$ is a scaler, then

$$|⟨x,ax⟩| = |a ⟨x,x⟩| = |a||x|^2 = |x| |ax|$$

Then

$$|⟨x,y_{(\parallel x)}⟩| = |x||y_{(\parallel x)}|| \leq |x||y|$$ $\square$

(d) The proof of the Cauchy–Schwarz inequality in part (c) refers to parts (a) and (b), part (a) refers to orthogonality, orthogonality refers to an angle, and as explained in the text, the fact that angles make sense depends on the Cauchy–Schwarz inequality. And so the proof in part (c) apparently relies on circular logic. Explain why the logic is in fact not circular.

Solution:

Note that when we define $\cos θ$, it is this:

$$\cos θ = \frac{⟨x,y⟩}{|x||y|}$$

But here, we define $y_{(\parallel x)}$ as

$$y_{(\parallel x)} = \frac{⟨x,y⟩}{|x|^2}x$$

If we think geometrically in $\mathbb{R}^2$ or $\mathbb{R}^3$, the projection of $y$ on $x$ should be $y \cos θ$.

The unit length vector along $x$ is $\frac{x}{|x|}$. Also the length of $y$ projection is

$$|y \cos θ| = \frac{⟨x,y⟩}{|x|}$$

So it's reasonable to define $y_{(\parallel x)}$ this way.

Now, since we does not rely on the fact that $|\cos θ| \leq 1$, it is not circular.

$\square$

2.2.16.

Given nonzero vectors $x_1,x_2,...,x_n$ in $\mathbb{R}^n$, the Gram–Schmidt process is to set

$$\begin{split} x_1' &= x_1 \\ x_2' &= x_2 - (x_2)_{(\parallel x_1')}\\ x_3' &= x_3 - (x_3)_{(\parallel x_2')} - (x_3)_{(\parallel x_1')} \\ \vdots \\ x_n' &= x_n - (x_n)_{(\parallel x_{n-1}')} - \cdots - (x_n)_{(\parallel x_1')} \\ \end{split}$$

Answer:

(a)What is the result of applying the Gram–Schmidt process to the vectors $x_1 = (1,0,0), x_2 = (1,1,0)$, and $x_3 = (1,1,1)$?

Solution:

$$x_1' = (1,0,0) \\ x_2' = (0,1,0) \\ x_3' = (0,0,1) \\ $$

$\square$

(b) Returning to the general case, show that $x_1',...,x_n'$ are pairwise orthogonal and that each $x_j'$ has the form

$$x_j'= a_{j,1} x_1 +a_{j,2}x_2 +···+a_{j,j−1}x_{j−1} +x_j.$$

Thus every linear combination of the new ${x_j'}$ is also a linear combination of the original ${x_j}$. The converse is also true and will be shown in Exercise 3.3.13.

Proof:

First, we want to prove a small lemma: Given $x, y, z$, if $x \perp y$, then we want to show

$$z_{(\parallel x)} \perp y$$

Note

$$⟨z_{(\parallel x)},y⟩ = ⟨\frac{⟨z,x⟩}{|x|^2}x,y⟩ = \frac{⟨z,x⟩}{|x|^2} ⟨x,y⟩ = 0$$

Then we can use induction. And assume $$x_1',...,x_{k-1}'$$

are orthogonal to each other. Assume $1 \leq j \leq k-1$,

$$⟨x_j',x_{k}'⟩ = ⟨x_j',x_k - {x_k}_{(\parallel x_j')}⟩ - \sum_{i \neq j} ⟨x_j', {x_k}_{(\parallel x_i')}⟩$$

$⟨x_j',x_k - {x_k}_{(\parallel x_j')}⟩ = 0$ is proved in exercise 2.2.15.

Also, since $⟨x_j',x_i'⟩=0$ for $j \neq i$, then $⟨x_j', {x_k}_{(\parallel x_i')}⟩ = 0$ as just proved in the lemma.

So $⟨x_j',x_{k}'⟩ = 0$.

For the 2nd part, we can still use induction.

And assume for $x_1',...,x_{k-1}'$, they are all linear combination of $x_1, x_2, x_3, \cdots, x_{k-1}$.

Then

$$x_k' = x_k - (x_k)_{(\parallel x_{k-1}')} - \cdots - (x_k)_{(\parallel x_1')}$$

Note $(x_k)_{(\parallel x_j')}$ is a scaler of $x_j', j < k$, thus a linear combination of $x_1, x_2, x_3, \cdots, x_{k-1}$.

Therefore, $x_k'$ is a linear combination of $x_1,...,x_{k}$.

$\square$

2.3 Analysis: Continuous Mappings

2.3.1.

For $A ⊂ \mathbb{R}^n$, partially verify that $\mathcal{M}(A,\mathbb{R}^m)$ is a vector space over $\mathbb{R}$ by showing that it satisfies vector space axioms (A4) and (D1).

Proof:

(A4) Addition is commutative

$$\begin{align*} (f+g)(x) &= f(x) + g(x) \quad &\text{by definition of “+” in } \mathcal{M}(A,\mathbb{R}^m)\\ &= g(x) + f(x) \quad &\text{by commutativity of “+” in }\mathbb{R}^m\\ &= (g+f)(x) &\text{by definition of “+” in } \mathcal{M}(A,\mathbb{R}^m) \end{align*}$$

Next,

(D1) Scalar multiplication distributes over scalar addition

$$\begin{align*} ((a+b)f)(x) &= (a+b)f(x) \quad &\text{by definition of “.” in } \mathcal{M}(A,\mathbb{R}^m)\\ &= af(x) + bf(x) \quad &\text{by D1 in }\mathbb{R}^m\\ &= (af)(x) + (bf)(x) \quad &\text{by definition of “.” in } \mathcal{M}(A,\mathbb{R}^m)\\ &= ((af) + (bf))(x) \quad &\text{by definition of “+” in } \mathcal{M}(A,\mathbb{R}^m)\\ \end{align*}$$

$\square$

2.3.2.

Define multiplication

$$* : \mathcal{M}(A, \mathbb{R}^m) \times \mathcal{M}(A, \mathbb{R}^m) \longrightarrow \mathcal{M}(A, \mathbb{R}^m)$$

Is $\mathcal{M}(A, \mathbb{R}^m)$ a field with $“+”$ from the section and this multiplication? Does it have a subspace that is a field?

Solution

Define:

$$(f * g)(x) = f(x) * g(x)$$

And $f(x) = 1$ is it's multiplication identity.

However, this function does not have inverse.

$$f(x) = \begin{cases} 0 &\text{if } x = 0 \\ 1 &\text{if } x \neq 0 \\ \end{cases}$$

Consider

$$V = \{f : f(x) = (r, \cdots, r) \in \mathbb{R}^m \text{ for all } x \in A\}$$

This is a field.

$\square$

2.3.3.

For $A ⊂ \mathbb{R}^n$ and $m ∈ \mathbb{Z}^+$ define a subspace of the space of mappings from $A$ to $\mathbb{R}^m$

$$\mathcal{C}(A, \mathbb{R}^m) = \{ f \in \mathcal{M}(A, \mathbb{R}^m) : f \text{ is continuous on } A \}.$$

Briefly explain how this section has shown that $\mathcal{C}(A, \mathbb{R}^m)$ is a vector space.

Solution:

It's stated in Proposition 2.3.7 (Vector space properties of continuity).

$\square$

2.3.4.

Define an inner product and a modulus on $\mathcal{C}([0,1], \mathbb{R})$ by

$$⟨f,g⟩ = \int_{0}^{1} f(t)g(t)dt, \quad |f| = \sqrt[]{⟨f,f⟩}.$$

Do the inner product properties (IP1),(IP2),and(IP3) (see Proposition2.2.2) hold for this inner product on $\mathcal{C}([0,1], \mathbb{R})$ How much of the material from Section2.2 on the inner product and modulus in $\mathbb{R}^n$ carries over to $\mathcal{C}([0,1], \mathbb{R})$? Express the Cauchy–Schwarz inequality as a relation between integrals.

Proof:

(IP1)

$$⟨f, f⟩ = \int_{0}^{1} f(t)f(t)dt \geq \int_{0}^{1} 0dt = 0$$

It achieves 0 only when $f(t) = 0$.

(IP2)

$$⟨f,g⟩ = \int_{0}^{1} f(t)g(t)dt = \int_{0}^{1} g(t)f(t)dt = ⟨g,f⟩$$

So IP2 holds.

(IP3)

First, we want to show

$$\int_{0}^{1} f (g+h) = \int_{0}^{1} fg + \int_{0}^{1} fh$$

Based on the definition of integral, give a partition $P$ on $[0,1]$.

$$L(fg, P) + L(fh, P) \leq L(fg+fh, P) \leq U(fg+fh, P) \leq U(fg, P) + L(fh, P)$$

So we have

$$\int_{0}^{1} f (g+h) = \int_{0}^{1} fg + \int_{0}^{1} fh.$$

Now

$$⟨f,g+h⟩ = \int_{0}^{1} f (g+h) = \int_{0}^{1} fg + \int_{0}^{1} fh = ⟨f,g⟩ + ⟨f,h⟩$$

Also

$$⟨af,g⟩ = \int_{0}^{1} (af)g = a(\int_{0}^{1} fg) = a ⟨f,g⟩$$

$\square$

All of the material from Section2.2 on the inner product and modulus in $\mathbb{R}^n$ carries over to $\mathcal{C}([0,1], \mathbb{R})$

Cauchy–Schwarz inequality

$$\left| \int_{0}^{1} f(t)g(t) dt \right| \leq \sqrt[]{\int_{0}^{1}f^2(t)dt} \cdot \sqrt[]{\int_{0}^{1}g^2(t)dt}$$

$\square$

2.3.5

Use the definition of convergence and the componentwise nature of nullness to prove the componentwise nature of convergence. (The argument is short.)

Proof:

$$\{(x_{1, ν}, \cdots, x_{n, ν})\} \longrightarrow p \\ \Longleftrightarrow \\ x_ν - p = \text{null} \\ \text{definition of convergence} \\ \Longleftrightarrow \\ x_{j, ν} - p_j = \text{null} \\ \text{Lemma 2.3.2 (Componentwise nature of nullness).} \\ \Longleftrightarrow \\ x_{j, ν} \longrightarrow p_j \\ \text{definition of convergence} \\ $$

$\square$

2.3.6.

Use the definition of continuity and the componentwise nature of convergence to prove the componentwise nature of continuity.

Proof:

$\Rightarrow$

Given any sequence $(x_{\nu}) \rightarrow p$, since $f$ is continuous at $p$, then $f(x_{\nu}) \rightarrow f(p)$.

From Proposition 2.3.5 Componentwise nature of convergence $f_i(x_{\nu}) \rightarrow f_i(p)$.

So $f_i$ is continuous at $p$.

$\Leftarrow$

Given any sequence $(x_{\nu}) \rightarrow p$, and since $f_i$ is continuous at $p$, then $f_i(x_{\nu}) \rightarrow f_i(p)$.

From Proposition 2.3.5 Componentwise nature of convergence $f(x_{\nu}) \rightarrow f(p)$.

So $f$ is continuous at $p$.

$\square$

2.3.7

Prove the persistence of continuity under composition.

Proof:

Given $p \in A$, and $\{x_ν\} \rightarrow p$. We have $\{f(x_ν)\} \rightarrow f(p)$.

Since $f(x_ν), f(p) \in B$, and $g$ is a continuous mapping, we have $\{g(f(x_ν))\} \rightarrow g(f(p))$.

Therefore, $g \circ f$ is continuous.

$\square$

2.3.8.

Define $f : \mathbb{Q} \longrightarrow \mathbb{R}$ by the rule

$$f(x) = \begin{cases} 1 &\text{if } x^2 < 2,\\ 0 &\text{if } x^2 > 2.\\ \end{cases}$$

Solution:

It is continuous. Given any $p \in \mathbb{Q}$, if $p^2 < 2$, we can always find a $U_{\delta}(p)$ such that if $q \in U_{\delta}(p)$, then $q^2 < 2$.

Same thing when $p^2 > 2$.

$\square$

2.3.9.

Which of the following functions on $\mathbb{R}^2$ can be defined continuously at $0$?

(a)

$$f(x, y) = \begin{cases} \frac{x^4 - y^4}{(x^2 + y^2)^2} &\text{if } (x,y) \neq 0 \\ b &\text{if } (x,y) = 0 \\ \end{cases}$$

Solution

$$\begin{split} \frac{x^4 - y^4}{(x^2 + y^2)^2} &= \frac{x^2 - y^2}{x^2 + y^2} \end{split}$$

When $(x,y)$ approaches $0$ with $y = mx$, then

$$\begin{split} \frac{x^2 - y^2}{x^2 + y^2} &= \frac{1 - m^2}{1 + m^2} \end{split}$$

So it cannot be defined continuously at $0$.

$\square$

(b)

$$g(x, y) = \begin{cases} \frac{x^2 - y^3}{x^2 + y^2} &\text{if } (x,y) \neq 0 \\ b &\text{if } (x,y) = 0 \\ \end{cases}$$

Solution:

$$\begin{split} \frac{x^2 - y^3}{x^2 + y^2} &= 1 - \frac{y^2 + y^3}{x^2 + y^2} \end{split}$$

When $(x,y)$ approaches $0$ with $y = mx$, then

$$\frac{y^2 + y^3}{x^2 + y^2} \\ = \frac{m^2 + m^3y}{1+m^2} \rightarrow \frac{m^2}{1+m^2}$$

So it cannot be defined continuously at $0$.

$\square$

(c)

$$h(x, y) = \begin{cases} \frac{x^3 - y^3}{x^2 + y^2} &\text{if } (x,y) \neq 0 \\ b &\text{if } (x,y) = 0 \\ \end{cases}$$

Solution:

$$\begin{split} \left| \frac{x^3 - y^3}{x^2 + y^2} \right| &= \left| (x-y) \frac{x^2 + xy + y^2}{x^2 + y^2} \right| \\ & \leq |x-y| + |x-y| \left| \frac{xy}{x^2 + y^2} \right| \\ & \leq |x-y| + |x-y| \left| \frac{|(x,y)||(x,y)|}{|(x,y)|^2} \right| \\ & = 2 |x - y| \\ & \leq 2 (|x| + |y|) \rightarrow 0 \end{split}$$

So it can be defined continuously at 0.

$\square$

(d)

$$k(x, y) = \begin{cases} \frac{xy^2}{x^2 + y^6} &\text{if } (x,y) \neq 0 \\ b &\text{if } (x,y) = 0 \\ \end{cases}$$

Solution:

When $(x,y)$ approaches $0$ with $y = x$, then

$$\frac{xy^2}{x^2 + y^6} = \frac{x^3}{x^2 + x^6} = \frac{x}{1 + x^4} \rightarrow 0$$

However, when $(x,y)$ approaches $0$ with $y = \sqrt[]{x}$, then

$$\frac{xy^2}{x^2 + y^6} = \frac{x^2}{x^2 + x^3} = \frac{1}{1 + x} \rightarrow 1$$

So it cannot be defined continuously at $0$.

$\square$

2.3.10.

Let $f(x,y) = g(xy)$, where $g : \mathbb{R} → \mathbb{R}$ is continuous. Is $f$ continuous?

Proof:

Let $h(x,y) = xy$, then $h(x,y)$ is a continuous function.

Then from Proposition 2.3.8 (Persistence of continuity under composition), $f = g \circ h$. So $f$ is also continuous.

$\square$

2.3.11.

Let $f,g \in \mathcal{M}(\mathbb{R}^n, \mathbb{R})$
such that $f +g$ and $fg$ are continuous. Are $f$ and $g$ necessarily continuous?

Proof:

No, it's not necessary. Consider the case when $n = 1$.

$$f(x) = \begin{cases} 0 &\text{if } x \in \mathbb{Q}\\ 1 &\text{if } x \in \mathbb{I}\\ \end{cases}$$

Then $g(x) = 1 - f(x)$ .

Then $f+g = 1, fg = 0$ are continuous. But $f, g$ are not continuous.

$\square$

2.4 Topology: Compact Sets and Continuity

2.4.1.

Are the following subsets of $\mathbb{R}^n$ closed, bounded, compact?

(a) $B(0,1)$: not closed, bounded, not compact.

(b) $\{(x,y) \in \mathbb{R}^2 : y - x^2 = 0\}$:

closed, not bounded, not compact.

(c) $\{(x,y,z) \in \mathbb{R}^3: x^2 +y^2 +z^2 −1 = 0 \}$

closed, bounded, compact.

(d) $\{x : f(x) = 0_m\}$, where $f \in \mathcal{M}(\mathbb{R}^n, \mathbb{R}^m)$ is continuous (this generalizes (b) and (c)),

Solution: Consider $\{x_ν\} \rightarrow p$, since $f$ is continuous, $\{f(x_ν)\} \rightarrow f(p)$. So $f(p) = 0$. So it's closed. It can be either bounded or unbounded, so either compact or not compact.

(f) $\{(x_1,...,x_n) : x_1 +···+x_n > 0\}$.

not closed, not bounded, not compact.

$\square$

2.4.2.

Give a set $A ⊂ \mathbb{R}^n$ and limit point $b$ of $A$ such that $b \not\in A$. Give a set $A ⊂ \mathbb{R}^n$ and a point $a ∈ A$ such that $a$ is not a limit point of $A$.

Solution:

Consider $n = 1$, let $A = (0, 1)$, $b = 0$.

Let $A = \mathbb{N}, a = 0$.

$\square$

2.4.3.

Let $A$ be a closed subset of $\mathbb{R}^n$ and let $f ∈ \mathcal{M}(A, \mathbb{R}^m)$. Define the graph of $f$ to be

$$G(f) = \{(a, f(a)) : a \in A\}$$

a subset of $\mathbb{R}^{n+m}$. Show that if $f$ is continuous then its graph is closed.

Proof:

Given $\{x_ν\} \rightarrow p$, $\{x_ν\} \in G(f)$.

Let $x_ν = (a_ν, f(a_ν))$. Then from Proposition 2.3.5 (Componentwise nature of convergence), $\{a_ν\} \rightarrow p_0$, and since $f$ is continuous $\{f(a_ν)\} \rightarrow f(p_0)$. Also since $A$ is closed, so $p_0 \in A$.

So $p = (p_0, f(p_0) \in G(f)$.

Then $G(f)$ is closed.

$\square$

2.4.4.

Prove the closed set properties: (1) the empty set $∅$ and the full space $\mathbb{R}^n$ are closed subsets of $\mathbb{R}^n$;

(2) every intersection of closed sets is closed;

(3) every finite union of closed sets is closed.

Proof:

(1) $∅$ does not have limit point, so $∅$ contains all it's limit point, which is nothing. $\mathbb{R}^n$'s limit points are in $\mathbb{R}^n$. So $\mathbb{R}^n$ is closed.

(2) Let $C = \cap C_i$, $p$ is a limit point of $C$. So we can find a $\{x_ν\} \rightarrow p$, where $x_ν \in C$. Then $x_ν \in C_i$. So $p \in C_i$. Then $p \in C$.

(3) Let $C = \cup_{i=1}^{n} C_i$, $p$ is a limit point of $C$. So we can find a $\{x_ν\} \rightarrow p$, where $x_ν \in C$. Then we can find at least one $i$, such that there are infinitely many $x_ν \in C_i$. It is a subsequence of $\{x_ν\}$ and also convergences to $p$. So $p \in C_i \subseteq C$.

$\square$

2.4.5.

Prove that every ball $B(p,ε)$ is bounded in $\mathbb{R}^n$.

Proof:

Assume $x \in B(p,ε)$, then

$$|x| \leq |x-p| + |p| < \epsilon + p$$

$\square$

2.4.6.

Show that $A$ is a bounded subset of $\mathbb{R}^n$ if and only if for each $j ∈ \{1,...,n\}$, the $j$th coordinates of its points form a bounded subset of $\mathbb{R}$.

Proof: We will need Proposition 2.2.7 (Size bounds)

$$A \text{ is a bounded } \\ \Longleftrightarrow \\ |x| < R, \forall x \in A \\ \Longleftrightarrow \\ |x_j| < R \\ \Longleftrightarrow \\ j\text{th coordinates of its points form a bounded subset of} \mathbb{R}$$

2.4.7.

Show by example that a closed set need not satisfy the sequential characterization of bounded sets, and that a bounded set need not satisfy the sequential characterization of closed sets.

Solutions:

Consider $A = [0, +\infty)$ which is a closed set. Let $\{x_ν\} = \{1, 2, 3, \cdots\}$, then it does not have a subsequence that convergences.

Consider $A = (0, 1)$ which is bounded. Let $\{x_ν\} = \{1, 1/2, 1/3, \cdots\}$, we have $\{x_ν\} \rightarrow 0 \not\in A$.

$\square$

2.4.8.

Show by example that the continuous image of a closed set need not be closed, that the continuous image of a closed set need not be bounded, that the continuous image of a bounded set need not be closed, and that the continuous image of a bounded set need not be bounded.

Solution:

Consider $f(x) = \frac{1}{x}, x \in [1, +\infty)$. $f(C) = (0, 1]$ which is not closed.

Consider $f(x) = x, x \in [1, +\infty)$. $f(C) = [1, +\infty)$ which is not bounded.

Consider $f(x) = x, x \in (0, 1)$. $f(C) = (0, 1)$ which is not closed.

Consider $f(x) = 1/x, x \in (0, 1)$. $f(C) = [1, +\infty)$ which is not bounded.

2.4.9.

A subset $A$ of $\mathbb{R}^n$ is called discrete if each of its points is isolated. (Recall that the term isolated was defined in this section.) Show or take for granted the (perhaps surprising at first) fact that every mapping whose domain is discrete must be continuous. Is discreteness a topological property? That is, need the continuous image of a discrete set be discrete?

Solution:

First we show a function $f : A \longrightarrow \mathbb{R}^n$ is continuous at any point $a \in A$.

Give any $\{x_ν\} \rightarrow a$, since $a$ is isolated, we must have $x_ν = a$ for $ν > \Nu$, then $\{f(x_ν)\} \rightarrow f(a)$, which means $f$ is continuous.

Consider

$$f(x) = \begin{cases} 0 &\text{if } x = 0\\ 1/x &\text{otherwise}\\ \end{cases}$$

and let $A = \mathbb{N}$. So $0$ is in $f(A)$, and $0$ is not isolated.

$\square$

2.4.10

A subset $A$ of $\mathbb{R}^n$ is called path-connected if for every two points $x,y ∈ A$, there is a continuous mapping

$$γ : [0,1] \longrightarrow A$$

such that $γ(0) = x$ and $γ(1) = y$. (This $γ$ is the path that connects $x$ and $y$.) Draw a picture to illustrate the definition of a path-connected set. Prove that path-connectedness is a topological property.

Proof:

Let $f : A \longrightarrow \mathbb{R}^m$ is a continuous map. Given $x, y \in A$, we want to show $f(x), f(y)$ is also path-connected.

From Proposition 2.3.8 (Persistence of continuity under composition), $f \circ γ : [0,1] \longrightarrow \mathbb{R}^m$ is a continuous map and

$$(f \circ γ)(0) = f(x) \\ (f \circ γ)(1) = f(y) \\ $$

So $f(x), f(y)$ is also path-connected.

$\square$