If you find any mistakes, please make a comment! Thank you.

Solution to Principles of Mathematical Analysis Chapter 1 Part B


Chapter 1 The Real and Complex Number Systems

Exercise 11

(By ghostofgarborg) There is a solution, and it is unique whenever $z \neq 0$. Assume $z = rw$, $r \geq 0$ and $|w|^2 = w \bar w = 1$. Then\[ r = \sqrt{r^2} = \sqrt{(rw)(\overline{rw})} = \sqrt{z \bar z} = |z|. \]If $r = 0$, we can take any $w$ with $|w| = 1$, e.g. $w = e^{i \theta}$. Otherwise \[ w = \frac{z}{r} = \frac{z}{|z|}.\]It is easy to check that the uniquely determined $r$ and $w$ have the desired properties.


Exercise 12

(By ghostofgarborg) Note that by the triangle inequality, $|z_1 + z_2| \leq |z_1| + |z_2|$. Assume the statement holds for $n-1$. Then
\[ |z_1 + \cdots + z_{n-1} + z_n| \leq |z_1 + \cdots + z_{n-1}| + |z_n| \leq |z_1| + \cdots + |z_n|, \]which establishes the claim by induction.


Exercise 13

(By ghostofgarborg) By the triangle inequality
\[ |x| = | (x – y) + y| \leq |x-y| + |y|\]so that\[ |x| – |y| \leq |x-y|.\]Interchanging the roles of $x$ and $y$ in the above, we also have\[ |y| – |x| \leq |x-y| \]so that
\[ ||x| – |y|| \leq |x-y|.\]


Exercise 14

(By ghostofgarborg) We have\begin{align*}
|1+z|^2 + |1-z|^2 &= (1+z)(\overline{1+z}) + (1-z)(\overline{1-z}) \\
&= (1 + z)(1 + \bar z) + (1 – z)(1 – \bar z) \\
&= (1 + z + \bar z + z \bar z) + (1 -z – \bar z + z \bar z) \\
&= 2 + 2z \bar z \\
&= 4.
\end{align*}


Exercise 15

(By ghostofgarborg) We observe that $(AB-|C|^2) > 0$ iff $S = \sum |Ba_j – Cb_j|^2 > 0$.

If $S=0$, then $a_j = – \frac C B b_j = zb_j$. Assume $a_j = zb_j$. Then $C = zB$, so that $S = \sum |Bzb_j – zBb_j|^2 = 0$. I.e., equality holds iff there is a $z \in \mathbb C$ such that $a_j = zb_j$.


Exercise 16

(By analambanomenos)

(a): To simplify the calculations, we may assume without losing generality that $\mathbf{x}=(0,\ldots,0)$ and $\mathbf{y}=(d,0,\ldots,0)$. (Translate $\mathbb R^k$ by $-\mathbf x$, then rotate it to place $\mathbf y$ on the positive x-axis. Both transformations preserve distances and angles.)

For $0\le\theta<1$ let $$\mathbf{z}_\theta=(d/2,\cos\theta\sqrt{4r^2-d^2}/4,\sin\theta\sqrt{4r^2-d^2}/2,0,\ldots,0).$$ Then $$|\mathbf{z}_\theta-\mathbf{x}|^2=|\mathbf{z}_\theta-\mathbf{y}|^2=
\frac{d^2}{4}+(\cos^2\theta+\sin^2\theta)\frac{4r^2-d^2}{4}=r^2.$$For $k=2$, the hyperplane of equidistant points is the line $(d/2,x_2)$, so there are two points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$, $(d/2,\pm\sqrt{4r^2-d^2}/2)$. For $k=1$, the hyperplane reduces to the single point $z=d/2$, so there are no points with distance $r>d/2$ from $\mathbf x$ and $\mathbf y$.

(b): If $d=2r$, then $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{x}|=|\mathbf{x}-\mathbf{y}|$, that is, we have equality in Theorem 1.37(f). Looking at the proof, this only happens if $$(\mathbf{z}-\mathbf{x})\cdot(\mathbf{z}-\mathbf{y})=|\mathbf{z}-\mathbf{x}| |\mathbf{z}-\mathbf{y}|.$$ By the solution to exercise 15, this only happens if $\mathbf{z}-\mathbf{x}=t(\mathbf{z}-\mathbf{y})$ for some real number $t$. Since $|\mathbf{z}-\mathbf{x}|=|\mathbf{z}-\mathbf{y}|$, $t$ must be $\pm 1$. If $t=1$, then $\mathbf{x}=\mathbf{y}$, so $t=-1$ and $\mathbf z$ is the midpoint $(\mathbf{x}-\mathbf{y})/2$.

(c): $|\mathbf{z}-\mathbf{x}|+|\mathbf{z}-\mathbf{y}|=2r<|\mathbf{x}-\mathbf{y}|$ contradicts Theorem 1.37(f).


Exercise 17

(By Matt frito Lundy)

This solution will use some linear algebra. In particular, for any inner product $\langle \cdot {,} \cdot \rangle$ the norm is $$| \cdot | = \sqrt{\langle \cdot {,} \cdot \rangle}$$ or
$$| \cdot |^2 = \langle \cdot {,} \cdot \rangle.$$ In this real vector space $R^k$, for any $x,y,z \in R^k$ we also have the following properties: $$\langle x + y , z \rangle = \langle x , z \rangle + \langle y , z\rangle $$ $$\langle x , y \rangle = \langle y , x \rangle.$$ Then
\begin{align*}
|x + y|^2 + |x – y|^2 &= \langle x , x \rangle + 2\langle x , y \rangle + \langle y , y \rangle
+ \langle x , x \rangle – 2\langle x , y \rangle +\langle y , y \rangle \\
&= 2\langle x , x \rangle + 2\langle y , y \rangle \\
&= 2|x|^2 + 2|y|^2
\end{align*}Geometrically, this says that for any parallelogram in Euclidian Space, the sum of the lengths of the diagonals is equal to the perimeter.


Exercise 18

(By Matt frito Lundy) This solution uses some linear algebra. If $x = 0$ then any non-zero $y$ will suffice, so assume that $x \neq 0$. Also choose any $z \in R^k$ such that $x$ and $z$ are linearly indepedent (this is possible because $k \geq 2$). Then let $$ y = z – \frac{\langle z, x \rangle}{\langle x , x \rangle} x.$$ Notice that $y \neq 0$ because $x$ and $z$ are linearly independent and $y$ is a linear combination of $x$ and $z$ with not all coefficients equal to zero. And so we have
\begin{align*}
\langle x ,y \rangle &= \langle x , z – \frac{\langle z, x \rangle}{\langle x , x \rangle} x \rangle\\
&= \langle x , z \rangle -\frac{\langle z ,x \rangle \langle x , x \rangle}{\langle x, x \rangle} \\
&= \langle x , z \rangle – \langle x , z \rangle \\
&= 0
\end{align*} There is trouble when $k = 1$ because any two vectors in $R^1$ are linearly dependent.


Exercise 19

(By analambanomenos)\begin{align*}
4|\mathbf{x}-\mathbf{b}|^2 &= |\mathbf{x}-\mathbf{a}|^2 \\
4|\mathbf{x}|^2-8\mathbf{x}\cdot\mathbf{b}+4|\mathbf{b}|^2 &= |\mathbf{x}|^2-2\mathbf{x}\cdot\mathbf{a}+|\mathbf{a}|^2 \\
|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) &= (1/3)|\mathbf{a}|^2-(4/3)|\mathbf{b}|^2 \\
|\mathbf{x}|^2-2\mathbf{x}\cdot\bigl((1/3)(4\mathbf{b}-\mathbf{a})\bigr) + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/3)|\mathbf{a}|^2 – (4/3)|\mathbf{b}|^2 + \big|(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(3|\mathbf{a}|^2-12\mathbf{b}^2+|\mathbf{a}|^2-8\mathbf{a}\cdot\mathbf{b}+16|\mathbf{b}|^2) \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (1/9)(4|\mathbf{a}|^2 – 8\mathbf{a}\cdot\mathbf{b} + 4|\mathbf{b}|^2) \\
\big|\mathbf{x}-(1/3)(4\mathbf{b}-\mathbf{a})\big|^2 &= (4/9)\big|\mathbf{a}-\mathbf{b}\big|^2
\end{align*}
This describes a sphere with center $(1/3)(4\mathbf{b}-\mathbf{a})$ and radius $(2/3)|\mathbf{a}-\mathbf{b}|$.


Exercise 20

(By Jack Gallagher)Proof is omitted whenever the proof in the book does not use property (III).

In this case, a cut is defined as any inhabited set $\alpha \subset Q$ such that

  1. $\alpha \neq Q$
  2. If $p \in \alpha$, $q \in Q$, and $q < p$, then $q \in \alpha$

I’ll adopt the same convention as the book, labelling rationals with Roman letters and cuts with Greek. The proof that $R’$ has the least-upper-bound property is the same as that given in the book.

The proofs of axioms (A1-3) are again the same. For (A4), we define
$$ 0_R = \left\{ n\ |\ n \in Q,\ n \leq 0 \right\}.$$(A4) For any $a \in \alpha$ and $s \in 0_r$, we have either that $$ a + s = a \text{ or } a + s < a.$$ In either case we have that $a + s \in \alpha$, either by equality or by (II). Thus $\alpha + 0_{R’} \subset \alpha$.

Similarly, if $b \in \alpha + 0_{R’}$, we have that $b = a + s$ for some $a \in \alpha$ and $s \in 0_{R’}$, and the same analysis applies.

Finally, to negate (A5), we shall first show that any addition with an open cut will produce another open cut.

Consider two cuts $\alpha \in R$, $\beta \in R’$. For any $r \in \alpha + \beta$, we have that $r = a + b$ for some $a \in \alpha$, $b \in \beta$. But, because $\alpha$ is open, we can pick some $a’ \in \alpha$ such that $a’ > a$, and therefore we have $a’ + b > a + b \in \alpha + \beta$.

(A5) fails! Suppose we had a valid negation operation. Then we would have
$$ 0^* – 0^* = 0_{R’} $$ But this leads to a contradiction, as with the given definition of addition we cannot add any number to the open $0^*$ to produce the closed $0_{R’}$.

Baby Rudin 数学分析原理完整第一章习题解答

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu