Problems for Chapter 1

(a) Statement. For an integer $𝑛\ge 2$, $\mathbb{Z}/𝑛\mathbb{Z}$ is a field if and only if $𝑛$ is prime (Proposition 1.1.24).

Proof. By Lemma 1.1.21, $\mathbb{Z}/𝑛\mathbb{Z}$ satisfies every field axiom except possibly the existence of multiplicative inverses, so it is a field exactly when every nonzero element has an inverse.

($\Rightarrow$) We prove the contrapositive. If $𝑛$ is not prime, write $𝑛=𝑟𝑠$ with $1<𝑟,𝑠<𝑛$. Then $𝑟$ and $𝑠$ are nonzero elements of $\mathbb{Z}/𝑛\mathbb{Z}$ with $𝑟{\cdot }_{𝑛}𝑠=𝑛mod𝑛=0$. A field has no zero divisors (Proposition 1.1.17), so $\mathbb{Z}/𝑛\mathbb{Z}$ is not a field.

($\Leftarrow$) Let $𝑛=𝑝$ be prime and $𝑎\in \{1,\dots ,𝑝-1\}$. The only positive divisors of $𝑝$ are $1$ and $𝑝$, and $𝑝\nmid 𝑎$, so $gcd(𝑎,𝑝)=1$. By Bézout’s identity, $𝑎𝑥+𝑝𝑦=1$ for some integers $𝑥,𝑦$. Let $𝑏=𝑥mod𝑝$. By Lemma 1.1.21(a),

so $𝑏$ is an inverse of $𝑎$.

(b) The Bézout proof computes inverses: run the Euclidean algorithm and substitute back. Here

so

Hence $17\cdot 6=102=101+1$, and ${17}^{-1}=6$ in ${𝔽}_{101}$. By Proposition 1.1.13, $17𝑥+3=5$ holds exactly when $17𝑥=2$, that is, $𝑥=6\cdot 2=12$. Check: $17\cdot$$12+$$3=$$207=$$2\cdot$$101+$$5$.

The counting proof of Proposition 1.1.24 shows that $17$ has an inverse without producing it; using it to find ${17}^{-1}$ would mean trying the candidates one by one.

(a) We use only (A1)–(A7) and facts proved from them. Cancellation (Proposition 1.2.30(b)) and uniqueness of negatives (Proposition 1.2.28) use only (A1)–(A4). The first proof of Proposition 1.2.33 uses (A7), (A3), and cancellation, so $𝑎\mathrm{𝟎}=\mathrm{𝟎}$ holds without (A8). Now

so $𝑎(-𝐰)$ is an additive inverse of $𝑎𝐰$, and $𝑎(-𝐰)=-(𝑎𝐰)$ by Proposition 1.2.28. Hence, by (A7),

(b) First,

so $(-𝑏)𝐯=-(𝑏𝐯)$ by Proposition 1.2.28. Then by (A8),

The proof uses (A8) twice directly, and once more through Proposition 1.2.32, whose proof starts from $(0+0)𝐯=0𝐯+0𝐯$. Besides (A8) it uses (A1)–(A4) through Proposition 1.2.28 and cancellation.

(c) By Example 1.2.25(b), ${ℝ}^2$ with its usual addition and $𝑎\odot (𝑥,𝑦)=$$(𝑎𝑥,𝑦)$ satisfies every axiom except (A8). There,

while $2\odot (0,1)-$$1\odot (0,1)=$$(0,1)-$$(0,1)=$$(0,0)$. So $(𝑎-𝑏)\odot 𝐯\ne$$𝑎\odot 𝐯-$$𝑏\odot 𝐯$ for $𝑎=2$, $𝑏=1$, $𝐯=(0,1)$. Every step of (b) that invokes (A8) fails here, and so do the derived facts $0\odot 𝐯=\mathrm{𝟎}$ and $(-𝑏)\odot 𝐯=-(𝑏\odot 𝐯)$: for instance $0\odot (0,1)=(0,1)$, and $(-1)\odot (0,1)=$$(0,1)\ne$$(0,-1)$. Part (a), whose proof avoided (A8), still holds in this structure, as it must.

(a) Statement. $\mathrm{ℒ︀}$ is linearly independent if every vector of $span(\mathrm{ℒ︀})$ is represented by exactly one linear combination of $\mathrm{ℒ︀}$ (Definition 1.5.1). The test at zero: $\mathrm{ℒ︀}$ is independent if and only if ${𝑎}_1{𝐯}_1+$$\cdots +$${𝑎}_{𝑘}{𝐯}_{𝑘}=$$\mathrm{𝟎}$ implies ${𝑎}_1=\cdots ={𝑎}_{𝑘}=0$.

Proof. If $\mathrm{ℒ︀}$ is independent, then $\mathrm{𝟎}\in span(\mathrm{ℒ︀})$ is represented by the trivial relation, and by nothing else, so every relation is trivial. Conversely, we prove the contrapositive: suppose $\mathrm{ℒ︀}$ is dependent, so that some $𝐰$ has two different representations, with coefficients ${𝑎}_{𝑖}$ and ${𝑏}_{𝑖}$. By Lemma 1.3.9 and $-𝐱=(-1)𝐱$ (Proposition 1.2.34),

and ${𝑎}_{𝑗}-{𝑏}_{𝑗}\ne 0$ for some $𝑗$. So the test at zero fails.

(b) Let ${𝑎}_1,\dots ,{𝑎}_{𝑘}$ be the unique representation of $𝐰$, and let ${𝑐}_1,\dots ,{𝑐}_{𝑘}$ be any linear relation of $\mathrm{ℒ︀}$. By Lemma 1.3.9, the coefficients ${𝑎}_{𝑖}+{𝑐}_{𝑖}$ represent $𝐰+\mathrm{𝟎}=𝐰$. By uniqueness, ${𝑎}_{𝑖}+{𝑐}_{𝑖}={𝑎}_{𝑖}$ for every $𝑖$, so ${𝑐}_{𝑖}=0$. Every relation is trivial, and $\mathrm{ℒ︀}$ is independent by (a).

So uniqueness at a single vector of the span forces uniqueness everywhere. This is the reason the test can be run at $\mathrm{𝟎}$: any vector of the span would do, and $\mathrm{𝟎}$ is the one whose representation we always know.

(c) Neither. In ${ℝ}^2$ let $𝐰=(0,1)$. The list $(1,0)$ is independent (Proposition 1.5.15(b)), the list $(1,0),(2,0)$ is dependent (Proposition 1.5.15(d)), and $𝐰$ lies in neither span, since both spans consist of vectors with second coordinate $0$. A vector outside the span says something about spanning, not about independence.

(a) Statement. If $\mathrm{ℒ︀}={𝐯}_1,\dots ,{𝐯}_{𝑘}$ is linearly dependent, then ${𝐯}_{𝑗}\in$$span({𝐯}_1,\dots ,{𝐯}_{𝑗-1})$ for some $𝑗$, and for every such $𝑗$, deleting ${𝐯}_{𝑗}$ from $\mathrm{ℒ︀}$ does not change the span (Lemma 1.5.10).

Proof. By Theorem 1.5.5 there is a relation ${𝑎}_1{𝐯}_1+$$\cdots +$${𝑎}_{𝑘}{𝐯}_{𝑘}=$$\mathrm{𝟎}$ with some ${𝑎}_{𝑙}\ne 0$. Let $𝑗$ be the largest index with ${𝑎}_{𝑗}\ne 0$. The terms with index above $𝑗$ are $\mathrm{𝟎}$, so ${𝑎}_1{𝐯}_1+$$\cdots +$${𝑎}_{𝑗}{𝐯}_{𝑗}=$$\mathrm{𝟎}$, and since ${𝑎}_{𝑗}\ne 0$,

where for $𝑗=1$ the sum is empty and the statement is ${𝐯}_1=\mathrm{𝟎}$. Now let $𝑗$ be any index with ${𝐯}_{𝑗}\in$$span({𝐯}_1,\dots ,{𝐯}_{𝑗-1})$, and let ${\mathrm{ℒ︀}}^{\prime }$ be $\mathrm{ℒ︀}$ without ${𝐯}_{𝑗}$. Then ${𝐯}_{𝑗}\in span({\mathrm{ℒ︀}}^{\prime })$, since ${𝐯}_1,\dots ,{𝐯}_{𝑗-1}$ is a sublist of ${\mathrm{ℒ︀}}^{\prime }$ (Corollary 1.3.20(b)). By Corollary 1.3.20(c), $span({\mathrm{ℒ︀}}^{\prime },{𝐯}_{𝑗})=$$span({\mathrm{ℒ︀}}^{\prime })$, and ${\mathrm{ℒ︀}}^{\prime },{𝐯}_{𝑗}$ is a reordering of $\mathrm{ℒ︀}$, so its span is $span(\mathrm{ℒ︀})$ by Corollary 1.3.20(a).

(b) With $𝑖$ the smallest index with ${𝑎}_{𝑖}\ne 0$, the terms before $𝑖$ vanish, so ${𝑎}_{𝑖}{𝐯}_{𝑖}+$${𝑎}_{𝑖+1}{𝐯}_{𝑖+1}+$$\cdots +$${𝑎}_{𝑘}{𝐯}_{𝑘}=$$\mathrm{𝟎}$, and

So a dependent list also has an entry in the span of the entries after it. The extremal choice decides the side: the largest index leaves earlier entries, the smallest leaves later ones.

(c) Write ${𝐯}_1,$${𝐯}_2,$${𝐯}_3=$$(1,0),$$(0,1),$$(0,2)$.

Earlier entries. ${𝐯}_1\ne \mathrm{𝟎}$, so $𝑗=1$ fails. $span({𝐯}_1)$ is the first axis, which does not contain ${𝐯}_2$, so $𝑗=2$ fails. $span({𝐯}_1,{𝐯}_2)={ℝ}^2$ contains ${𝐯}_3$, so $𝑗=3$ works.

Later entries. $span({𝐯}_2,{𝐯}_3)$ is the second axis, which does not contain ${𝐯}_1$, so $𝑖=1$ fails. ${𝐯}_2=(1/2){𝐯}_3$, so $𝑖=2$ works. $span(())=\{\mathrm{𝟎}\}$ and ${𝐯}_3\ne \mathrm{𝟎}$, so $𝑖=3$ fails.

The two versions of the lemma pick out different redundant entries, here ${𝐯}_3$ and ${𝐯}_2$. Deleting either one leaves a basis of ${ℝ}^2$.

(a) By Theorem 1.4.25, the lines are the sets $span(𝐯)$ with $𝐯\ne \mathrm{𝟎}$, and $span(𝐯)=$$span(𝑐𝐯)$ for $𝑐\ne 0$ (Exercise 1.3.28(b)). If $𝐯=(𝑎,𝑏)$ with $𝑎\ne 0$, then $span(𝐯)=$$span((1,𝑏/𝑎))$; if $𝑎=0$, then $𝑏\ne 0$ and $span(𝐯)=$$span((0,1))$. The lines $span((1,𝑐))$, $𝑐\in {𝔽}_5$, are different from each other and from $span((0,1))$, since the only multiple of $(1,𝑐)$ with first coordinate $1$ is $(1,𝑐)$ itself, and $span((0,1))$ contains no vector with first coordinate $1$. So the six lines are

For $(2,3)$: $3/2=3\cdot 2^{-1}=3\cdot 3=9=4$, so $(2,3)=2(1,4)$ lies on $span((1,4))$.

(b) For $(𝑥,𝑦)\in {𝔽}_5^2$, the equation $𝑎(1,2)+$$𝑏(1,3)=$$(𝑥,𝑦)$ reads $𝑎+𝑏=𝑥$ and $2𝑎+3𝑏=𝑦$. Subtracting twice the first equation from the second gives $𝑏=𝑦-2𝑥$, and then $𝑎=𝑥-𝑏=3𝑥-𝑦$. Conversely these values satisfy both equations: $𝑎+𝑏=𝑥$, and $2𝑎+$$3𝑏=$$6𝑥-$$2𝑦+$$3𝑦-$$6𝑥=$$𝑦$. So every $(𝑥,𝑦)$ has exactly one representation, $\mathrm{ℬ︀}$ is a basis (Definition 1.6.1), and by (1.6.1),

For $(2,3)$: $3\cdot 2-3=3$ and $3-4=-1=4$, so ${[(2,3)]}_{\mathrm{ℬ︀}}=(3,4)$. Check: $3(1,2)+$$4(1,3)=$$(7,18)=$$(2,3)$ in ${𝔽}_5^2$.

(c) A list $𝐮,𝐯$ in ${𝔽}_5^2$ with $𝐮\ne \mathrm{𝟎}$ spans ${𝔽}_5^2$ exactly when $𝐯\notin span(𝐮)$ (Exercise 1.3.37(b)), and then it is also independent by Proposition 1.5.14, since the list $𝐮$ is independent (Proposition 1.5.15(b)). So it is a basis exactly when $𝐮\ne \mathrm{𝟎}$ and $𝐯\notin span(𝐮)$ (Proposition 1.6.2). The first entry is one of the $4$ nonzero vectors of $span((1,2))$, and its span is that line, which has $5$ elements. So there are $25-5=20$ choices for the second entry, and $4\cdot 20=80$ bases in all.

Write $𝑝=$$𝑎+$$𝑏𝑥+$$𝑐{𝑥}^2+$$𝑑{𝑥}^3$. By Definition 1.2.17, $𝑝(1)=𝑎+𝑏+𝑐+𝑑$ and $𝑝(-1)=$$𝑎-$$𝑏+$$𝑐-$$𝑑$.

(a) If $𝑎+𝑐=0$ and $𝑏+𝑑=0$, then $𝑝(1)=$$(𝑎+𝑐)+$$(𝑏+𝑑)=$$0$ and $𝑝(-1)=$$(𝑎+𝑐)-$$(𝑏+𝑑)=$$0$. Conversely, if $𝑝(1)=𝑝(-1)=0$, then adding and subtracting gives $(1+1)(𝑎+𝑐)=0$ and $(1+1)(𝑏+𝑑)=0$, so $𝑎+𝑐=0$ and $𝑏+𝑑=0$ because $1+1\ne 0$ (Proposition 1.1.17). Hence $𝑝\in 𝑈\cap 𝑊$ exactly when $𝑐=-𝑎$ and $𝑑=-𝑏$, that is,

So $𝑈\cap$$𝑊=$$span(1-{𝑥}^2,𝑥-{𝑥}^3)$. The two polynomials are nonzero of degrees $2$ and $3$, so they are independent (Proposition 1.5.24), and they form a basis of $𝑈\cap 𝑊$ (Proposition 1.6.2).

For $𝑈+𝑊$: the polynomials $𝑥-1$, ${𝑥}^2-1$, ${𝑥}^3-1$ vanish at $1$ and lie in $𝑈$, and $𝑥+1$ vanishes at $-1$ and lies in $𝑊$. The sum $𝑈+𝑊$ is a subspace containing $𝑈$ and $𝑊$ (Proposition 1.4.24), so it contains

hence $1={(1+1)}^{-1}(1+1)$, and then $𝑥=(𝑥-1)+1$, ${𝑥}^2=({𝑥}^2-1)+1$, and ${𝑥}^3=({𝑥}^3-1)+1$. These span ${𝐹[𝑥]}_{\le 3}$ (Proposition 1.6.7), so $𝑈+𝑊={𝐹[𝑥]}_{\le 3}$ by Proposition 1.4.14(b).

(b) Over ${𝔽}_2$, $-1=1$, so $𝑝(-1)=𝑝(1)$ for every $𝑝$, and $𝑈=𝑊$. Hence $𝑈\cap 𝑊=𝑈+𝑊=𝑈$. Since $-(𝑏+𝑐+𝑑)=$$𝑏+$$𝑐+$$𝑑$ over ${𝔽}_2$, $𝑝\in 𝑈$ exactly when $𝑎=𝑏+𝑐+𝑑$, that is,

So $𝑈=$$span(1+𝑥,1+{𝑥}^2,1+{𝑥}^3)$, and this list is a basis of $𝑈$ by distinct degrees. The subspace $𝑈+𝑊=𝑈$ is proper, since $1\notin 𝑈$. Over ${𝔽}_2$, the two conditions are one condition, and the computation in (a) broke exactly where it divided by $1+1$.

(a) The list spans $𝑉$ by definition. Suppose $𝑎+$$𝑏cos𝑡+$$𝑐sin𝑡=$$0$ for all $𝑡$. At $𝑡=0$ this gives $𝑎+𝑏=0$, at $𝑡=𝜋$ it gives $𝑎-𝑏=0$, so $𝑎=𝑏=0$, and at $𝑡=𝜋/2$ it gives $𝑎+𝑐=0$, so $𝑐=0$. The list is independent (Theorem 1.5.5), hence a basis of $𝑉$ (Proposition 1.6.2).

(b) By (1.6.1), it suffices to write each function as a combination of $1,cos,sin$, valid for every $𝑡$.

• $cos(𝑡+𝜋/3)=$$cos(𝜋/3)cos𝑡-$$sin(𝜋/3)sin𝑡=$$(1/2)cos𝑡-$$(\sqrt{3}/2)sin𝑡$, with coordinates $(0,1/2,-\sqrt{3}/2)$.
• $cos𝑡=$$1-$$2{sin}^2(𝑡/2)$, so ${sin}^2(𝑡/2)=$$1/2-$$(1/2)cos𝑡$, with coordinates $(1/2,-1/2,0)$.
• ${(cos(𝑡/2)+sin(𝑡/2))}^2=$${cos}^2(𝑡/2)+$${sin}^2(𝑡/2)+$$2sin(𝑡/2)cos(𝑡/2)=$$1+$$sin𝑡$, with coordinates $(1,0,1)$.

(c) No. Suppose ${cos}^2𝑡=$$𝑎+$$𝑏cos𝑡+$$𝑐sin𝑡$ for all $𝑡$. At $𝑡=0$ and $𝑡=𝜋$ we get $1=𝑎+𝑏$ and $1=𝑎-𝑏$, so $𝑎=1$ and $𝑏=0$. At $𝑡=𝜋/2$ we get $0=𝑎+𝑐$, so $𝑐=-1$. So the only candidate is $1-sin𝑡$. But at $𝑡=-𝜋/2$, ${cos}^2𝑡=0$ while $1-sin𝑡=2$. Hence $𝑡\mapsto {cos}^2𝑡$ is not in $𝑉$. Three evaluations determined the only possible coefficients, and a fourth refuted them.

(a) ${𝐾}^{𝑛}$ contains $\mathrm{𝟎}$, since $0\in 𝐾$. For $𝐱,𝐲\in {𝐾}^{𝑛}$ and $𝑎\in 𝐾$, the coordinates ${𝑥}_{𝑖}+{𝑦}_{𝑖}$ and $𝑎{𝑥}_{𝑖}$ lie in $𝐾$, because $𝐾$ is closed under the operations of $𝐹$ (Definition 1.1.32). So ${𝐾}^{𝑛}$ is a subspace of the $𝐾$-vector space ${𝐹}^{𝑛}$ by Theorem 1.4.3. Over $𝐹$ it is not closed under scalar multiplication: choose $𝑐\in 𝐹$ with $𝑐\notin 𝐾$; then ${𝐞}_1\in {𝐾}^{𝑛}$, but $𝑐{𝐞}_1=$$(𝑐,0,\dots ,0)\notin$${𝐾}^{𝑛}$.

(b) $\{\mathrm{𝟎}\}$ is such a subspace. Let $𝑈$ be a subspace of the $𝐹$-vector space ${𝐹}^{𝑛}$ with $𝑈\subseteq {𝐾}^{𝑛}$, and suppose for contradiction that $𝑈$ contains some $𝐯\ne \mathrm{𝟎}$. Choose $𝑖$ with ${𝑣}_{𝑖}\ne 0$. For any $𝑐\in 𝐹$, the vector $(𝑐{𝑣}_{𝑖}^{-1})𝐯$ lies in $𝑈$, hence in ${𝐾}^{𝑛}$, and its $𝑖$-th coordinate is $𝑐{𝑣}_{𝑖}^{-1}{𝑣}_{𝑖}=𝑐$. So $𝑐\in 𝐾$ for every $𝑐\in 𝐹$, that is, $𝐹=𝐾$, a contradiction.

(c) $𝑈\cap {𝐾}^{𝑛}$ contains $\mathrm{𝟎}$. Let $𝐱,𝐲\in 𝑈\cap {𝐾}^{𝑛}$ and $𝑎\in 𝐾$. Then $𝐱+𝐲\in 𝑈$ and $𝑎𝐱\in 𝑈$, because $𝑈$ is closed under sums and under multiples by scalars of $𝐹$, and $𝑎\in 𝐾\subseteq 𝐹$. Also $𝐱+𝐲\in {𝐾}^{𝑛}$ and $𝑎𝐱\in {𝐾}^{𝑛}$, by the computation in (a). So $𝑈\cap {𝐾}^{𝑛}$ is a subspace of the $𝐾$-vector space ${𝐾}^{𝑛}$ by Theorem 1.4.3.

Now let $𝐹=ℂ$, $𝐾=ℝ$, $𝑛=2$. For $𝑈=span((1,𝑖))$, a vector $𝑐(1,𝑖)=$$(𝑐,𝑐𝑖)$ lies in ${ℝ}^2$ only if $𝑐\in ℝ$ and $𝑐𝑖\in ℝ$. Writing $𝑐𝑖=0+𝑐𝑖$ (Definition 1.1.29), the second condition forces $𝑐=0$. So $𝑈\cap {ℝ}^2=\{\mathrm{𝟎}\}$, although $𝑈$ is a line in ${ℂ}^2$. For $𝑈=span((1,1))$, the intersection $𝑈\cap {ℝ}^2$ contains $(1,1)$; in fact it is $\{(𝑐,𝑐)|𝑐\in ℝ\}$, the real line through $(1,1)$.

By (b), $𝑈\cap {ℝ}^2$ is never all of $𝑈$ unless $𝑈=\{\mathrm{𝟎}\}$.

(a) The zero vector is $1$, since $𝑥\oplus 1=𝑥\cdot 1=𝑥$ for all $𝑥\in 𝑃$. So $1,2$ is dependent: it contains the zero vector (Proposition 1.5.15(c)). Next, $2\odot 2=2^2=4$, so $4$ is a scalar multiple of $2$, and $2,4$ is dependent (Proposition 1.5.15(d)). Finally, $({log}_{2}3)\odot 2=$$2^{{log}_{2}3}=$$3$, so $2,3$ is dependent as well. In fact any two vectors $𝑥,𝑦$ of $𝑃$ are dependent over $ℝ$: if $𝑥=1$, the list contains the zero vector, and if $𝑥\ne 1$, then $𝑦=(ln𝑦/ln𝑥)\odot 𝑥$ is a multiple of $𝑥$.

(b) A linear combination over $\mathbb{Q}$ represents $𝑎\odot 2\oplus 𝑏\odot 3=2^{𝑎}{3}^{𝑏}$. Suppose $2^{𝑎}{3}^{𝑏}=1$ with $𝑎,𝑏\in \mathbb{Q}$. Write $𝑎=𝑚/𝑁$ and $𝑏=𝑛/𝑁$ with integers $𝑚,𝑛$ and $𝑁\ge 1$. Raising to the power $𝑁$ gives $2^{𝑚}{3}^{𝑛}=1$, that is,

where ${𝑚}^{+}=max(𝑚,0)$ and ${𝑚}^{-}=max(-𝑚,0)$, and similarly for $𝑛$. Both sides are positive integers, so by unique factorization the exponents of $2$ agree and the exponents of $3$ agree: ${𝑚}^{+}={𝑚}^{-}$ and ${𝑛}^{+}={𝑛}^{-}$. One of ${𝑚}^{+},{𝑚}^{-}$ is always $0$, so $𝑚=0$, and likewise $𝑛=0$. Hence $𝑎=𝑏=0$, and $2,3$ is independent over $\mathbb{Q}$ by Theorem 1.5.5.

On the other hand, $1\odot 2\oplus$$1\odot 3\oplus$$(-1)\odot 6=$$2\cdot$$3\cdot$$6^{-1}=$$1$, the zero vector, so $2,3,6$ is dependent over $\mathbb{Q}$. And $2\odot 2\oplus$$(-1)\odot 4=$$4\cdot$$4^{-1}=$$1$, so $2,4$ is dependent over $\mathbb{Q}$ too.

(c) Suppose $5=2^{𝑎}{3}^{𝑏}$ with $𝑎=𝑚/𝑁$, $𝑏=𝑛/𝑁$ as before. Raising to the power $𝑁$ gives $5^{𝑁}=2^{𝑚}{3}^{𝑛}$, so $5^{𝑁}{2}^{{𝑚}^{-}}{3}^{{𝑛}^{-}}=$$2^{{𝑚}^{+}}{3}^{{𝑛}^{+}}$. The prime $5$ divides the left side and not the right side, contradicting unique factorization. So $5\notin span(2,3)$ over $\mathbb{Q}$.

In this structure $1$ is the zero vector, so “a list containing $1$” is the analogue of a list containing $\mathrm{𝟎}$. Under the bijection $ln:𝑃\to ℝ$ the problem becomes a statement about $ln2$, $ln3$, and $ln5$ over $\mathbb{Q}$, which is Exercise 1.5.39.

(a) True. By Proposition 1.5.15(d), if $𝐮,𝐰$ were dependent, one would be a multiple of the other. If $𝐮=𝑐𝐰$, then $𝐮\in 𝑊$, since $𝑊$ is closed under scalar multiplication; if $𝐰=𝑐𝐮$, then $𝐰\in 𝑈$. Both contradict the hypotheses.

(b) False. Over ${𝔽}_2$, let $𝑉={𝔽}_2^2$, $𝑈=span({𝐞}_1)$, $𝑊=span({𝐞}_2)$, $𝐮={𝐞}_1$, $𝐰={𝐞}_2$. The hypotheses hold, but $𝐮-𝐰=𝐮+𝐰$ because $-1=1$, so the list repeats an entry and is dependent (Proposition 1.5.15(c)). (If $1+1\ne 0$, the statement is true, by (a) and Exercise 1.5.8.)

(c) True. If $\mathrm{ℒ︀}$ is independent, then by Proposition 1.5.14, $\mathrm{ℒ︀},𝐯$ is dependent exactly when $𝐯\in span(\mathrm{ℒ︀})$, so the set is $span(\mathrm{ℒ︀})$, a subspace (Proposition 1.4.14). If $\mathrm{ℒ︀}$ is dependent, then $\mathrm{ℒ︀},𝐯$ has the dependent sublist $\mathrm{ℒ︀}$ for every $𝐯$, so it is dependent (Corollary 1.5.17), and the set is $𝑉$.

(d) False. The list $1$ spans $ℂ$ over $ℂ$, but over $ℝ$ its span is $ℝ\ne ℂ$ (Example 1.6.5).

(e) False. For $\mathrm{ℒ︀}=\mathrm{𝟎}$, the list of length one, $span(\mathrm{ℒ︀})=\{\mathrm{𝟎}\}$, so $span(\mathrm{ℒ︀})\cap 𝑈=\{\mathrm{𝟎}\}$ for every $𝑈$, but $\mathrm{ℒ︀}$ is dependent (Proposition 1.5.15(b)).

(f) True. $𝑈+𝑊$ is a subspace containing $𝑈$ and $𝑊$ (Proposition 1.4.24), so it contains every entry of $\mathrm{ℒ︀}$, and hence $span(\mathrm{ℒ︀})$ by Proposition 1.4.14(b).

(a) The error is “these are the entries of $\mathrm{ℒ︀}$”. The entries before ${𝐰}_{𝑗}$ in $\mathrm{ℒ︀},\mathrm{ℳ︀}$ are ${𝐮}_1,\dots ,{𝐮}_{𝑘}$ and ${𝐰}_1,\dots ,{𝐰}_{𝑗-1}$, and the span of these need not lie in $𝑈$. For a counterexample, take $𝑉={ℝ}^2$, $𝑈=span({𝐞}_1)$ with basis $\mathrm{ℒ︀}={𝐞}_1$, and $𝑊={ℝ}^2$ with basis $\mathrm{ℳ︀}={𝐞}_2,{𝐞}_1+{𝐞}_2$. Sifting ${𝐞}_1,{𝐞}_2,{𝐞}_1+{𝐞}_2$ keeps ${𝐞}_1$ (it is nonzero), keeps ${𝐞}_2$ (it is not a multiple of ${𝐞}_1$), and deletes ${𝐞}_1+$${𝐞}_2\in$$span({𝐞}_1,{𝐞}_2)$. But ${𝐞}_1+{𝐞}_2\notin 𝑈$, so the deleted entry is not in $𝑈\cap 𝑊=𝑈$.

(b) For $𝑗\le 𝑘$, the entry ${𝐮}_{𝑗}$ is preceded only by ${𝐮}_1,\dots ,{𝐮}_{𝑗-1}$. Since $\mathrm{ℒ︀}$ is independent, ${𝐮}_{𝑗}\notin$$span({𝐮}_1,\dots ,{𝐮}_{𝑗-1})$ by Corollary 1.5.13, so ${𝐮}_{𝑗}$ is kept. By Theorem 1.6.10 (third proof), the kept entries form an independent list with the same span as $\mathrm{ℒ︀},\mathrm{ℳ︀}$, which is $span(\mathrm{ℒ︀})+$$span(\mathrm{ℳ︀})=$$𝑈+$$𝑊$ by Exercise 1.3.36. So they form a basis of $𝑈+𝑊$ (Proposition 1.6.2).

(c) If ${𝐰}_{𝑗}$ is deleted, then ${𝐰}_{𝑗}\in$$span({𝐮}_1,\dots ,{𝐮}_{𝑘},{𝐰}_1,\dots ,{𝐰}_{𝑗-1})$, say

Let $𝐲=$${\sum }_{𝑙<𝑗}{𝑏}_{𝑙}{𝐰}_{𝑙}$. Then ${𝐰}_{𝑗}-$$𝐲=$${\sum }_{𝑖}{𝑎}_{𝑖}{𝐮}_{𝑖}$ lies in $𝑈$, and it lies in $𝑊$ because ${𝐰}_{𝑗}$ and $𝐲$ do. It is nonzero: otherwise ${𝐰}_{𝑗}-$${\sum }_{𝑙<𝑗}{𝑏}_{𝑙}{𝐰}_{𝑙}=$$\mathrm{𝟎}$ would be a nontrivial relation of $\mathrm{ℳ︀}$ (the coefficient of ${𝐰}_{𝑗}$ is $1$), contradicting the independence of $\mathrm{ℳ︀}$.

So each deleted entry produces a nonzero vector of $𝑈\cap 𝑊$, though not necessarily the entry itself. Showing that these vectors form a basis of $𝑈\cap 𝑊$ gives one route to the formula of Section 4.4 relating the sizes of bases of $𝑈$, $𝑊$, $𝑈+𝑊$, and $𝑈\cap 𝑊$.