Linear Algebra Done Right: Chapter 1
07.01.2019

This post works though the exercises for Chapter 1 of Axler's Linear Algebra Done Right. The summary review can be found here.

Chapter 1s are usually gentle introductions and that is no less true for this text than any other. After the briefest introduction of complex numbers and their arithmetic, and an introduction of the concept of a field to almost the minimal extent possible, Axler presents the concepts of vector spaces, vector subspaces, and sums and direct sums. The latter is perhaps the most surprising of anything introduced in this chapter; most undergraduate texts tend to skip this entirely.

Axler introduces fields but tells the reader to ignore them for the most part, and seems to present the concept so that he can use $$\mathbb{F}$$ to represent $$\mathbb{R}$$ or $$\mathbb{C}$$. He moves the reader towards abstract concepts but develops very little foundation as to why, how, or for what benefit. Likewise, he introduces abstract polynomial vector spaces, $$\mathcal{P}(\mathbb{F})$$ with seemingly little guidance.

Strangely, he presumes familiarity with set theory and set builder notation; I've had far more advanced texts spend at least a couple early pages on this to ensure completeness. No theorems, and one or two useful propositions, are presented and proved. His writing style does not beget clarity; most texts, even ones that strive for more abstraction, will present things like properties and definitions in some coherent, explicit manner. Axler does not, and you have to do the mental bookkeeping by extracting what he wants from his prose.

1. Suppose $$a$$ and $$b$$ are real numbers, not both $$0$$. Find real numbers $$c$$ and $$d$$ such that $$1 / (a + bi) = c + di$$.

We have

\begin{align} 1 / (a + bi) &= c + di \\ 1 &= (a + bi)(c + di) \\ &= ac - bd + (bc + ad)i. \end{align}

This implies

\begin{align} ac - bd &= 1, \\ bc + ad &= 0. \end{align}

Let us proceed by cases. Suppose $$a = 0.$$ Then by hypothesis, $$b \neq 0$$. From $$ac - bd = 1$$ we get $$d = -\frac{1}{b}$$ and from $$bc + ad = 0$$ we get $$bc = 0 \implies c = 0$$.

Suppose $$b = 0,$$ then by hypothesis $$a \neq 0$$ and we reduce to real-number division.

Finally, suppose neither $$a$$ nor $$b$$ are zero. Then $$bc = -ad \implies c = -\frac{ad}{b}$$. Substituting, we find, $$-\frac{a^2d}{b} - bd = 1$$, which reduces to $$d(a^2+b^2) = -b$$. This gives us the general result

\begin{align} c &= \frac{a}{a^2+b^2}, \\ d &= -\frac{b}{a^2+b^2}. \end{align}

2. Show that $$\frac{-1+\sqrt{3}i}{2}$$ is a cube root of $$1.$$

We verify through computation:

\begin{align} \left(\frac{-1 + \sqrt{3}i}{2}\right)^3 &= \left(-\frac12+\frac{\sqrt3}{2}i\right) \left( -\frac12 - \frac{\sqrt3}{2}i\right) \\ &= \frac14 + \frac34 \\ &= 1. \end{align}

3. Prove that $$-(-v) = v$$ for every $$v \in V.$$

From $$-v + v = 0$$, we have that $$-v \equiv (-1)v$$, so $$-(-v) = (-1)(-v) = (-1)((-1)v).$$ By scalar multiplication, this becomes $$-(-v) = (-1)(-1)v = 1v = v$$.

4. Prove that if $$a \in \mathbb{F}, v \in V$$, and $$av = 0$$, then $$a = 0$$ or $$v = 0.$$

Suppose not. Then $$a \neq 0$$ and $$v \neq 0$$. Since $$av = (av_1, av_2, \ldots)$$ and since $$v \neq 0$$ implies that there exists some $$j$$ for which $$v_j \neq 0$$, this implies that $$av_j = 0$$ for non-zero $$a$$ and $$v_j$$. However, since $$\mathbb{F}$$ is a field, it has no non-zero zero divisors, so this proves the contradiction.1

5. For each of the following subsets of $$\mathbb{F}^3$$, determine whether it is a subspace of $$\mathbb{F}^3$$:

(a.) $$\{(x_1, x_2, x_3) \in \mathbb{F}^3\ :\ x_1 + 2x_2 + 3x_3 = 0\}$$

(b.) $$\{(x_1, x_2, x_3) \in \mathbb{F}^3\ :\ x_1 + 2x_2 + 3x_3 = 4\}$$

(c.) $$\{(x_1, x_2, x_3) \in \mathbb{F}^3\ :\ x_1x_2x_3 = 0\}$$

(d.) $$\{(x_1, x_2, x_3) \in \mathbb{F}^3\ :\ x_1 = 5x_3\}$$

We can proceed by checking against the subspace axioms. For (a), we have $$0 + 2\cdot 0 + 3\cdot 0 = 0$$ and $$ax_1 + 2ax_2 + 3ax_3 = a(x_1+2x_2+3x_3) = a0 = 0$$ so we must only do work to verify closure by addition. Let $$x, y$$ be two vectors satisfying the subset condition. Take $$x_1+y_1 + 2(x_2 + y_2) + 3(x_3 + y_3) = x_1 + 2x_2 + 3x_3 + y_1 + 2y_2 + 3y_3 = 0 + 0 = 0.$$ Therefore, the set is closed under vector addition, so the space is a subspace.

For (b), we have an immediate counterexample: $$0 + 2\cdot 0 + 3\cdot 0 \neq 4,$$ so the subset is not a subspace.

For (c), consider $$x = (1, 0, 0)$$ and $$y = (0, 1, 1)$$. Then, $$x+y = (1, 1, 1)$$ is not in the set, so it is not closed under addition and cannot be a subspace.

For (d), we have that $$0 = 5\cdot 0$$. If $$x_1 = 5x_3$$, then $$ax_1 = 5ax_3$$, so all that is left is to check closure of addition. Once again, if $$x$$ and $$y$$ meet the criteria, then $$x_1 = 5x_3$$ and $$y_1 = 5y_3$$ so $$x_1 + y_1 = 5x_3 + 5y_3 \implies (x_1 + y_1) = 5(x_3 + y_3)$$ so we do in fact have a vector subspace.

6. Give an example of a nonempty subset $$U$$ of $$\mathbb{R}^2$$ such that $$U$$ is closed under addition and under taking additive inverses, but $$U$$ is not a subspace of $$\mathbb{R}^2$$.

This is the first thought-provoking exercise of the book. We can construct what we must see in the counter-example from what is given. If $$U$$ is closed under addition (satisfying one axiom), and is closed under additive inverses, then it must also contain the identity element (satisfying another axiom). The remaining axiom is closure under scalar multiplication. This makes it easy at this point. Let

$U = \left\{ (x, y)\ :\ x, y \in \mathbb{Z}\right\},$

then for any $$\alpha \in \mathbb{R} \setminus \mathbb{Z}$$ it is clear that $$\alpha (x,y) \notin U$$.

7. Give an example of a nonempty subset $$U$$ of $$\mathbb{R}^2$$ such that $$U$$ is closed under scalar multiplication, but $$U$$ is not a subspace of $$\mathbb{R}^2$$.

Like the previous example, we can use the axioms to derive an example. Consider the following:

$U = \left\{ (x, 0)\ :\ x \in \mathbb{R} \right\} \cup \left\{ (0, y)\ :\ y \in \mathbb{R} \right\}.$

Evidently, $$U \subset \mathbb{R}^2.$$ For any $$v \in U$$, we have $$\alpha v$$ implies $$(\alpha x, 0) \in U$$ or $$(0, \alpha y) \in U$$, so we satisfy closure under scalar multiplication. However, $$(x, y) \notin U$$ for $$x, y$$ both nonzero, so we do not satisfy closure under addition, and therefore we have not created a subspace.

8. Prove that the intersection of any collections of subspaces of $$V$$ is a subspace of $$V.$$

Let $$U_i$$ be a subspace of $$V$$ for all $$i$$, and let $$U = \bigcap_i U_i$$. Suppose $$U$$ is not a subspace of $$V$$. Then, at least one of the following statements must be true:

• There is a $$U_j$$ such that $$0 \notin U_j$$.
• Take $$x, y \in U$$. Then, there is a $$U_k$$ such that $$x, y \in U_k$$ but $$x + y \notin U_k$$.
• Take $$x \in U$$. Then, there is a $$U_l$$ such that $$x \in U_l$$ but $$\alpha x \notin U_l$$.

Each of these conditions violates the hypotheses that for all $$j$$, $$U_j$$ is a subspace. Therefore, $$U$$ must be a subspace.

9. Prove that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces is contained in the other.

Let $$V_1$$ and $$V_2$$ be subspaces of $$V$$ and assume that $$U = V_1 \cup V_2$$ is a subspace of $$V$$. Take $$x \in V_1 \setminus V_2$$ and $$y \in V_2 \setminus V_1$$. Evidently, $$x, y \in U$$, so $$x + y \in U$$. Since $$U = V_1 \cup V_2$$, then $$x + y \in V_1$$ or $$x + y \in V_2$$, necessarily. By closure under addition, this implies that either $$y \in V_1$$ or $$x \in V_2$$, contradicting the assumption that both subspaces can have unique elements.

For the converse, assume $$V_1 \subset V_2$$. Then $$V_1 \cup V_2 = V_2$$, which is a subspace by hypothesis.

10. Suppose $$U$$ is a subspace of $$V$$. What is $$U + U$$?

Since $$U$$ is a subspace, it is closed under addition. Since every element of $$U+U$$ is the sum of two elements of $$U$$, and since $$U$$ is closed under addition, then $$U + U = U$$.

11. Is the operation of addition on the subspaces of $$V$$ commutative? Associative?

Let $$U_1, U_2, U_3$$ be subspaces of $$V$$. We have:

\begin{align} U_1 + U_2 &= \left\{ x + y\ :\ x \in U_1, y \in U_2 \right\} \\ &= \left\{ (x_1, x_2, \ldots) + (y_1, y_2, \ldots)\ :\ x \in U_1, y \in U_2 \right\} \\ &= \left\{ (x_1 + y_1, x_2 + y_2, \ldots)\ :\ x \in U_1, y \in U_2 \right\} \\ &= \left\{ (y_1 + x_1, y_2 + x_2, \ldots)\ :\ x \in U_1, y \in U_2 \right\} \\ &= \left\{ (y_1, y_2, \ldots) + (x_1, x_2, \ldots)\ :\ x \in U_1, y \in U_2 \right\} \\ &= U_2 + U_1. \end{align}

Furthermore,

\begin{align} (U_1 + U_2) + U_3 &= \left\{ (x + y) + z\ :\ x \in U_1, y \in U_2, z \in U_3 \right\} \\ &= \left\{ x + (y + z)\ :\ x \in U_1, y \in U_2, z \in U_3 \right\} \\ &= U_1 + (U_2 + U_3). \end{align}

12. Does the operation of addition on subspaces of $$V$$ have an additive identity? Which subspaces have additive inverses?

Consider a subset containing the zero vector and nothing more; this is trivially a subspace. Moreover, since every subspace by definition contains the zero vector, adding this trivial subspace to any other nontrivial subspace yields precisely the non-trivial subspace.

Suppose a subspace is an additive inverse of another subspace. Then, by the definition of subspace addition, every combination of sums of elements must be the zero vector. But since every subspace contains the zero vector, this can only be true if every element of the subspace and its additive inverse is the zero vector. Therefore, only the trivial subspace has an additive inverse.

13. Prove or give a counterexample: if $$U_1, U_2, W$$ are subspaces of $$V$$ such that $$U_1 + W = U_2 + W$$, then $$U_1 = U_2$$.

This is not generally true. Let $$U_1 = W$$ and let $$U_2$$ be the trivial subspace. Then since $$W + W = W$$, we have $$U_1 + W = U_2 + W$$ but by definition $$U_1 \neq U_2$$.

14. Suppose $$U$$ is the subspace of $$\mathcal{P}(\mathbb{F})$$ consisting of all polynomials $$p$$ of the form

$p(z) = az^2 + bz^5,$

where $$a, b \in \mathbb{F}$$. Find a subspace $$W$$ of $$\mathcal{P}(\mathbb{F})$$ such that $$\mathcal{P}(\mathbb{F}) = U \oplus W$$.

Proposition 1.9 of the text tells us that $$\mathcal{P}(\mathbb{F}) = U \oplus W$$ if and only if $$\mathcal{P}(\mathbb{F}) = U + W$$ and $$U \cap W = \{0\}$$. Therefore, we can construct $$W$$ in such a way as to avoid the "powers" contained in $$U$$; namely:

$W = \left\{ \sum_{i \neq 2, 5} c_i z^i\ :\ c_i \in \mathbb{F} \right\}.$

Let us verify that $$W$$ has all the desired properties:

• $$0 \in W$$: let $$c_i = 0$$ for all $i$;
• closed under addition: let $$x, y \in W$$, then by the properties of fields, we find that $$x + y \in W$$;
• closed under scalar multiplication: $$\alpha c_i \in \mathbb{F}$$ for all $$\alpha \in \mathbb{F}$$ implies that $$\alpha x \in W$$ for all $$x \in W$$;
• $$U \cap W = \{0\}$$: by construction;
• $$U + W = \mathcal{P}(\mathbb{F})$$: by construction.

Therefore, $$\mathcal{P}(\mathbb{F}) = U \oplus W$$.

15. Prove or give a counterexample: if $$U_1, U_2, W$$ are subspaces of $$V$$ such that $$V = U_1 \oplus W$$ and $$V = U_2 \oplus W$$, then $$U_1 = U_2$$.

Let $$V = \left\{ \left( x_1, x_2 \right) \ :\ x_1, x_2 \in \mathbb{F} \right\}$$. Take $$U_1 = \left\{ \left( x_1, 0 \right) \ :\ x_1 \in \mathbb{F} \right\}$$ and $$U_2 = \left\{ \left(0, x_2 \right) \ :\ x_2 \in \mathbb{F} \right\}$$. Then, clearly we have $$U_1, U_2 \subset V$$ and that $$U_1, U_2$$ satisfy the vector space conditions. Hence, they are subspaces of $$V$$.

Now take $$W = \left\{ \left( x_1, x_1 \right) \ :\ x\in \mathbb{F} \right\}$$. Again, it is clear that $$W \subset V$$ and the vector space axioms hold. So $$W$$ is a subspace of $$V$$ as well.

Now, it is clear that $$U_1 \neq U_2$$, and that $$U_1 \cap W = \{0\} = U_2 \cap W,$$ by definition. So we must simply show that $$V = U_1 + W = U_2 + W$$. Take $$\left(x_1, x_2\right) = v \in V$$, and let $$x_1 < x_2$$. Then, choose $$\left(x_1, x_1\right) = w_1 \in W$$ and then choose $$\left(x_2 - x_1, 0\right) \in U_1$$. Likewise, if $$x_2 > x_1$$, we merely reverse the indices. This demonstrates that $$V = U_1 + W = U_2 + W$$, so $$V = U_1 \oplus W = U_2 \oplus W$$, and a counterexample is derived.

1. Axler does not introduce this property of fields at this stage of the text. Implicitly, by the books convention, we have $$\mathbb{F}$$ to be assumed as either $$\mathbb{R}$$ or $$\mathbb{C}$$, but this restriction herein adds very little except extra baggage to account for. [return]

Posted: 07.01.2019

Built: 11.08.2022

Updated: 29.12.2021

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