Solutions

1.1 Fields and Subfields ¶

Solution: Exercise.

It is known that $ℚ \subset ℝ$ . We can show that $ℚ$ is a subfield of $ℝ$ by verifying that $ℚ$ satisfies the three conditions in Theorem 2.

Existence of identities. The rational numbers $ℚ$ contain the additive identity $0_{ℝ}$ and the multiplicative identity $1_{ℝ}$ .
Closure under subtraction. Suppose $𝑎, 𝑏 \in ℚ$ . Then there exist integers $𝑚, 𝑛, 𝑝, 𝑞$ with $𝑛 \neq 0$ and $𝑞 \neq 0$ such that $𝑎 = 𝑚 / 𝑛$ and $𝑏 = 𝑝 / 𝑞$ . We have:
$𝑎 - 𝑏 = \frac{𝑚}{𝑛} - \frac{𝑝}{𝑞} = \frac{𝑚 𝑞 - 𝑛 𝑝}{𝑛 𝑞},$
which is a rational number since $𝑚 𝑞 - 𝑛 𝑝$ is an integer and $𝑛 𝑞 \neq 0$ .
Closure under division. Suppose $𝑎, 𝑏 \in ℚ$ with $𝑏 \neq 0$ . Then there exist integers $𝑚, 𝑛, 𝑝, 𝑞$ with $𝑛 \neq 0$ and $𝑞 \neq 0$ such that $𝑎 = 𝑚 / 𝑛$ and $𝑏 = 𝑝 / 𝑞$ . We have:
$𝑎 / 𝑏 = \frac{𝑚}{𝑛} \cdot {(\frac{𝑝}{𝑞})}^{- 1} = \frac{𝑚}{𝑛} \cdot \frac{𝑞}{𝑝} = \frac{𝑚 𝑞}{𝑛 𝑝},$
which is a rational number since $𝑚 𝑞$ is an integer and $𝑛 𝑝 \neq 0$ .

Thus, by Theorem 2, $ℚ$ is a subfield of $ℝ$ . ∎

Solution: Exercise.

Let $𝐵 = {0, 1}$ be the Boolean algebra defined by:

\begin{matrix} 0 \oplus 0 = 0, 0 \oplus 1 = 1, 1 \oplus 0 = 1, 1 \oplus 1 = 0, & (XOR) \\ 0 \cdot 0 = 0, 0 \cdot 1 = 0, 1 \cdot 0 = 0, 1 \cdot 1 = 1 . & (AND) \end{matrix}

To show that $(𝐵, \oplus, \cdot)$ is a field, we need to verify that it satisfies the field axioms:

1.2 Vector Spaces and Subspaces ¶

1.3 Span and Linear Independence ¶

1.4 Basis and Dimension ¶

2.1 Introduction to Linear Maps ¶

Solution: Exercise.

Suppose

𝑇 : 𝑉 \to 𝑊

is a linear map as defined in Definition 26.

Solution: Exercise.

Suppose $𝑉$ is a finite-dimensional vector space, and let $𝑇 : 𝑉 \to 𝑉$ be a linear map such that $𝑇^{2} = 𝑇$ , i.e. $𝑇 \circ 𝑇 = 𝑇$ .

Lemma 1:

\ker (𝑇) \cap im (𝑇) = {𝟎}

Suppose $𝐯 \in \ker (𝑇) \cap im (𝑇)$ . Then $𝐯 \in \ker (𝑇)$ and $𝐯 \in im (𝑇)$ :

By the definition of kernel, $𝑇 (𝐯) = 𝟎$ .
By the definition of image, $\exists 𝐱 \in 𝑉$ such that $𝐯 = 𝑇 (𝐱)$ .

We can deduce:

\begin{matrix} 𝑇 (𝐯) & = 𝟎 & by definition of kernel \\ 𝑇 (𝑇 (𝐱)) & = 𝟎 & by substitution of 𝐯 \\ 𝑇 (𝐱) & = 𝟎 & because 𝑇 \circ 𝑇 = 𝑇 \\ 𝐯 & = 𝟎 & by substitution \end{matrix}

Since assuming $𝐯 \in \ker (𝑇) \cap im (𝑇)$ leads to $𝐯 = 𝟎$ , we have $\ker (𝑇) \cap im (𝑇) \subseteq {𝟎}$ .

Also, suppose $𝐯 \in {𝟎}$ , i.e. $𝐯 = 𝟎$ . Then:

$𝑇 (𝐯) = 𝑇 (𝟎) = 𝟎$ , so $𝐯 \in \ker (𝑇)$ .
$𝟎 \in im (𝑇)$ because $𝟎$ is in every vector space.

Thus, $𝐯 \in \ker (𝑇) \cap im (𝑇)$ , so ${𝟎} \subseteq \ker (𝑇) \cap im (𝑇)$ .

By double containment, $\ker (𝑇) \cap im (𝑇) = {𝟎}$ .

Lemma 2:

\ker (𝑇) + im (𝑇) = 𝑉

Let $𝐰$ be an arbitrary vector in $𝑉$ . Choose $𝐮 = 𝑇 (𝐰) \in im (𝑇)$ and $𝐯 = 𝐰 - 𝐮$ , such that $𝐰 = 𝐮 + 𝐯$ . It follows that:

\begin{matrix} 𝑇 (𝐯) & = 𝑇 (𝐰 - 𝐮) & by definition of 𝐯 \\ 𝑇 (𝐯) & = 𝑇 (𝐰) - 𝑇 (𝐮) & by linearity of 𝑇 \\ 𝑇 (𝐯) & = 𝑇 (𝐰) - 𝑇 (𝑇 (𝐰)) & by substitution of 𝐮 \\ 𝑇 (𝐯) & = 𝑇 (𝐰) - 𝑇 (𝐰) & because 𝑇 \circ 𝑇 = 𝑇 \\ 𝑇 (𝐯) & = 𝟎 & by definition of additive inverse \end{matrix}

Thus, $𝐯 \in \ker (𝑇)$ . Since $𝐰 = 𝐮 + 𝐯$ , we have $𝐰 \in \ker (𝑇) + im (𝑇)$ , so $𝑉 \subseteq \ker (𝑇) + im (𝑇)$ .

Now, suppose $𝐰 \in \ker (𝑇) + im (𝑇)$ . Then there exist $𝐮 \in im (𝑇)$ and $𝐯 \in \ker (𝑇)$ such that $𝐰 = 𝐮 + 𝐯$ . Since $im (𝑇)$ and $\ker (𝑇)$ are subspaces of $𝑉$ , it follows that $𝐮 \in 𝑉$ and $𝐯 \in 𝑉$ . $𝑉$ is closed under vector addition, so $𝐰 = 𝐮 + 𝐯 \in 𝑉$ , which implies $\ker (𝑇) + im (𝑇) \subseteq 𝑉$ .

By double containment, $\ker (𝑇) + im (𝑇) = 𝑉$ .

By Theorem 10, $\ker (𝑇) \oplus im (𝑇) = 𝑉$ . ∎

2.2 Introduction to Matrices ¶

6.2 Orthogonality and Projections ¶

Solution: Exercise.

Let $𝑉$ be an inner product space, and let ${𝐪_{1}, 𝐪_{2}, \dots, 𝐪_{𝑘}} \subseteq 𝑉$ be an orthonormal set with $𝑘 \geq 2$ . Let $𝐯 \in {(span {𝐪_{1}, 𝐪_{2}, \dots, 𝐪_{𝑘}})}^{⟂}$ such that $𝐯 \neq 𝟎$ .

Choose any two distinct vectors $𝐯 + 𝐪_{𝑖}$ and $𝐯 + 𝐪_{𝑗}$ in the set ${𝐯 + 𝐪_{1}, 𝐯 + 𝐪_{2}, \dots, 𝐯 + 𝐪_{𝑘}}$ with $𝑖 \neq 𝑗$ . Notice that $𝐯 ⟂ 𝐪_{𝑖}$ and $𝐯 ⟂ 𝐪_{𝑗}$ because $𝐯 \in {(span {𝐪_{1}, 𝐪_{2}, \dots, 𝐪_{𝑘}})}^{⟂}$ and $𝐪_{𝑖}, 𝐪_{𝑗} \in span {𝐪_{1}, 𝐪_{2}, \dots, 𝐪_{𝑘}}$ . We have:

\begin{matrix} ⟨ 𝐯 + 𝐪_{𝑖}, 𝐯 + 𝐪_{𝑗} ⟩ & = ⟨ 𝐯, 𝐯 + 𝐪_{𝑗} ⟩ + ⟨ 𝐪_{𝑖}, 𝐯 + 𝐪_{𝑗} ⟩ & by linearity in the first argument \\ = ⟨ 𝐯, 𝐯 ⟩ + ⟨ 𝐯, 𝐪_{𝑗} ⟩ + ⟨ 𝐪_{𝑖}, 𝐯 ⟩ + ⟨ 𝐪_{𝑖}, 𝐪_{𝑗} ⟩ & by sesquilinearity \\ = ⟨ 𝐯, 𝐯 ⟩ + ⟨ 𝐯, 𝐪_{𝑗} ⟩ + ⟨ 𝐪_{𝑖}, 𝐯 ⟩ + 0 & because 𝐪_{𝑖} ⟂ 𝐪_{𝑗} \\ = ⟨ 𝐯, 𝐯 ⟩ + 0 + 0 + 0 & because 𝐯 ⟂ 𝐪_{𝑖} and 𝐯 ⟂ 𝐪_{𝑗} \\ = ⟨ 𝐯, 𝐯 ⟩ \neq 0 & because 𝐯 \neq 𝟎 (positive-definiteness). \end{matrix}

Thus, $⟨ 𝐯 + 𝐪_{𝑖}, 𝐯 + 𝐪_{𝑗} ⟩ \neq 0$ , which implies that

. Therefore, no two distinct vectors in the set ${𝐯 + 𝐪_{1}, \dots, 𝐯 + 𝐪_{𝑘}}$ are orthogonal. ∎