Clarify: by looking at the real and imaginary parts of the left side of the equation 𝑎(1+𝑖)+𝑏(1−𝑖) = 0. above, we see that 𝑎 + 𝑏 = 0 and 𝑎 − 𝑏 = 0, which implies 𝑎 = 𝑏 = 0. Hence  --- Commentary: $\textbf{Exercise 7.}$ (a) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list $1 + i$, $1 - i$ is linearly independent. (b) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then the list $1 + i$, $1 - i$ is linearly dependent. $\textbf{Solution 7.}$ (a) Think of $\mathbb{C}$ as a vector space over $\mathbb{R}$. Suppose $a, b \in \mathbb{R}$ and $a(1 + i) + b(1 - i) = 0.$ By looking at the real and imaginary parts of the left side of the equation above, we see that $a + b = 0$ and $a - b = 0$, which implies $a = b = 0$. Hence the list $(1 + i, 1 - i)$ is linearly independent. (b) Think of $\mathbb{C}$ as a vector space over $\mathbb{C}$. Then $1 - i = a(1 + i),$ where $a = \frac{1-i}{1+i}$. Thus the list $(1 + i, 1 - i)$ is linearly dependent. $\textit{Commentary:}$ This exercise highlights the importance of specifying the field over which a vector space is defined. The same set can have different linear independence properties when viewed as a vector space over different fields. In part (a), we're viewing $\mathbb{C}$ as a 2-dimensional vector space over $\mathbb{R}$. The vectors $1 + i$ and $1 - i$ are linearly independent because they point in different directions in the complex plane. In part (b), we're viewing $\mathbb{C}$ as a 1-dimensional vector space over $\mathbb{C}$. Now, any two non-zero vectors are linearly dependent, because one is always a complex scalar multiple of the other. This exercise also provides practice in working with the definition of linear independence, particularly in the context of the complex numbers. $\textit{Example:}$ If we think of $\mathbb{Q}_p(\sqrt{p})$ as a vector space over $\mathbb{Q}_p$, then the list $1$, $\sqrt{p}$ is linearly independent. However, if we think of $\mathbb{Q}_p(\sqrt{p})$ as a vector space over itself, then the list $1$, $\sqrt{p}$ is linearly dependent, because $\sqrt{p} = \sqrt{p} \cdot 1$. Similarly, if we think of $\mathbb{F}_4 = {0, 1, \alpha, \alpha^2}$ as a vector space over $\mathbb{F}_2$, then the list $1$, $\alpha$ is linearly independent. But if we think of $\mathbb{F}_4$ as a vector space over itself, then the list $1$, $\alpha$ is linearly dependent, because $\alpha = \alpha \cdot 1$. These examples illustrate the same principle in the context of extensions of $p$-adic fields and finite fields. The choice of the base field can drastically change the linear independence properties of a given set of vectors. --- $\textbf{Exercise 7.}$ (a) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{R}$, then the list $1 + i$, $1 - i$ is linearly independent. (b) Show that if we think of $\mathbb{C}$ as a vector space over $\mathbb{C}$, then the list $1 + i$, $1 - i$ is linearly dependent. $\textbf{Solution 7.}$ (a) Think of $\mathbb{C}$ as a vector space over $\mathbb{R}$. Suppose $a, b \in \mathbb{R}$ and $a(1 + i) + b(1 - i) = 0.$ By looking at the real and imaginary parts of the left side of the equation above, we see that $a + b = 0$ and $a - b = 0$, which implies $a = b = 0$. Hence the list $(1 + i, 1 - i)$ is linearly independent. (b) Think of $\mathbb{C}$ as a vector space over $\mathbb{C}$. Then $1 - i = a(1 + i),$ where $a = \frac{1-i}{1+i}$. Thus the list $(1 + i, 1 - i)$ is linearly dependent. $\textit{Commentary:}$ This exercise highlights the importance of specifying the field over which a vector space is defined. The same set can have different linear independence properties when viewed as a vector space over different fields. In part (a), we're viewing $\mathbb{C}$ as a 2-dimensional vector space over $\mathbb{R}$. The vectors $1 + i$ and $1 - i$ are linearly independent because they point in different directions in the complex plane. In part (b), we're viewing $\mathbb{C}$ as a 1-dimensional vector space over $\mathbb{C}$. Now, any two non-zero vectors are linearly dependent, because one is always a complex scalar multiple of the other. This exercise also provides practice in working with the definition of linear independence, particularly in the context of the complex numbers. $\textit{Additional Examples:}$ If we think of $\mathbb{Q}_p(\sqrt{p})$ as a vector space over $\mathbb{Q}_p$, then the list $1$, $\sqrt{p}$ is linearly independent. However, if we think of $\mathbb{Q}_p(\sqrt{p})$ as a vector space over itself, then the list $1$, $\sqrt{p}$ is linearly dependent. Similarly, if we think of $\mathbb{F}_4 = {0, 1, \alpha, \alpha^2}$ as a vector space over $\mathbb{F}_2$, then the list $1$, $\alpha$ is linearly independent. But if we think of $\mathbb{F}_4$ as a vector space over itself, then the list $1$, $\alpha$ is linearly dependent. If we think of the space of complex polynomials $\mathbb{C}[x]$ as a vector space over $\mathbb{R}$, then the list $1$, $i$, $x$, $ix$ is linearly independent. But if we think of $\mathbb{C}[x]$ as a vector space over $\mathbb{C}$, then this list is linearly dependent. If we think of the space of complex-valued continuous functions $C([0,1], \mathbb{C})$ as a vector space over $\mathbb{R}$, then the list $1$, $i$, $e^x$, $ie^x$ is linearly independent. But if we think of $C([0,1], \mathbb{C})$ as a vector space over $\mathbb{C}$, then this list is linearly dependent. These examples illustrate the same principle in the context of extensions of $p$-adic fields, finite fields, polynomial spaces, and function spaces. The choice of the base field can drastically change the linear independence properties of a given set of vectors.