Linear Algebra 题目选做

Consider the set of complex numbers

$$G = \{a + bi \mid a, b \in \mathbb{Q}\}.$$

(The $G$ stands for Gauss; these numbers might be called Gaussian rational numbers, although I don’t know if they actually are.) Is $G$ a field (with the same addition and multiplication operations as in $\mathbb{C}$)? For a question like this, you should either explain why all the axioms for a field are satisfied (you can assume that they hold for $\mathbb{C}$), or else explain why one of the axioms fails. A few sentences could be enough to write.

Answer

$G$ is a field.

• Closure under addition and multiplication:
  • Assume $x = a+bi, y=c+di\in G$
  • So $a, b, c, d \in \mathbb{Q}$
  • So $a + c, b + d, (ac - bd, ad + bc)\in \mathbb{Q}$
  • $x + y=(a + c) + (b + d) i\in \mathbb{Q}$
  • $xy = (ac - bd) + (ad + bc)i\in \mathbb{Q}$
• Community of addition and multiplication:
  • $x + y = (a + c) + (b + d)i = (c + a) + (d + b)i = y + x$
  • $xy = (ac - bd) + (ad + bc)i = (ca - db) + (cb + da)i = yx$
• Existence of additive and multiplicative identity:
  • $0 = 0 + 0i\in G$
  • $1 = 1 + 0i\in G$
• Existence of additive inverse:
  • $-x = -a - bi\in G$
• Existence of multiplicative inverse:
  • Assume $x = a + bi\in G\setminus\{0\}$ and $x^{-1}=c+di$
  • We have $(a + bi)\cdot (d + di)=1$
  • So $\begin{cases} ac - bd = 1 \\ ad + bc = 0 \end{cases}$
  • Which is $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}\begin{bmatrix} c \\ d \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$
  • We get the inverse matrix $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}^{-1}=\frac{1}{a^2+b^2}\begin{bmatrix} a & b \\ -b & a \end{bmatrix}$
  • So $\begin{bmatrix} c \\ d \end{bmatrix}=\frac{1}{a^2+b^2}\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} 1 \\ 0 \end{bmatrix}=\frac{1}{a^2+b^2}\begin{bmatrix} a \\ -b \end{bmatrix}$
  • So $c, d\in \mathbb{Q}$, and $x^{-1}=c+di\in G$
• Distributivity of multiplication over addition:
  • Follows from the distributivity of multiplication over addition in $\mathbb{C}$

Consider the set of complex numbers

$$M = \left\{r_0 + r_1 e^{\frac{\pi i}{2}} \mid r_0, r_1 \in \mathbb{Q}\right\}$$

Is $M$ a field?

Answer

$M$ is a field.

Consider that $e^{\frac{\pi i}{2}} = i$, we have $M = \{r_0 + r_1 i \mid r_0, r_1 \in \mathbb{Q}\}=G$.

Since $G$ is a field, $M$ is also a field.

Consider the set of complex numbers

$$P = \left\{re^{2\pi i\theta} \mid r, \theta \in \mathbb{Q}\right\}$$

Is $P$ a field?

Answer

$P$ is not a field.

Consider $x = e^{\frac{\pi i}{4}}, y=e^{-\frac{\pi i}{4}}\in P$, we have:

$$x + y = \sqrt{2}\notin P$$

So $P$ is not closed under addition.

The set $W = \left\{(x, y, z, w) \in \mathbb{R}^4 \mid x + y + z + w = 0\right\}$ is a subspace of $\mathbb{R}^4$. Find a basis of $W$.

Answer

We firstly add $(1, -1, 0, 0), (1, 0, -1, 0), (1, 0, 0, -1)$ into the basis of $W$, they are linearly independent.

Then we could find that if we add $(0, 1, 0, 2)$ in to the set, it will still be linearly independent, and