Algebra I -- partial summary 1

2023-11-12 本文已影响0人好之者不如乐之者

Problem 1 -- bases and dimensions

Suppose that $W$ is a subspace of $V = F^n$ with $\dim(W) = n-1$ . Fix a basis $x_i = (x_{i, 1}, \dots, x_{i, n}), 1 \leq i \leq n-1$ , of $W$ (so $x_{i, j} \in F$ ). Define a function $T: F^n \rightarrow F^{n-1}$ by
$\begin{align} T(a_1, \dots, a_n) &= (a_1x_{1, 1} + \dots + a_nx_{1, n}, \dots, a_1x_{n-1, 1} + \dots + a_nx_{n-1, n}) \\ &= (\sum_{j=1}^{n}a_jx_{1, j}, \sum_{j=1}^{n}a_jx_{2, j}, \dots, \sum_{j=1}^{n}a_jx_{n-1, j}). \end{align}$

(a) Show that $T$ is linear.
(b) Use the dimension theorem to deduce that there exists a nonzero $(a_1, \dots, a_n) \in F^n$ such that $W \subset \{(c_1, \dots, c_n) \in F^n: a_1c_1 + \dots + a_nc_n = 0\}$ . (1)
(c) Deduce that eqaulity holds in (1).

$\textbf{Proof}$

(a)

Take $c \in F, a = (a_1, \dots, a_n), b = (b_1, \dots, b_n)\in F^n$ , then consider

$\begin{align} T(c \cdot a + b) &= T((c \cdot a_1, \dots, c \cdot a_n) + (b_1, \dots, b_n)) \\ &= T(ca_1 + b_1, \dots, ca_n + b_n) \\ &= (\sum_{j = 1}^n{(ca_j + b_j) x_{1, j}}, \dots, \sum_{j = 1}^n{(ca_j + b_j) x_{n - 1, j}}) \\ &= (\sum_{j = 1}^n{ca_j x_{1, j}} + \sum_{j = 1}^n{b_j x_{1, j}}, \dots, \sum_{j = 1}^n{ca_j x_{n - 1, j}} + \sum_{j = 1}^n{b_j x_{n - 1, j}}) \\ &= (c \cdot \sum_{j = 1}^n{a_j x_{1, j}} + \sum_{j = 1}^n{b_j x_{1, j}}, \dots, c \cdot \sum_{j = 1}^n{a_j x_{n - 1, j}} + \sum_{j = 1}^n{b_j x_{n - 1, j}}). \end{align}$

and

$\begin{align} c \cdot T(a) + T(b) &= c \cdot T(a_1, \dots, a_n) + T(b_1, \dots, b_n) = c \cdot (\sum_{j = 1}^n{a_j x_{1, j}}, \dots, \sum_{j = 1}^n{a_j x_{n - 1, j}}) + (\sum_{j = 1}^n{b_j x_{1, j}}, \dots, \sum_{j = 1}^n{b_j x_{n - 1, j}}) \\ &= (c \cdot \sum_{j = 1}^n{a_j x_{1, j}}, \dots, c \cdot \sum_{j = 1}^n{a_j x_{n - 1, j}}) + (\sum_{j = 1}^n{b_j x_{1, j}}, \dots, \sum_{j = 1}^n{b_j x_{n - 1, j}}) \\ &= (c \cdot \sum_{j = 1}^n{a_j x_{1, j}} + \sum_{j = 1}^n{b_j x_{1, j}}, \dots, c \cdot \sum_{j = 1}^n{a_j x_{n - 1, j}} + \sum_{j = 1}^n{b_jx_{n-1, j}}). \end{align}$

Based on above analysis, $T(c\cdot a + b) = c\cdot T(a) + T(b)$ ; therefore, $T$ is linear.

(b)

Based on dimension theorem, $\dim N(T) + \dim R(T) = \dim V$ ; in this problem, based on Replacement Theorem (Steinitz Exchange Lemma), we have $\dim V = \dim F^n = n, R(T) \subset F^{n-1} \Rightarrow n = \dim V = \dim N(T) + \dim R(T) \leq \dim N(T) + (n - 1) \Rightarrow \dim N(T) \geq n - (n-1) = 1.$ Thus, we can choose a nonzero $(a_1, \dots, a_n) \in N(T) \subset V = F^n.$
Write $W^{'} = \{(c_1, \dots, c_n) \in F^n: a_1c_1 + \dots + a_nc_n = 0\}.$
Then
$\begin{align} 0 &= T(a_1, \dots, a_n) \\ &= (\sum_{j = 1}^n{a_j x_{1, j}}, \dots, \sum_{j = 1}^n{a_j x_{n - 1, j}}). \end{align}$
$i.e.$
$\sum_{j = 1}^n{a_j x_{i, j}} = 0, \ \ \ \ \ i = 1, \dots, n,$
$i.e. x_1, \dots, x_n \in W^{'}$ , since $W^{'}$ is a subspace, $W = span \{x_1, \dots, x_n\} \subset W^{'}.$

(c)

First, we will analyze the dimension of $W^{'}.$
Since $(a_1, \dots, a_n) \neq (0, \dots, 0)$ , we have $a_i \neq 0$ for some $1 \leq i \leq n.$ Then
$a_1c_1 + \dots + a_nc_n = 0$
if and only if
$\begin{align} c_i &= -a_i^{-1}(a_1c_1 + \dots + a_{i-1}c_{i-1} + a_{i+1}c_{i+1} + \dots + a_nc_n) \\ &= -a_i^{-1}\sum_{j \neq i}a_jc_j. \end{align}$
Then $\forall (c_1, \dots, c_n) \in W^{'}$ can be written as
$\begin{align} (c_1, \dots, c_n) &= (c_1, \dots, c_{i-1}, -a_i^{-1}\sum_{j \neq i}a_jc_j, c_{i+1}, \dots, c_n) \\ &= \sum_{j \neq i}{(e_j - a_i^{-1}a_je_i)c_j}. \end{align}$
Thus, $\beta = \{e_j - a_i^{-1}a_je_i: j \neq i \}$ is a spanning set of $W^{'}$ .
And $\beta$ is apparently linearly independent. So $\beta$ is a basis of $W^{'}$ ; therefore, $\dim W^{'} = n-1$ , so $\{x_i\}$ is a basis of $W^{'}$ by RT Cor 2(i). Thus, $W = span \{x_i\} = W^{'}$ .

$\textbf{Note:}$
Interesting theory about linearly independent set and spanning (generating) set is below.

Let $S = \{v_1, \dots, v_n\} \subset V$ - a vector space. Show that there exists a linearly independent subset $B \subset S$ such that $spanB = spanS$ .
$\textbf{Proof}$
TBD.

Problem 2 -- cardinality of finite fields

Show that if $F$ is any finite field, then $F$ has cardinality $p^n$ for some prime $p$ and some $n \geq 1.$

$\textbf{Proof}$
First, let the charatetistic of $F$ is a prime $p$ .
Second, we can show that $E = \mathbb{Z}/m\mathbb{Z} \subset F$ because $0, 1 \in F$ and $F$ has characteristics $p$ .
Finally, demonstrate $F$ can be a vector space over $E$ (fulfilling all vector space axioms); therefore, cardinality of $F$ must be $p^n$ .

Construction TBD.

Problem 3 -- relationship between dimensions of subspaces

If $W_1, W_2$ are finite-dimensional subspaces of $V$ , show that $W_1 + W_2$ and $W_1 \cap W_2$ are finite-dimensional and that $\dim (W_1 + W_2) + \dim (W_1 \cap W_2) = \dim (W_1) + \dim (W_2).$

$\textbf{Proof}$
First, apparently, $W_1 + W_2$ and $W_1 \cap W_2$ are subspaces; therefore, $W_1 \cap W_2 \subset W_1$ is finite. For $W_1 + W_2$ , we can find a finite basis as below.
Second, suppose $\alpha = \{x_1, \dots, x_n\}$ is a basis of $W_1 \cap W_2$ . Then we can extend $\alpha$ into a basis of $W_1$ , $\beta = \{x_1, \dots, x_n, y_1, \dots, y_m\}$ , and a basis of $W_2$ , $\gamma = \{x_1, \dots, x_n, z_1, \dots, z_l\}$ . Thus, basis of $W_1 + W_2$ is $\theta = \{x_1, \dots, x_n, y_1, \dots, y_m, z_1, \dots, z_l\}$ , which is apparently spanning. Now, we will prove that $\theta$ is linearly independent.

Suppose $\theta$ is linearly dependen, there exists $a_1, \dots, a_n, b_1, \dots, b_m, c_1, \dots, c_l \in F$ such that $a_1, \dots, a_n, b_1, \dots, b_m, c_1, \dots, c_l$ are not all $0_F$ and $a_1x_1 + \dots + a_nx_n + b_1y_1 + \dots + b_my_m + c_1z_1 + \dots + c_lz_l = 0_V$ . Because $\gamma$ is linearly independent, we have $c_1z_1 + \dots + c_lz_l \neq 0_V$ here, and $-(c_1z_1 + \dots + c_lz_l) = a_1x_1 + \dots + a_nx_n + b_1y_1 + \dots + b_my_m \neq 0_V \Rightarrow -(c_1z_1 + \dots + c_lz_l) \neq 0_V \in W_1 \cap W_2$ ; however, if a linear combination of $\gamma$ is an element in $W_1 \cap W_2$ , then $c_1z_1 + \dots + c_lz_l = 0_V$ (can be easily deduced), which deduces contradiction. Thus, $\theta$ is linearly independent.

In conclusion, the formula is correct.

Corollary

Suppose $V$ is finite-dimensional. Show that $\dim (W_1 + \dots + W_n) \leq \dim (W_1) + \dots + \dim(W_n)$ for any subspaces $W_i$ of $V.$
This can be done easily by induction and formula above.

Knowledge 1 -- Lagrange Interpolation

Take $c_0, \dots, c_n \in F$ ,
then $f_i(x) = \prod_{k = 0, k \neq i}^n \frac{x - c_k}{c_i - c_k}$ and $\{f_0, \dots, f_n\}$ is a basis of $P_n(F)$ (to prove that it is linearly independent, consider $\sum_{j = 1}^n a_jf_i(c_j)$ ).
Construct $g_i(x) = \sum_{j = 1}^nb_jf_j(x)$ ,
then $g_j(c_j) = b_j$ .

Knowledge 2 -- Linear Transformations and Matrics

$T$ is a linaer transformation such that $T: L \rightarrow W$ . $\beta$ is a basis of $V$ , $\gamma$ is a basis of $W$ , and $v, w$ are elements in $V, W$ such that $T(v) = w.$ Then under matrix representation, $v = \beta\ [v]_{\beta}, w = \gamma\ [w]_{\gamma}$ ; moreover, we have $[T]_{\beta}^{\gamma}\ [v]_{\beta} = [w]_{\gamma}$ .