📂Linear Algebra

몫공간의 기저와 차원

Theorem¹

Let $V$ be a $n$-dimensional vector space, and let $W \le V$ be a $k (\lt n)$-dimensional subspace. Let $\left\{ u_{1}, \dots, u_{k} \right\}$ be a basis of $W$. And let $\left\{ u_{1}, \dots, u_{k}, u_{k+1}, \dots, u_{n} \right\}$ be a basis of $V$ obtained by extending this basis. Then

• $\left\{ u_{k+1} + W, \dots, u_{n} + W \right\}$ is a basis of the quotient space $V/W$.

• $\dim(V) = \dim(V/W) + \dim(W)$

Explanation

The result on dimensions can also be obtained by another proof. This theorem gives an intuitive reason why the quotient space is called a quotient and why it is denoted as $V / W$. The dimension of a vector space is written in exponential form as $\mathbb{R}^{n}$, and this shows that the dimension of the quotient space is computed by subtracting exponents—analogous to division in the laws of exponents. If we write $V = \mathbb{R}^{5}$ and $W = \mathbb{R}^{2}$, then

$$ V/W = \dfrac{V}{W} = \dfrac{\mathbb{R}^{5}}{\mathbb{R}^{2}} = \mathbb{R}^{5-2} = \mathbb{R}^{3} $$

Proof

Let $\beta = \left\{ u_{k+1} + W, \dots, u_{n} + W \right\}$.

• $\beta$ is linearly independent.

  Since the zero vector of $V/W$ is $W$, we need to show that the only solution to the following equation is $a_{k+1} = \cdots = a_{n} = 0$.

  $$ a_{k+1}(u_{k+1} + W) + \cdots + a_{n}(u_{n} + W) = W $$

  According to the addition defined in $V/W$ (../3359),

  $$ \begin{align*} a_{k+1}(u_{k+1} + W) + \cdots + a_{n}(u_{n} + W) &= (a_{k+1}u_{k+1} + \cdots + a_{n}u_{n}) + W \\ &= W \end{align*} $$

  Lemma

  (b) For $v_{1}, v_{2} \in V$, being $v_{1} + W = v_{2} + W$ is equivalent to being $v_{1} - v_{2} \in W$.

  (c) $V/W$ is a vector space, and its zero vector is $0_{V} + W = W$. ($0_{V}$ is the zero vector of $V$.)

  Then by the above lemma we obtain:

  $$ (a_{k+1}u_{k+1} + \cdots + a_{n}u_{n}) + W = 0_{V} + W $$ $$ \implies a_{k+1}u_{k+1} + \cdots + a_{n}u_{n} - 0_{V} = a_{k+1}u_{k+1} + \cdots + a_{n}u_{n} \in W \tag{1} $$

  By definition, $u_{k+1}, \dots, u_{n}$ is clearly not an element of $W$.

  $$ u_{k+1}, \dots, u_{n} \notin W \tag{2} $$

  However, since $u_{k+1}, \dots, u_{n}$ is linearly independent, to satisfy both $(1)$ and $(2)$ simultaneously one must have $a_{k+1}u_{k+1} + \cdots + a_{n}u_{n} = 0_{V} = 0_{W}$. Hence we obtain:

  $$ a_{k+1} = \cdots = a_{n} = 0 $$

• $\span \beta = V/W$

  For $w \in W$ we have $w + W = W$, so for any $v \in V$,

  $$ \begin{align*} v + W &= (a_{1}u_{1} + \cdots + a_{k}u_{k} + a_{k+1}u_{k+1} + \cdots + a_{n}u_{n}) + W \\ &= (a_{k+1}u_{k+1} + \cdots + a_{n}u_{n}) + W \\ &= (a_{k+1}u_{k+1} + W) + \cdots + (a_{n}u_{n} + W) \\ \end{align*} $$

Therefore $\beta$ is a basis of $V/W$. Hence

$$ \begin{align*} \dim(V) &= \dim(V/W) + \dim(W) \\ n &= (n-k) + k \end{align*} $$

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