📂Linear Algebra

몫공간의 기저와 차원

Proposition¹

Let $V$ be a $n$-dimensional vector space, and $W \le V$ be a $k (\lt n)$-dimensional subspace. Let the basis of $W$ be $\left\{ u_{1}, \dots, u_{k} \right\}$. Furthermore, let the basis of $V$, extended with this basis, be $\left\{ u_{1}, \dots, u_{k}, u_{k+1}, \dots, u_{n} \right\}$. Then:

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

• $\dim(V) = \dim(V/W) + \dim(W)$

Explanation

The result concerning dimension can also be obtained through alternative proofs.

Proof

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

• $\beta$ is linearly independent.

  Since the zero vector in $V/W$ is $W$, we must show that the only solution that satisfies 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 addition defined in $V/W$,

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

  Lemma

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

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

  From the lemma above, we obtain:

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

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

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

  However, since $u_{k+1}, \dots, u_{n}$ is linearly independent, $(1)$ and $(2)$ must be simultaneously satisfied, which implies $a_{k+1}u_{k+1} + \cdots + a_{n}u_{n} = 0_{V} = 0_{W}$. Therefore, we have:

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

• $\span \beta = V/W$

  For $w \in W$ such that $w + W = W$, for any given $v \in V$:

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

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

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

■