📂Linear Algebra

Basis and Dimension of Quotient Space

Theorem¹

Let $V$ be a $n$-dimensional vector space, $W \le V$ be a $k$-dimensional subspace. Let the basis of $W$ be $\left\{ u_{1}, \dots, u_{k} \right\}$. And let the basis of $V$, which is an extension of 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 of the quotient space $V/W$.

• $\dim(V) = \dim(V/W) + \dim(W)$

Explanation

The results about dimensions can also be derived from another proof.

Proof

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

• $\beta$ is linearly independent.

  Since the zero vector of $V/W$ is $W$, there must only exist the solution $a_{k+1} = \cdots = a_{n} = 0$ for the following equation to hold.

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

  According to the addition defined in $V/W$,

  $$ \begin{align*} a_{k+1}(u_{k+1} + W) + \cdots + a_{n}(u_{n} + W) &= (a_{k+1}u_{k+1} + \cdots + a_{n}u_{n}) + W \\ &= 0_{V} + W \end{align*} $$

  At this point, $0_{V}$ is the zero vector of $V$. By assumption, since $u_{k+1}, \dots, u_{n}$s are linearly independent,

  $$ a_{k+1}u_{k+1} + \cdots + a_{n}u_{n} = 0_{V} \implies a_{k+1} = \cdots = a_{n} = 0 $$

• $\span \beta = V/W$

  Since $w \in W$ implies $w + W = W$ 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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