I seek to show that the dimension of the finite-dimensional vector space $V$ modulo $W$ (subspace) is the equal to the difference of the dimension of each space, i.e. dim$(V / W) =$ dim$(V) - $ dim$(W)$.
I start by letting $W \leqslant V$ and stating that
$$V / W = \{v+W \, \big| \, v \in V \}$$
Thus, $V/W$ objects are the collection of left $W-$cosets. I understand that the dimension of a vector space is equivalent to the cardinality of its basis. So my suspicion here is to first find a basis for $V / W$ and then show that its cardinality is equal to dim$(V)$ $-$ dim$(W)$. But I have a couple of questions:
How does one find the basis for a quotient space? ($V / W$'s objects are sets, not vectors anymore)
Is this the proper approach or is there a simpler manner of dealing with dim$(V / W)$?
3 Answers
$\begingroup$Another proof that may be circular depending on how you proved the Rank-Nullity theorem is as follows: Let $\pi:V\to V/W$ be the canonical projection. Notice that it is linear and surjective and the kernel by definition is $W$ hence:
$$\dim(V)=\dim(V/W)+\dim(W)$$
And your equality follows.
$\endgroup$ $\begingroup$Suppose $\{ w_1, w_2, \dots, w_r \}$ is a basis of $W$. You can complete it to get a basis $\{ w_1, \dots, w_r, v_{r+1}, \dots, v_n \}$ of the whole space. The clases $\{ v_{r+1} + W, \dots, v_n + W \}$ are a basis of the quotient space (Why?) A proof of the dimension now follows easily.
Since you ask for another proof. If you have studied First Isomorphism Theorem you can apply it to the map $\phi: V \to V\W$ where $\phi(x) = x + W$.
$\endgroup$ $\begingroup$Let $\{w_i\}$ be a basis for $W$ and extend by $\{v_j\}$ to get a basis for $V$. Then you can show $\{v_j + W\}$ is a basis for the quotient.
$\endgroup$
