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:
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How does one find the basis for a quotient space? ($V / W$'s objects are sets, not vectors anymore)
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Is this the proper approach or is there a simpler manner of dealing with dim$(V / W)$?
Best Answer
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.