Talk:Cardinality of Generator of Vector Space is not Less than Dimension/Proof 1

Is $\mathcal B$ supposed to be a basis? --GFauxPas 22:18, 1 April 2012 (EDT)

Not necessarily, I don't think. --prime mover 01:11, 2 April 2012 (EDT)

Then I'm having a hard time with the last line:

So $\left\{{y_1, y_2, \ldots, y_n}\right\}$ is linearly dependent.

The result follows by definition of dimension.

How does the result follow? --GFauxPas 08:06, 2 April 2012 (EDT)

Any $\left\{{y_1, y_2, \ldots, y_n}\right\}$ where $n > m$ is linearly dependent, so $n$ must be greater than the dimension. --prime mover 16:58, 2 April 2012 (EDT)