Say I have a vector space $V$ which has a basis $B = \langle\vec{b}_1,...,\vec{b}_n\rangle$.
Now, I want to prove that an alternative basis, $D = \langle\vec{d}_1,...,\vec{d}_k\rangle$, has the same number of basis vectors. That is to say, $k = n$. Every basis of $V$ will have a minimum number of vectors which corresponds to the number of linearly independent vectors needed to span all of $V$. So, it makes intuitive sense to me that $k \geq n$. But do I constraint this statement further to deduce that $k = n$?
