Talk:Change of Basis is Invertible

The following preceded to proof as it is now:

Let $\mathbf P = \left[{\alpha}\right]_{n}$ be the matrix $\left[{I_M; \left \langle {a_n} \right \rangle, \left \langle {b_n} \right \rangle}\right]$ corresponding to the change of basis from $\left \langle {b_n} \right \rangle$ to $\left \langle {a_n} \right \rangle$.

Then:
 * $\displaystyle \forall j \in \left[{1 \,.\,.\, n}\right]: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$

Thus the matrix corresponding to the change of basis from $\left \langle {a_n} \right \rangle$ to $\left \langle {b_n} \right \rangle$ is also the matrix:
 * $\left[{v; \left \langle {a_n} \right \rangle}\right]$

where $v$ is the endomorphism of $M$ which satisfies $\forall k \in \left[{1 \,.\,.\, n}\right]: v \left({a_k}\right) = b_k$.

This looks more like part of a proof for the (yet to come) Product of Change of Basis Matrices. Also, the fact that this defines $\mathbf P$ differently from the definition in the theorem statement, made me take that part away. (I'm working on cleaning up Matrix Corresponding to Change of Basis under Linear Transformation) --barto (talk) 09:14, 2 May 2017 (EDT)