Talk:Change of Basis is Invertible

The following preceded to proof as it is now:

Let $\mathbf P = \sqbrk \alpha_n$ be the matrix $\sqbrk {I_M; \sequence {a_n}, \sequence {b_n} }$ corresponding to the change of basis from $\sequence {b_n}$ to $\sequence {a_n}$.

Then:

$\ds \forall j \in \closedint 1 n: b_j = \sum_{i \mathop = 1}^n \alpha_{i j} a_i$

Thus the matrix corresponding to the change of basis from $\sequence {a_n}$ to $\sequence {b_n}$ is also the matrix:

$\sqbrk {v; \sequence {a_n} }$

where $v$ is the endomorphism of $M$ which satisfies $\forall k \in \closedint 1 n: \map v {a_k} = b_k$.

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. --barto (talk) 09:14, 2 May 2017 (EDT)