Formula for the change of basis
Task number: 2567
Prove that \([id]_{AB}=([id]_{BK})^{-1}[id]_{AK}\).
Hint
Use the theorem on the matrix of the composed map.
Resolution
To the formula \([g\circ f]_{XZ}=[g]_{YZ}[f]_{XY}\) substitute \(f=g=id\), \(X=A\), \(Y=K\) and \(Z=B\), and get \([id]_{AB}=[id]_{KB}[id]_{AK}\).
The matrix of the change of basis fom any basis to the same basis is the identity matrix \([id]_{BB}=\mathbf I_n\), since the vectors are not changed, and hence also the coordinates.
We substitute \(f=g=id\), \(X=Z=B\) and \(Y=K\) and get \([id]_{KB}[id]_{BK}=[id]_{BB}=\mathbf I_n\). Therefore \([id]_{KB}=([id]_{BK})^{-1}\), since both matrices are square (both bases have the same cardinality).
Substitute \([id]_{KB}=([id]_{BK})^{-1}\) into \([id]_{AB}=[id]_{KB}[id]_{AK}\) and get the desired formula.
