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.

Difficulty level: Easy task (using definitions and simple reasoning)
Proving or derivation task
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