## Formula for the change of basis

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.
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