Change of Basis is Invertible

Theorem

Let $R$ be a ring with unity.

Let $M$ be a free $R$-module of finite dimension $n > 0$.

Let $\AA$ and $\BB$ be ordered bases of $M$.

Let $\mathbf P$ be the change of basis matrix from $\AA$ to $\BB$.

Then $\mathbf P$ is invertible, and its inverse $\mathbf P^{-1}$ is the change of basis matrix from $\BB$ to $\AA$.

Proof

From Product of Change of Basis Matrices and Change of Basis Matrix Between Equal Bases:

$\sqbrk {I_M; \AA, \BB} \sqbrk {I_M; \BB, \AA} = \sqbrk {I_M; \AA, \AA} = I_n$

$\sqbrk {I_M; \BB, \AA} \sqbrk {I_M; \AA, \BB} = \sqbrk {I_M; \BB, \BB} = I_n$

Hence the result.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices