Matrix of Bilinear Form Under Change of Basis
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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 $\mathbf M_{\AA, \BB}$ be the change of basis matrix from $\AA$ to $\BB$.
Let $f : M \times M \to R$ be a bilinear form.
Let $\mathbf M_{f, \AA}$ be its matrix relative to $\AA$.
Then its matrix relative to $\BB$ equals:
- $\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$
Proof
Let $m \in M$, and let $\sqbrk m_\AA$ and $\sqbrk m_\BB$ denote its coordinate vectors relative to $\AA$ and $\BB$.
We have:
\(\ds \map f m\) | \(=\) | \(\ds \sqbrk m_\AA^\intercal \cdot \mathbf M_{f, \AA} \cdot \sqbrk m_\AA\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB}^\intercal \cdot \mathbf M_{f, \AA} \cdot \paren {\mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB}\) | Change of Coordinate Vector Under Change of Basis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqbrk m_\BB^\intercal \cdot \mathbf M_{\AA, \BB}^\intercal \cdot \mathbf M_{f, \AA} \cdot \mathbf M_{\AA, \BB} \cdot \sqbrk m_\BB\) | Transpose of Matrix Product |
Thus $\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$.
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$\blacksquare$