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 $\mathbf M_{f, \BB} = \mathbf M_{\AA, \BB}^\intercal \mathbf M_{f, \AA} \mathbf M_{\AA, \BB}$.



$\blacksquare$