[proofplan] We rewrite the same vector $v \in V$ using its coordinate column in the basis $\mathcal C$, then convert those coordinates into $\mathcal B$-coordinates using the given change-of-coordinates matrix. The defining formula for the matrix $A$ of $q$ in the basis $\mathcal B$ then expresses $q(v)$ as a quadratic polynomial in the $\mathcal C$-coordinates. After moving the transpose through the product, the coefficient matrix in $\mathcal C$-coordinates is exactly $P_{\mathcal B \leftarrow \mathcal C}^{\top}AP_{\mathcal B \leftarrow \mathcal C}$. [/proofplan] [step:Write the $\mathcal B$-coordinates in terms of the $\mathcal C$-coordinates] Fix $v \in V$. Define the coordinate column \begin{align*} x := [v]_{\mathcal C} \in F^n. \end{align*} By the defining property of the change-of-coordinates matrix $P_{\mathcal B \leftarrow \mathcal C}$, we have \begin{align*} [v]_{\mathcal B}=P_{\mathcal B \leftarrow \mathcal C}x. \end{align*} [/step] [step:Substitute the coordinate change into the quadratic expression] Since $A$ is the matrix of $q$ in the ordered basis $\mathcal B$, the defining property of $A$ gives \begin{align*} q(v)=[v]_{\mathcal B}^{\top}A[v]_{\mathcal B}. \end{align*} Substituting $[v]_{\mathcal B}=P_{\mathcal B \leftarrow \mathcal C}x$ into this expression yields \begin{align*} q(v)=(P_{\mathcal B \leftarrow \mathcal C}x)^{\top}A(P_{\mathcal B \leftarrow \mathcal C}x). \end{align*} By the transpose identity $(P_{\mathcal B \leftarrow \mathcal C}x)^{\top}=x^{\top}P_{\mathcal B \leftarrow \mathcal C}^{\top}$ and associativity of matrix multiplication over $F$, this becomes \begin{align*} q(v)=x^{\top}P_{\mathcal B \leftarrow \mathcal C}^{\top}AP_{\mathcal B \leftarrow \mathcal C}x. \end{align*} [guided] The goal is to express $q(v)$ entirely in terms of the coordinate column of $v$ in the basis $\mathcal C$. We have already defined \begin{align*} x := [v]_{\mathcal C} \in F^n. \end{align*} The hypothesis on the change-of-coordinates matrix says exactly that this column becomes the $\mathcal B$-coordinate column after multiplication by $P_{\mathcal B \leftarrow \mathcal C}$: \begin{align*} [v]_{\mathcal B}=P_{\mathcal B \leftarrow \mathcal C}x. \end{align*} Now use the defining property of the matrix $A$ of the [quadratic form](/page/Quadratic%20Form) $q$ in the basis $\mathcal B$. For every vector $w \in V$, and hence for our fixed vector $v \in V$, this definition gives \begin{align*} q(v)=[v]_{\mathcal B}^{\top}A[v]_{\mathcal B}. \end{align*} Replacing the two occurrences of $[v]_{\mathcal B}$ by $P_{\mathcal B \leftarrow \mathcal C}x$ gives \begin{align*} q(v)=(P_{\mathcal B \leftarrow \mathcal C}x)^{\top}A(P_{\mathcal B \leftarrow \mathcal C}x). \end{align*} The only algebraic simplification needed is the transpose rule for a matrix times a column vector. Since $P_{\mathcal B \leftarrow \mathcal C} \in F^{n \times n}$ and $x \in F^n$, the product $P_{\mathcal B \leftarrow \mathcal C}x$ is a column in $F^n$, and by the definition of transpose, \begin{align*} (P_{\mathcal B \leftarrow \mathcal C}x)^{\top}=x^{\top}P_{\mathcal B \leftarrow \mathcal C}^{\top}. \end{align*} Substituting this identity and using associativity of matrix multiplication over the field $F$, we obtain \begin{align*} q(v)=x^{\top}P_{\mathcal B \leftarrow \mathcal C}^{\top}AP_{\mathcal B \leftarrow \mathcal C}x. \end{align*} This is the desired expression of $q(v)$ as a quadratic expression in the $\mathcal C$-coordinate column $x=[v]_{\mathcal C}$. [/guided] [/step] [step:Identify the resulting coefficient matrix in the basis $\mathcal C$] Define \begin{align*} M := P_{\mathcal B \leftarrow \mathcal C}^{\top}AP_{\mathcal B \leftarrow \mathcal C} \in F^{n \times n}. \end{align*} The previous step shows that for every $v \in V$, with $x=[v]_{\mathcal C}$, one has \begin{align*} q(v)=x^{\top}Mx=[v]_{\mathcal C}^{\top}M[v]_{\mathcal C}. \end{align*} By the definition of the matrix of a quadratic form in an ordered basis, $M$ is the matrix of $q$ in the basis $\mathcal C$. Therefore the matrix of $q$ in the basis $\mathcal C$ is \begin{align*} P_{\mathcal B \leftarrow \mathcal C}^{\top}AP_{\mathcal B \leftarrow \mathcal C}. \end{align*} [/step]