Let \(f\) be the bilinear form on \(\RR^2\) defined by \[ f((x_1, x_2), (y_1, y_2)) = x_1y_1 + x_2y_2. \] Find the matrix of \(f\) in each of the following bases: \[ \{(1, 0), (0, 1)\}, \quad \{(1, -1), (1, 1)\}, \quad \{(1, 2), (3, 4)\}. \]
The printed exercise groups the coordinates as \((x_1,y_1)\) and \((x_2,y_2)\) while retaining the dot-product formula \(x_1y_1+x_2y_2\). Taken literally, that expression is not bilinear in the two displayed vector arguments. The preceding dot-product example makes the intended grouping \((x_1,x_2)\) and \((y_1,y_2)\) clear; that corrected statement is used here.
Solution
If \(\mathfrak B=(v_1,v_2)\) is an ordered basis and \(P\) is the matrix whose columns are \(v_1,v_2\) in the standard basis, then the matrix of the standard dot-product form in \(\mathfrak B\) is \[ [f]_{\mathfrak B}=P^tIP=P^tP. \]
1. Compute the matrix in the standard basis.
For \[ \mathfrak B_1=((1,0),(0,1)), \] the change-of-basis matrix is \(P_1=I\). Therefore \[ \boxed{ [f]_{\mathfrak B_1} =\begin{pmatrix}1&0\\0&1\end{pmatrix}.} \]
2. Compute the matrix in the basis \(((1,-1),(1,1))\).
Here \[ P_2= \begin{pmatrix} 1&1\\ -1&1 \end{pmatrix}. \] Hence \[ [f]_{\mathfrak B_2} =P_2^tP_2 =\begin{pmatrix} 2&0\\ 0&2 \end{pmatrix}. \] Thus \[ \boxed{[f]_{\mathfrak B_2}=2I_2.} \]
3. Compute the matrix in the basis \(((1,2),(3,4))\).
The change-of-basis matrix is \[ P_3= \begin{pmatrix} 1&3\\ 2&4 \end{pmatrix}. \] Therefore \[ \begin{aligned} [f]_{\mathfrak B_3} &=P_3^tP_3\\ &= \begin{pmatrix} 1&2\\ 3&4 \end{pmatrix} \begin{pmatrix} 1&3\\ 2&4 \end{pmatrix}\\ &= \begin{pmatrix} 5&11\\ 11&25 \end{pmatrix}. \end{aligned} \] Hence \[ \boxed{ [f]_{\mathfrak B_3} =\begin{pmatrix}5&11\\11&25\end{pmatrix}.} \]