Let \(f\) be the 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 previous transcription swapped the second coordinate of the first vector with the first coordinate of the second vector, producing a formula that was not bilinear. The displayed statement restores the source’s coordinate pairs.
Solution
For an ordered basis \(\mathfrak B=(v_1,v_2)\), the matrix of \(f\) is the Gram matrix \[ [f]_{\mathfrak B} =\begin{pmatrix} f(v_1,v_1)&f(v_1,v_2)\\ f(v_2,v_1)&f(v_2,v_2) \end{pmatrix}. \] Here \(f\) is the standard dot product on \(\mathbb R^2\).
1. Standard basis.
For \[ v_1=(1,0),\qquad v_2=(0,1), \] the basis is orthonormal. Hence \[ \boxed{[f]_{(v_1,v_2)}= \begin{pmatrix}1&0\\0&1\end{pmatrix}.} \]
2. Basis \(((1,-1),(1,1))\).
Let \[ v_1=(1,-1),\qquad v_2=(1,1). \] Then \[ v_1\cdot v_1=2, \qquad v_2\cdot v_2=2, \qquad v_1\cdot v_2=0. \] Therefore \[ \boxed{[f]_{(v_1,v_2)}= \begin{pmatrix}2&0\\0&2\end{pmatrix}.} \]
3. Basis \(((1,2),(3,4))\).
Let \[ v_1=(1,2),\qquad v_2=(3,4). \] Their pairwise dot products are \[ v_1\cdot v_1=5, \qquad v_1\cdot v_2=11, \qquad v_2\cdot v_2=25. \] Thus \[ \boxed{[f]_{(v_1,v_2)}= \begin{pmatrix}5&11\\11&25\end{pmatrix}.} \]