Which of the following functions \(f\), defined on vectors \(\alpha = (x_1, x_2)\) and \(\beta = (y_1, y_2)\) in \(\RR^2\), are bilinear forms?
\(f(\alpha, \beta) = 1\).
\(f(\alpha, \beta) = (x_1 - y_1)^2 + x_2y_2\).
\(f(\alpha, \beta) = (x_1 + y_1)^2 - (x_1 - y_1)^2\).
\(f(\alpha, \beta) = x_1y_2 - x_2y_1\).
Solution
Goal: Test each function for bilinearity in both slots.
1. (a) \(f(\alpha,\beta)=1\) is not bilinear.
1.1. Bilinearity in the first slot requires \(f(0,\beta)=0\) for all \(\beta\).
1.2. Here \(f(0,\beta)=1\neq0\), so (a) fails.
2. (b) \(f(\alpha,\beta)=(x_1-y_1)^2+x_2y_2\) is not bilinear.
2.1. Expand:\[ (x_1-y_1)^2+x_2y_2=x_1^2-2x_1y_1+y_1^2+x_2y_2.\]
2.2. The terms \(x_1^2\) and \(y_1^2\) are quadratic in one variable and violate linearity.
2.3. Therefore (b) fails bilinearity.
3. (c) \(f(\alpha,\beta)=(x_1+y_1)^2-(x_1-y_1)^2\).
3.1. Expand:\[ (x_1+y_1)^2-(x_1-y_1)^2=4x_1y_1.\]
3.2. This has the form \(4\,x_1y_1\), linear in each variable.
3.3. Hence (c) is bilinear, represented by matrix \(\begin{pmatrix}4&0\\0&0\end{pmatrix}\).
4. (d) \(f(\alpha,\beta)=x_1y_2-x_2y_1\).
4.1. This is the determinant form \(\det(\alpha,\beta)\).
4.2. It is linear in each entry separately:\[f(\alpha+\alpha',\beta)=f(\alpha,\beta)+f(\alpha',\beta),\qquad f(\lambda\alpha,\beta)=\lambda f(\alpha,\beta).\]
4.3. Symmetric checks for \(\beta\) are analogous.
4.4. Therefore (d) is bilinear, represented by matrix \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\).