Solution to Linear Algebra Hoffman & Kunze Chapter 9.2 Exercise 9.2.1
Solution:
(a) No. Since $f(0,\beta)\ne 0$.
(b) No. Since $f((0,0),(1,0))\ne 0$.
(c) Yes. Since $f(\alpha,\beta)=4x_1\bar y_1$.
(d) No. Because of $\bar x_2$ there, it is not linear on $\alpha$.
