## Elementary Linear Algebra 7th Edition

(a) the set is not orthogonal. (b) the set is not orthonormal. (c) the set is a basis for $R^2$.
Let $u=(-4,6)$ and $v=(5,0)$, then we have (a) since $u\cdot v=-20+0=-20$, then the set is not orthogonal. (b) since the set is not orthogonal then it is not orthonormal. (c) to check if it is a basis for $R^2$, consider the combination $$a(-4,6)+b(5,0)=(0,0)$$ then we have the system $$-4a+5b=0, \quad 6a=0.$$ It is clear that the above system has the solution $a=0, b=0$. Then the set is linearly independent and $R^2$ has dimension $2$, then it is a basis for $R^2$.