Answer
See solution
Work Step by Step
a) Let z be in $W^\perp$ and u be any element in W. Where c is a scalar, cz is orthogonal to u because $cz\cdot u=c(z\cdot u)=c\cdot0=0$, which means cz is in $W^\perp$. This means $W^\perp$ is closed under scalar multiplication.
b) Let $z_1$ and $z_2$ be in $W^\perp$ and u be any element in W. Then, $(z_1+z_2)u=z_1u+z_2u=0+0=0$ because u is orthogonal to any element in $W^\perp$. This means $z_1+z_2$ is in $W^\perp$ and $W^\perp$ is closed under vector addition.
c) The zero vector is in $W^\perp$ because the dot product of any vector by the zero vector is 0, which means the three conditions of a subspace are fulfilled.
