Find basis of $U ^\bot$, Given $U = \{ x \in R^5 | 3x_1 - 3x_2 + x_3 - 4x_4 + x_5 = 0 \}$.
I thought about saying that the vector normal $b$ is $[3,-3,1,-4,1]^T$ is orthogonal to $U$, meaning it is orthogonal to every vector in $U$. But it's not enough to say that $b$ is basis now (I think).
So, if it's not enough, do I need to find Basis of $U$, show that it is spanned by 4 vectors in $R^5$, and then say that basis of $U^\bot$ must have one vector because $U \oplus U^\bot = R^5$.
What do u think?
It is easy to verify that $U$ is a subspace of $\mathbb R^5$. Since $$\{(-1,0,0,0,3),(0,1,0,0,3),(0,0,-1,0,1),(0,0,0,1,4)\}\subset U$$ is linearly independent, $\dim U\geqslant 4$. Since $(0,0,0,0,1)\notin U$, $U\neq\mathbb R^5$, and so $\dim U=4$. As $\mathbb R^5=U\oplus U^\perp$ it follows that $\dim U^\perp=1$. So any nonzero element of $U^\perp$ is a basis for $U^\perp$, including the normal vector $(3,-3,1,-4,1)$.