Let $W=\{(x,y,z):x-2y+z=0\}$ I wanna find:
- The basis of W
- Projection of u=(1,1,2) over W
- The dimension of W
- The orthogonal complement of W ($W^{\bot}$)
- The dimension of $W^{\bot}$
- Basis of $W^{\bot}$
So, I did the following, but not sure about it.
- Find an L.I. set of solutions for the equation. That gives me (2,1,0) and (0, 1, -2) $\in \mathbb{R}^3$ and $S=\{(2,1,0) ; (0, 1, 2)\}$
- I don't know how to project a vector over a space. I thought that I could use one of the basis and project u over each basis. How should I do it?
- Since I have only a set of two basis for W, then I think the dim(W)=2. Am I right?
- I am not sure either how to do it, but makes sense to me that every vector in $W^{\bot}$ when you do a inner product with a vector in W should result in 0. So I did $(x,y,z) \cdot (2,1,0)^{\bot}$ and that gave me (-1, 2, z) and $(x,y,z) \cdot (0,-1,2)^{\bot}$ that gave me (x, 2, 1) but then I figured out that this is the answer for item number 6 and would answer the item number 5 since there are two basis and them dim($W^{\bot}$)=2. So, how should I do it?
Also, what is a Nullspace, Column Space and Row Space in that case? EDIT Corrected the second basis.
I think you made an arithmetic error. For the first vector you're fine, but for the second you get $x-2y+z=0-2-2=-4 \neq 0$.
The general formula for the projection onto the span of a collection of vectors is $Px=A(A^T A)^{-1}A^Tx$, where the columns of $A$ are a basis for the space in question. If your basis is orthogonal, then this is equivalent to projecting onto each element of the basis and adding up the result.
Correct; your basis for $W$ along with any other basis has two elements.
4-6. Notice that
$$x-2y+z = \begin{bmatrix} x \\ y \\ z \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} = 0$$
What does this suggest about the orthogonal complement of $W$? (The result being explored in 4-6 is sometimes called the Fundamental Subspaces Theorem, which is a refinement of the rank-nullity theorem.)