$\vec{u}=\vec{w_{1}}+\vec{w_{2}}, w_1\in W$ and $w_2\in W^\perp$

Question

Let $S[(x_1,y_1,z_1)(x_2,y_2,z_2)]=3x_1x_2+x_2y_1+x_1y_2+y_1y_2+z_1z_2$ be an inner product,

$W=${$(x,y,z)\in\mathbb{R}^3|x+z=0$} a subspace and $\vec u=(1,0,0)$

Find $w_1\in W,$ $w_2\in W^\perp$ (orthogonal in respect of $S$) such that $\vec u=\vec w_1+\vec w_2$

Here's what I've done so far:

An orthogonal basis of W in respect of the inner product $S$ is $B=${$(1,0,-1),(-\frac{1}{4},1,\frac{1}{4}$)}

If $w=(x,y,z)\in W^\perp$ then $S[(x,y,z),(1,0,-1)]=0$ and $S[(x,y,z),(-\frac{1}{4},1,\frac{1}{4})]=0$

So $3x+y-z=0$ and $\frac{1}{4}x+\frac{3}{4}y+\frac{1}{4}z=0\Rightarrow x+3y+z=0$

Solving the above system we get that the basis for $W^\perp$ in respect of $S$ is $B'=\left \{(1,-1,0),(0,0,1) \right \}$

I don't know how to proceed from here..

Also, are the $\vec w_1$, $\vec w_2:\vec u=\vec w_1+\vec w_2$ unique?

Answer 1

You have found the basis for $W^\perp$ wrongly as those elements do not satisfy $3x+y-z=0$. To compute a solution for

$$3x+y-z=0$$

$$x+3y+z=0$$

We can simply compute the cross produdct:

$$\begin{bmatrix} 3 \\ 1 \\ -1\end{bmatrix} \times \begin{bmatrix} 1 \\ 3 \\ 1\end{bmatrix}=\begin{bmatrix}4 \\ -4 \\ 8 \end{bmatrix}=4 \begin{bmatrix} 1 \\ -1 \\ 2\end{bmatrix}$$

Hence a basis of $W^\perp = \left\{ ( 1 , -1 ,2)\right\}$

Let's project $u$ onto $W^\perp$. $$S[(x_1,y_1,z_1)(x_2,y_2,z_2)]=3x_1x_2+x_2y_1+x_1y_2+y_1y_2+z_1z_2$$ $$S[(1,-1,2),(1,-1,2)]=3(1)(1)+(1)(-1)+(1)(-1)+(-1)(-1)+(2)(2)=6$$ $$S[(1,0,0),(1,-1,2)]=3(1)(1)+(1)(0)+(1)(-1)+(0)(-1)+(0)(2)=2$$ $$w_2=\frac{2}{6}(1,-1,2)=(\frac13,-\frac13,\frac23)$$

$$w_1=w-w_2=(1,0,0)-(\frac13,-\frac13,\frac23)=(\frac23,\frac13,-\frac23)$$

Yes, the decomposition is unique.

Answer 2

With

$$ \begin{array}{rcl} \vec{u} & = & (1,0,0)\\ \vec{w}_{1} & = & (w_{11},w_{12},w_{13})\\ \vec{w}_{2} & = & (w_{21},w_{22},w_{23}) \end{array} $$

The inner product is weighted by

$$ M=\left[\begin{array}{ccc} 3 & 1 & 0\\ 1 & 1 & 0\\ 0 & 0 & 1 \end{array}\right] $$

The conditions are

$$ \left\{ \begin{array}{rcl} \vec u & = & \vec{w}_1+\vec{w}_2\\ \left\langle \vec{u},\vec{w}_{1}\right\rangle _{M} & = & \left\Vert \vec{w}_{1}\right\Vert _{M}^{2}\\ \left\langle \vec{u},\vec{w}_{2}\right\rangle _{M} & = & \left\Vert \vec{w}_{2}\right\Vert _{M}^{2}\\ w_{11}+w_{13} & = & 0 \end{array}\right. $$

Solving for $\vec{w}_1, \vec{w}_2$ we obtain the solutions

$$ \begin{array}{lccccc} \vec{w}_1 & = & \frac{1}{7} \left(4+\sqrt{2}\right) & \frac{1}{7} \left(2-3 \sqrt{2}\right) & \frac{1}{7} \left(-4-\sqrt{2}\right) \\ \vec{w}_2 & = & \frac{1}{7} \left(3-\sqrt{2}\right) & \frac{1}{7} \left(-2+3 \sqrt{2}\right) & \frac{1}{7} \left(4+\sqrt{2}\right) \end{array} $$

and

$$ \begin{array}{lccccc} \vec{w}_1 & = & \frac{1}{7} \left(4-\sqrt{2}\right) & \frac{2}{7}+\frac{3 \sqrt{2}}{7} & \frac{1}{7} \left(-4+\sqrt{2}\right) \\ \vec{w}_2 & = & \frac{1}{7} \left(3+\sqrt{2}\right) & \frac{1}{7} \left(-2-3 \sqrt{2}\right) & \frac{1}{7} \left(4-\sqrt{2}\right) \end{array} $$

NOTE

Here $< \cdot, \cdot >_M$ is the inner product $\vec x_1^{\top}M \vec x_2$ and $||\cdot ||_M$ indicated the norm $\sqrt{\vec x^{\top}M\vec x}$