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In $\mathbb{R}^4$, consider the subspace $W = Span(u_1, u_2,u_3)$ with

$$u_1 = (-1,1,0,0)$$

$$u_2 = (-1,0,1,0)$$

$$u_3 = (-1,0,0,1)$$

$(a)$ Use the Gram-Schmidt Process to construct an orthonormal basis for $W$.

(b) Which of the vectors $a=(1,1,0,-1)$, $b=(1,1,1,1)$ and $c=(0,0,-1,1)$ is a basis for $W^{\perp}$?

(c) Express the vector $x=(0,0,0,4)$ as a sum of a vector $x_1 \epsilon W$ and $x_2 \epsilon W^{\perp}$.

I already did part $(a)$. The orthonormal basis is:

$$\left\{\frac{1}{\sqrt{2}}(-1,1,0,0),\frac{1}{\sqrt{6}}(-1,-1,2,0),\frac{1}{\sqrt{12}}(-1,-1,-1,3)\right\}$$

How do I do part $(b)$ and $(c)$ though?

For part $(b)$, my idea was to put $u_1,u_2,u_3$ into a matrix augmented with $0$ since $W^{\perp}$ means checking the null space. Alternatively, my second idea was to take the dot product of $a,b,c$ with $u_1,u_2,u_3$ to see which one results in zero. Is that the right idea?

For part $(c)$, am I just supposed to express $x$ as a linear combination of some vector in $W$ and $W^{\perp}$?

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Yes, the vectors in $W^\perp$ are orthogonal to every vector of $W$. So the basis vector of $W^\perp$ must be orthogonal to each basis vector of $W$. Only $\bf b$ satisfies this condition.

For part $(b)$, note that the projection of $\bf x$ on $W^\perp$ is given by $\displaystyle\mathbf{x_2}=\frac{\langle\mathbf x,\mathbf b\rangle}{\langle\mathbf b,\mathbf b\rangle}\mathbf b=(1,1,1,1)$. So $\mathbf x_1=\mathbf{x-x_2}=(-1,-1,-1,3)$ lies in $W$. $$\mathbf x=(-1,-1,-1,3)+(1,1,1,1)$$

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(b) It's the vector $b$: it's the anloy of the theree vectors which is orthogonal to every $u_i$. Besides, $\dim W=3\implies\dim W^\perp=1$.

(c) If$$v_1=\frac{1}{\sqrt{2}}(-1,1,0,0),\ v_2=\frac{1}{\sqrt{6}}(-1,-1,2,0),\text{ and }v_3=\frac{1}{\sqrt{12}}(-1,-1,-1,3)$$then$$\langle x,v_1\rangle v_1+\langle x,v_2\rangle v_2+\langle x,v_3\rangle v_3=(-1,-1,-1,3).$$So$$x=\overbrace{x-(-1,-1,-1,3)}^{\in W^\perp}+\overbrace{(-1,-1,-1,3)}^{\in W}.$$

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