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 $W^{\perp}$?
Best Answer
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 $W$. $$\mathbf x=(-1,-1,-1,3)+(1,1,1,1)$$