Let $\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\in\mathbb{R}^n,$ where $\{\overrightarrow{v},\overrightarrow{u}\}$ is linearly independent. Let $\overrightarrow{n}=\overrightarrow{u}\times \overrightarrow{v}$ and $P=\mathrm{Span}\{\overrightarrow{v},\overrightarrow{u}\}$. Determine, with proof, which of the following statements are true.
- If $\{\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\}$ is linearly dependent, then $\mathrm{proj}_P (\overrightarrow w)=\overrightarrow w$.
- $\mathrm{proj}_{\overrightarrow{v}} \overrightarrow{w} \neq \mathrm{proj}_{\overrightarrow{u}}\overrightarrow{w}$
- $\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}=\overrightarrow{n}-\mathrm{proj}_P\overrightarrow{w}$
- $\{\overrightarrow{v},\overrightarrow{u},\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}\}$ is linearly independent.
Statement $1$ is the only true statement. Suppose $\{\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\}$ is linearly dependent. We have that $\mathrm{proj}_P{\overrightarrow{w}} =\mathrm{perp}_{\overrightarrow{n}}\overrightarrow{w}=\overrightarrow{w}-\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}.$
Since $\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}=\dfrac{\overrightarrow{n}\cdot\overrightarrow{w}}{\Vert \overrightarrow{n}\Vert^2}\cdot\overrightarrow{n}$ and $\{\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\}$ is linearly dependent, we have that $(\overrightarrow{u}\times \overrightarrow{v})\cdot \overrightarrow{w}=\overrightarrow{n}\cdot \overrightarrow{w} = 0\Rightarrow \mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}=\overrightarrow{0}.$ Hence $\mathrm{proj}_P\overrightarrow{w} = \overrightarrow{w}-\overrightarrow{0}=\overrightarrow{w}.$
To show that $2$ is false, consider $\overrightarrow{v} = (1,0,0),\overrightarrow{u}=(0,0,1),\overrightarrow{w}=(0,1,0).$ Since $\overrightarrow{v}$ and $\overrightarrow{u}$ are not scalar multiples of each other, $\{\overrightarrow{v},\overrightarrow{u}\}$ is linearly independent. However, $\mathrm{proj}_{\overrightarrow{v}}\overrightarrow{w} = \mathrm{proj}_{\overrightarrow{u}}\overrightarrow{w} = \overrightarrow{0}$, which contradicts the statement. In fact, choosing any two linearly independent vectors for $\overrightarrow{u}$ and $\overrightarrow{v}$ and $\overrightarrow{w}=\overrightarrow{u}\times\overrightarrow{v}$ will work.
$3$ is false. Consider when $w\in\mathrm{Span}\{\overrightarrow{u},\overrightarrow{v}\}.$ Then $\{\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\}$ is linearly dependent. From $1, \mathrm{proj}_P\overrightarrow{w}=\overrightarrow{w}$. Since $\overrightarrow{w}\in\mathrm{Span}\{\overrightarrow{u},\overrightarrow{v}\}, \mathrm{proj}_{\overrightarrow{n}}{\overrightarrow{w}}=\overrightarrow{0}$. However, $\overrightarrow{w}\neq \overrightarrow{n}$ as otherwise $\{\overrightarrow{v},\overrightarrow{u},\overrightarrow{w}\}$ would be linearly independent, a contradiction. Hence $\overrightarrow{n}-\mathrm{proj}_P\overrightarrow{w}\neq \overrightarrow{0}=\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}$ in this case, disproving the statement.
$4$ is false as $\{\overrightarrow{v},\overrightarrow{u},\mathrm{proj}_{\overrightarrow{n}}\overrightarrow{w}\}$ is linearly dependent whenever $\overrightarrow{w}\in\mathrm{Span}\{\overrightarrow{u},\overrightarrow{v}\}$.
Just wondering, is $3$ actually true?