# Thread: Vec subspace of R^3

2. ## Re: Vec subspace of R^3 Originally Posted by steelmaste Here is suggestion: not everyone reads French so you might consider translating for all.
Moreover mathematical notation is not all standard. The usual def of projection is $\rm{proj}_{\vec{u}}\vec{v}=\dfrac{\vec{v}\cdot \vec{u}}{\|\vec{u}\|^2}\vec{u}$
In your posted image $W=\left\{u\in\mathbb{R}^3| \rm{proj}_{\vec{u}}\vec{v}=0\right\}$ which is odd.
The only way a projection is the zero vector is for $\displaystyle \vec{v} \bot \vec{u}$, Why not just say $W=\left\{\vec{u}\in\mathbb{R}^3| \vec{u}\cdot\vec{v}=0\right\}~?$

To show that $W$ is a subspace you must show:
1) $\vec{O}\in W$
2) ($\forall \alpha\in\mathbb{R}~\&~\forall \{\vec{a},\vec{b}\}\subset W)[\alpha\vec{a}+\vec{b}\in W]$ that shows closure with/respect to addition & scalar multiplication.

3. ## Re: Vec subspace of R^3 my teacher actually gave me the answer and I still dont understand for the proj part why is <u,v> written in the form of scalar when it should just be ((v*u)/(u*u)) * u.

4. ## Re: Vec subspace of R^3

You seem to be mistaken as to basic definitions. The inner product, <u, v>, is a number. It is the square root of u*v. (u*v)/(u*u) u is the projection of v on u. That is NOT denoted by "<u, v>".