Why the projection formula of $x$ onto $v$ is given by $\frac{\langle x,v\rangle}{\langle v,v\rangle}v$?

Question

Can somebody explain the formula for the projection of a vector $x$ onto a vector $v$?

The formula is given by $\frac{(x,v)}{(v,v)}v$.

Thanks!

Answer 1

Given the vector $x$, you can see that the projection of $x$ in the direction of $v$ is just the component parallel to $v$.

To find this vector, we need the direction and the module:

Find the direction:

The direction is the same of $v$, so we can take a unit vector $\frac{v}{\|v\|}$

Find the length:

The length is geometrically the same as the component of $x$ given by cosine. Length = $\|x\|\cdot \cos(\theta)$

What is $\theta$ ? It's the angle between those vectos. How do we calculate it? with the inner product: $$\langle x, v\rangle = \|x\|\cdot\|v\|\cos\theta$$

So, finally the vector is : $$\operatorname{Proj}_vx = \text{length}\cdot\text{direction} = \|x\|\cdot \cos(\theta) \cdot \frac{v}{\|v\|}$$ $$ = \|x\|\cdot \frac{\langle x, v\rangle}{\|x\|\cdot\|v\|} \cdot \frac{v}{\|v\|} $$

Since $\|v\||^2 = \langle v,v\rangle$ :

$$\operatorname{Proj}_vx = \frac{\langle x,v \rangle \cdot v}{\|v\|^2} = \frac{\langle x,v \rangle \cdot v}{\langle v,v\rangle} $$

Answer 2

By definition, the projection of $v$ along $w$, where $w\neq 0$, is the vector $p = \alpha w$ such that $\langle v - p,w\rangle = 0$. Based on it, one has that \begin{align*} \langle v-p,w\rangle = \langle v - \alpha w,w\rangle = 0 & \Longleftrightarrow \langle v,w\rangle - \alpha\langle w,w\rangle = 0 \Longleftrightarrow \alpha = \frac{\langle v,w\rangle}{\langle w,w\rangle} \end{align*} and we are done. Hopefully this helps.