I have a list of points $S$ in the form of $(p, q)$:
$$\begin{align}
S = &(43, 58), (44, 60), (40, 60), (41, 61), \\
&(46, 60), (40, 57), (53, 62), (50, 61)
\end{align}$$
And I wish to center them on the origin $(0, 0)$. I would do this by subtracting from them the midpoints ($(\bar{p}, \bar{q})$) for each dimension:
$$\begin{align}
\bar{p} &= \frac{p_1 + p_2 + \dots + p_n}{n} \\
\bar{q} &= \frac{q_1 + q_2 + \dots + q_n}{n}
\end{align}$$
I find $\bar{p} = 44.625$ and $\bar{q} = 59.875$. I find my new $S$ to be:
$$\begin{align}
S_{\text{new}} = &(-1.625, -1.875), (-0.625, 0.125), (-4.625, 0.125), (-3.625, 1.125), \\
&(1.375, 0.125), (-4.625, -2.875), (8.375, 2.125), (5.375, 1.125)
\end{align}$$
Using linear regression, I've found the line of best fit for this data set which crosses the origin to be $y = 0.26x + 0$. This is the line in which I want to project points of data onto from right angles.
My question is, how do I find these projected points (marked as red dots)? Taking point $(1.375, 0.125)$, I can make a triangle with vertices at the origin, the point, and the projected point like so:
I know the slope of $c$ ($0.26$), the position of vertex $ba$ ($(1.375, 0.125)$), and position of vertex $ca$ ($(0, 0)$), but how do I find the position of vertex $cb$?
This is for principal component analysis. To find the eigenvalue, I need the sum of squared distances from projected points to the origin. I've already found the eigenvector to be $\begin{bmatrix}0.96 \\ 0.25\end{bmatrix}$.
Best Answer
One way to do this is by calculating the euclidean vector of the blue line, in this case it is $\begin{bmatrix} 1 \\ 0.26 \end{bmatrix}$, you want its norm to be $1$ so you divide it by its norm to get: $v = \begin{bmatrix} 0.97 \\ 0.25 \end{bmatrix}$.
Then see every point as vector and to get the coordinates of point A one the blue line you just have to calculate $(A \cdot v) \cdot v$.
For example, for the point $(-1.625, −1.875)$, you would find:
$$\left( \begin{bmatrix} -1.625 \\ -1.875 \end{bmatrix} \cdot \begin{bmatrix} 0.97 \\ 0.25\end{bmatrix}\right) \cdot \begin{bmatrix} 0.97 \\ 0.25 \end{bmatrix} = \begin{bmatrix} -1.98 \\ -0.51 \end{bmatrix}$$