Vtyjkyuk

I'm trying to project a set of 13-dimensional vectors to n (n = 1,2,3) dimensional space for visualization purposes.

Assume vector $v = (1,2,3,4,5,6,7,8,9,10,11,12,13)$ as one such 13-dimensional data point which needs to me projected to the $XYZ plane$.

Going by the definitions I should project $v$ into $A$ where $A^T$ is

$A^T = left( beginarrayccc
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \
endarray right)$

So using the following equation $ p = A(A^TA)^-1A^Tv$ I end up with $ p = (1,2,3,0,0,0,0,0,0,0,0,0,0)$

So the projection simply converts the values of the vector $v$ from the $4^th$ position onward to zero.

I need to confirm whether the projection I have done above is correct.

If the projection must be linear, but you have the freedom to choose a projection, and you want to preserve as much of the variance in the data as possible, you're probably looking for principal components analysis.

Edit: From this comment, it sounds like you're looking for a linear projection that you can use as an initial estimate, from which you will iteratively optimize Sammon's error. I'm not familiar with Sammon projection, but Wikipedia says that PCA may be used as an initial estimate, citing this article (PDF). So, go ahead and give PCA a shot! Unless you know something else about the data, I wouldn't simply project onto the first 3 out of 13 coordinates.

To widen Henning's comment a bit, any $A := beginbmatrix e_i & e_j & e_k endbmatrix$, for $i,j,k in 1,2,dots,13$ and $i ne j ne k ne i$, will give you an orthogonal to the 3D subspace induced by the vectors $e_i, e_j, e_k$ of the canonical base for $mathbbR^13$, disregarding all info of the other dimensions.

More generally, if you pick any $3$ orthonormal vectors $x_1, x_2, x_3$, then

$$A := X X^T, quad X := beginbmatrix x_1 & x_2 & x_3 endbmatrix,$$

will be an orthogonal projection of $mathbbR^13$ on the subspace spanned by $x_1, x_2, x_3$. This way, you can orthogonally project on any 3D subspace of $mathbbR^13$, as long as you know its orthonormal basis.

For perspective projections this is relatively simple but allowing for multiple variations. We can use parametric equations:

$$ x_1 = a_1*t$$
$$ x_2 = a_2*t$$
$$...$$
$$ x_n-1 = a_n-1*t$$
$$ x_n = a_n*t$$
Where $n$ is the dimension and $a_n$ is the coordinate at that dimension.

To project this to $n-1$ dimensions we can simply set $x_n$ as $1$ or some other value for a specific plane $x_n-1$ and then solve for $t$.

Example: (The 1,2,...13 is your example point )

$$x_1 = 1*t$$
$$x_2 = 2*t$$
$$...$$
$$x_13 = 13*t$$

We want $x_13 = 1$ so we solve for $t$ . This gives us $t= frac113$

Now we plug this value of $t$ into the other equations and get our values from $x_1$ to $x_12$ scaled correctly for $x_13=1$. Then we do the same again but for $x_12=1$ and so on until we get $x_4...13=1$ (or what ever specific plane you want for every dimension). The $Bbb R^3$ projection of your $Bbb R^13$ point will be $(x_1, x_2, x_3 )$.

Of course you could set any value for $x_n$ you like to get a projection onto a specific variation of a plane. However, setting the value to $0$ will not work as every value will become $0$. This is because the focal point is $0$ in every dimension, allowing for easier calculations. Also, this technique relies upon the concept of drawing a line from the point to the origin and solving for $t$ at a specific point which is the reason for the parametric equations. The technique can be applied to n-dimensional points with relative ease. These are very simple parametric equations and it is easy to understand so why do people not think of this way when projecting to lower dimensions?