FtYou writes

Hello everyone ! There is a concept I have a hard time getting my head wrap around. If you have a Vector Space V and a subspace W, I understand that you can find the least square vector approximation from any vector in V to a vector in W. And this correspond to the projection of V to the subspace W. Now , for data fitting ... Let's suppose you have a bunch of points (xi, yi) where you want to fit a set a regressors so you can approximate yi by a linear combination of the regressors lets say ( 1, x, x2 ... ). What Vector space are we talking about ? If we consider the Vector space of function R -> R, in what subspace are we trying to map these vectors ? I have a hard time merging these two concepts of projecting to a vector space and fitting the data. In the latter case what vector are we using ? The functions ? If so I understand the choice of regressors ( which constitute a basis for the vector space ) But what's the role of the (xi,yi) ? I want to point out that I understand completely how to build the matrices to get Y = AX and solving using least square approx. What I miss is the big picture. The linear algebra picture. Thanks for any help !

We'll go over this by closely examining and understanding an example. Suppose we have the data points ${(x_i, y_i)}$

$\displaystyle \begin{cases} (x_1, y_1) = (-1,8) \\ (x_2, y_2) = (0,8) \\ (x_3, y_3) = (1,4) \\ (x_4, y_4) = (2,16) \end{cases}, $

and we have decided to try to find the best fitting quadratic function. What do we mean by best-fitting? We mean that we want the one that approximates these data points the best. What exactly does that mean? We'll see that before the end of this note - but in linear algebra terms, we are projecting on to some sort of vector space - we claim that projection is the ''best-fit'' possible.

So what do we do? A generic quadratic function is ${f(t) = a + bt + ct^2}$. Intuitively, we apply what we know. Then the points above become

$\displaystyle \begin{cases} f(-1) = a - b + c = 8 \\ f(0) = a = 8 \\ f(1) = a + b + c = 4 \\ f(2) = a + 2b + 4c = 16 \end{cases}, $

and we want to find the best ${[a b c]}$ we can that ''solves'' this. Of course, this is a matrix equation:

$\displaystyle \begin{pmatrix} 1 & -1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 8 \\ 8 \\ 4 \\ 16 \end{pmatrix}. $

And so you see how the algorithm would complete this. But now let's get down the ''linear algebra picture,'' as you say.

We know that quadratic polynomials ${f(t) = a + bt + ct^2}$ are a three dimensional vector space (which I denote by ${P_2)}$ spanned by ${1, t, t^2}$. We know we have four data points, so we will define a linear transformation ${A}$ to be the transformation taking a quadratic polynomial ${f}$ to ${\mathbb{R}^4}$ by evaluating ${f}$ at ${-1, 0, 1, 2}$ (i.e. the ${x_i}$). In other words,

$\displaystyle A : P_2 \longrightarrow \mathbb{R}^4 $

where

$\displaystyle A(f) = \begin{pmatrix} f(-1) \\ f(0) \\ f(1) \\ f(2) \end{pmatrix}. $

We interpret ${f}$ as being given by three coordinates, ${a, b, c \in \mathbb{R}^3}$, so we can think of ${A}$ as a linear transformation from ${\mathbb{R}^3 \longrightarrow \mathbb{R}^4}$. In fact, ${A}$ is nothing more than the matrix we wrote above.

Then a solution to

$\displaystyle A^T A \begin{pmatrix} a \\ b \\ c \end{pmatrix} = A^T \begin{pmatrix} 8 \\ 8 \\ 4 \\ 16 \end{pmatrix} $

is the projection of the space of quadratic polynomials on ${\mathbb{R}^4}$ (which in this case is the space of evaluations of quadratic polynomials at four different points). If ${f^* }$ is the found projection, and I denote the ${y_i}$ coordinate vector as ${y^ *}$, then this projection minimizes

$\displaystyle || y^* - Af^*||^2 = (y_1 - f^*(x_1))^2 + \ldots + (y_4 - f^*(x_4))^2, $

and it is in this sense that we mean we have the ''best-fit.'' (This is roughly interpreted as the distances between the ${y_i}$ and ${f^*(x_i)}$ are minimized; really, it's the sum of the squares of the distances - hence ''Least-Squares'').

So in short: ${A}$ is a matrix evaluating quadratic polynomials at different points. The columns vectors correspond to a basis for the space of quadratic polynomials, ${1, t, t^2}$. The codomain is ${\mathbb{R}^4}$, coming from the evaluation of the input polynomial at the four different ${x_i}$. The projection of the set of quadratic polynomials onto their evaluation space minimizes the sum of the squares of the distances between ${f(x_i)}$ and ${y_i}$.

Does that make sense?

### Leave a comment

## Info on how to comment

To make a comment, please send an email using the button below. Your email
address **won't be shared** (unless you include it in the body
of your comment). If you don't want your real name to be used next to your
comment, please specify the name you would like to use. If you want your name
to link to a particular url, include that as well.

**bold**, *italics*, and plain text are allowed in
comments. A reasonable subset of markdown is supported, including lists,
links, and fenced code blocks. In addition, math can be formatted using
`$(inline math)$`

or `$$(your display equation)$$`

.

**Please use plaintext email** when commenting. See
Plaintext Email and
Comments on this site for more. Note also that
**comments are expected to be open, considerate, and
respectful.**

Comments (1)2021-03-18 TomekGod bless you David. Great explanation!