Regression with a Vandermonde matrix.

3.7k Views Asked by At

From what I understand, when doing a least squares regression with a Vandermonde matrix, you're essentially solving the equation

$y=Xa$

Where $y$ is a vector of $y$-values, $X$ is the Vandermonde matrix, and $a$ is a vector of coefficients.

When you solve this equation for $a$,

$a=(X^TX)^{-1}X^Ty$

You get the above expression for $a$.

My understanding is that this should be a solution to the set of equations. However, it is possible to fit $n$ points of data to a $k$-th degree polynomial, where $n>k$. This would imply that there are more equations than unknowns, which results in no possible solutions. However, the vector a can be calculated. This resulting vector does not completely perfectly satisfy

$y=Xa$

But instead is a good approximation, as with regression.

Why can we get a value for $a$? As I do not believe this could be possible if we were dealing with $n$ equations, and $k$ unknowns. Where does the matrix solution differ from solving $k$ unknowns with $n$ equations?