# Optimization & singular Hessian matrix

## Main Question or Discussion Point

I am trying to figure out how the least squares formula is derived.

With the error function as

Ei = yi - Ʃj xij aj

the sum of the errors is

SSE = Ʃi Ei2

so the 1st partial derivative of SSE with respect to aj is

∂SSE / ∂aj = Ʃi 2 Ei ( ∂Ei / ∂aj )

with the 1st partial derivative of Ei with respect to aj being

∂Ei / ∂aj = - Ʃ xij

so the 2nd derivative of SSE with respect to a single aj is

2SSE / ∂aj2 = 2 Ʃi { ( ∂Ei / ∂aj ) ( ∂Ei / ∂aj ) + Ei ( ∂2Ei / ∂aj2 ) }

and a double partial derivative (i.e., to aj & ak) is

2SSE / ( ∂aj ∂ak ) = 2 Ʃi { ( ∂Ei / ∂aj ) ( ∂Ei / ∂ak ) + Ei ( ∂2Ei / ( ∂aj ∂ak ) ) }

and with the 2nd partial derivative of Ei with respect to aj or the double partial derivative (i.e., to aj & ak) being 0, since the 1st partial is a constant (i.e., in aj )

2Ei / ∂aj2 = ∂2Ei / ( ∂aj ∂ak ) = 0

the 2nd derivatives reduce to

2SSE / ∂aj2 = 2 Ʃi { ( ∂Ei / ∂aj ) ( ∂Ei / ∂aj ) }

2SSE / ( ∂aj ∂ak ) = 2 Ʃi { ( ∂Ei / ∂aj ) ( ∂Ei / ∂ak ) }

which after the substitution of ∂Ei / ∂aj becomes

2SSE / ( ∂aj ∂ak ) = - 2 Ʃi xi( 2 j )

2SSE / ( ∂aj ∂ak ) = - 2 Ʃi xi( j + k )

so the Hessian matrix (e.g., 3 x 3) is

[ H ] =

[ n Ʃ x Ʃ x2 ]

[ Ʃ x Ʃ x2 Ʃ x3 ]

[ Ʃ x2 Ʃ x3 Ʃ x4 ]

which is the normal product of the Vandemonde matrix

[ V ] =

[ 1 x0 x02 ]

[ 1 x1 x12 ]

[ 1 x2 x22 ]

H = [ V ]T [ V ]

While the diagonal terms are sums of the even powers of x, and therefore always positive, it seems that the 2nd derivative test requires that the determinant of the Hessian matrix be positive. Is there any way to prove that this normal product (or even just the Vandermonde matrix itself, since the determinant of itself would merely be the square root of the determinant of the normal product) is guaranteed to have a positive determinant?

So how to determine that indeed the critical point (which is the solution to the coefficients) is a minimum?

Last edited:

haruspex