Understanding the Jacobian output of scipy.optimize.minimize - scipy

Question

I'm working with scipy.optimize.minimize to find the minimum of the RSS for a custom nonlinear function. I'll provide a simple linear example to illustrate what I am doing:
import numpy as np
from scipy import optimize
def response(X, b0, b1, b2):
return b2 * X[1]**2 + b1 * X[0] + b0
def obj_rss(model_params, y_true, X):
return np.sum((y_true - response(X, *model_params))**2)
x = np.array([np.arange(0, 10), np.arange(10, 20)])
r = 15. * x[1]**2 - 32. * x[0] + 10.
init_guess = np.array([0., 50., 10.])
res = optimize.minimize(obj_rss, init_guess, args=(r, x))
print res
This yields the results:
fun: 3.0218799331864133e-08
hess_inv: array([[ 7.50606278e+00, 2.38939463e+00, -8.33333575e-02],
[ 2.38939463e+00, 8.02462363e-01, -2.74621294e-02],
[ -8.33333575e-02, -2.74621294e-02, 9.46969972e-04]])
jac: array([ -3.31359843e-07, -5.42022462e-08, 2.34304025e-08])
message: 'Optimization terminated successfully.'
nfev: 45
nit: 6
njev: 9
status: 0
success: True
x: array([ 10.00066577, -31.99978062, 14.99999243])
And we see that the fitted parameters 10, -32, and 15 are equivalent to those used to generate the actuals data. That's great. Now my question:
I have the understanding that the Jacobian should be an m x n matrix where m is the number of records from the X input and n is the number of parameters. Clearly I don't have that in the results object. The results object yields an array that is referred to as the Jacobian in the documentation (1 and 2), but is only one-dimensional with a number of elements equal to the number of parameters.
Further confusing the matter, when I use method='SLSQP', the Jacobian that is returned has one more element than that returned by other minimization algorithms.
. . .
My larger goal here is to be able to calculate either confidence intervals or standard errors, t-, and p-values for the fitted parameters, so if you think I'm way off track here, please let me know.
EDIT:
The following is intended to show how the SLSQP minimization algorithm yields different results in the Jacobian than the default minimization algorithm, which is one of BFGS, L-BFGS-B, or SLSQP, depending on if the problem has constraints (as mentioned in the documentation). The SLSQP solver is intended for use with constraints.
import numpy as np
from scipy import optimize
def response(X, b0, b1, b2):
return b2 * X[1]**2 + b1 * X[0] + b0
def obj_rss(model_params, y_true, X):
return np.sum((y_true - response(X, *model_params))**2)
x = np.array([np.arange(0, 10), np.arange(10, 20)])
r = 15. * x[1]**2 - 32. * x[0] + 10.
init_guess = np.array([0., 50., 10.])
res = optimize.minimize(obj_rss, init_guess, method='SLSQP', args=(r, x))
print res
r_pred = response(x, *res.x)
Yields results:
fun: 7.5269461938291697e-10
jac: array([ 2.94677643e-05, 5.52844499e-04, 2.59870917e-02,
0.00000000e+00])
message: 'Optimization terminated successfully.'
nfev: 58
nit: 10
njev: 10
status: 0
success: True
x: array([ 10.00004495, -31.9999794 , 14.99999938])
One can see that there is an extra element in the Jacobian array that is returned from the SLSQP solver. I am confused where this comes from.