Calculation of gradient and hessian

Question

Let the function on $E = R ^ 2$ by $j (x, y) = x^2-xy+y^2+3x-2y+1$. Calculate the associated gradient and hessian. Deduce the value of a local minimal point is this a strict minimum. To answer this simplistic example, i begin by calculating the derivatives.

Gradient: $D'<f(x, y)/x> = 2x-y+3 $

$D'<f(x, y)/y> = 2y-x-2 $

Hessian:

$D''<f(x, y)/x^2> = 2 $

$D''<f(y)/y^2>= 2 $

And the value of a local minimal is a = (1,1) ?

Where comes from the point $a=(1,1)$? The minimum ( if it exists) is a point where the gradient is null. — Emilio Novati, Apr 23 '17 at 15:32
You can calculate the value of a local minimum point and what is a strict minimumn thanks you. — Paul-henri Guet, Apr 23 '17 at 15:36

score 1 · Answer 1 · answered Jun 14 '17 at 01:42

Let: $$J(x,y) = x^2-xy+y^2+3x-2y+1$$ Then: $$\partial_x J = 2x-y+3 \;\;\;\;\&\;\;\;\;\partial_yJ=-x+2y-2$$ Then: $\nabla J=(2x-y+3, 2y-x-2)$. The minimum must be at $\nabla J=0$. Thus: \begin{align} 2x+3 &= y\\ x+2 &= 2y \end{align} This implies $x+2=4x+6$, so $-3x = 4$. Thus, $x=-4/3$, and then $y = 2(-4/3)+3$. This is an extremal point.

The type of extremum is determined by the Hessian, which is $$ \mathcal{H}[J] = \begin{bmatrix} \partial_{xx} J & \partial_{xy} J\\ \partial_{yx} J & \partial_{yy} J \end{bmatrix} = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix} $$
The second derivative test works like this: if $\mathcal{H}[J]$ is positive definite, we are at a minimum; negative definite, a maximum; otherwise, a saddle-point. In 2D, this reduces to just looking at the determinant (eigenvalue product) and trace (eigenvalue sum) of the Hessian. Here: $$ \det(\mathcal{H}[J])=4-1=3 \;\;\;\;\&\;\;\;\; \text{tr}(\mathcal{H}[J])=4 $$ The sum and product of the eigenvalues are both positive; hence, the matrix is positive definite at the extremal point and therefore a minimum.

Calculation of gradient and hessian

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