Jacobians & Hessians

Drag the point around the bowl. Below it sits a small grid — the Jacobian of a different, vector-valued function evaluated at that same spot. Watch its four numbers shift as you move; unlike the single gradient arrow, this matrix has one row per output.

For a function $F:\mathbb{R}^2\to\mathbb{R}^2$, the Jacobian stacks the gradient of each output as a row:

• $F_1, F_2$ — the two output components of $F$.
• $J_F(x,y)$ — the Jacobian: every first partial derivative of every output, evaluated at $(x,y)$.

For a single scalar function $f:\mathbb{R}^2\to\mathbb{R}$, the Hessian plays the analogous role for second derivatives — the matrix of every way of differentiating twice:

• $H_f(x,y)$ — the Hessian of the scalar function $f$ at $(x,y)$: its matrix of second partials.

1. The Jacobian generalizes the gradient

   For a scalar function ($F$ has one output), the Jacobian is the gradient, just written as a row instead of a vector — nothing new, only more rows once there's more than one output.

2. The Hessian generalizes the second derivative

   Just as $f''(x)>0$ meant "curving upward" in one dimension, the Hessian's sign pattern says the same thing in two: a positive definite Hessian (both the top-left entry and the determinant positive) means $f$ curves upward in every direction — a genuine bowl.

3. The Jacobian's determinant measures local stretching

   Where $\det J_F \neq 0$, $F$ is locally invertible — near that point, $F$ doesn't collapse two different inputs onto the same output. Where $\det J_F = 0$, that guarantee breaks down.

The Hessian of $f$ here never changes — $f$ is a fixed quadratic bowl, so its curvature is the same everywhere. The Jacobian of $F$ genuinely depends on where you are, which is exactly why it needs to be recomputed at every point instead of stated once.

Let $F(x,y) = (x^2+y,\ x+y^2)$ and $f(x,y) = x^2+xy+y^2$.

1. Evaluate F and its Jacobian at (2, 3)

   Differentiate each output with respect to each input:

   • $\partial F_1/\partial x = 2x$
   • $\partial F_1/\partial y = 1$
   • $\partial F_2/\partial x = 1$
   • $\partial F_2/\partial y = 2y$

   That gives the general Jacobian

   $$J_F(x,y) = \begin{bmatrix}2x & 1 \\ 1 & 2y\end{bmatrix}$$

   Now evaluate both $F$ and $J_F$ at $(2,3)$:

   • $F(2,3) = (4+3,\ 2+9) = (7, 11)$
   • $J_F(2,3) = \begin{bmatrix}2(2) & 1 \\ 1 & 2(3)\end{bmatrix} = \begin{bmatrix}4 & 1 \\ 1 & 6\end{bmatrix}$
2. Its determinant

   $$\det J_F(2,3) = (4)(6) - (1)(1) = 23$$
3. The Hessian of f is constant

   First partials:

   • $\partial f/\partial x = 2x+y$
   • $\partial f/\partial y = x+2y$

   Differentiate each of those again:

   • $\partial^2f/\partial x^2 = \partial/\partial x[2x+y] = 2$
   • $\partial^2f/\partial x\partial y = \partial/\partial y[2x+y] = 1$
   • $\partial^2f/\partial y\partial x = \partial/\partial x[x+2y] = 1$
   • $\partial^2f/\partial y^2 = \partial/\partial y[x+2y] = 2$

   giving

   $$H_f = \begin{bmatrix}2 & 1\\1 & 2\end{bmatrix}$$

   everywhere — with no $x,y$ left in any entry, since $f$ is a fixed quadratic. Its determinant is $4-1=3>0$ and its top-left entry is $2>0$, so $f$ is positive definite: a true bowl with one minimum, at the origin.