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We are about to look at an important type of matrix in multivariable calculus known as Hessian Matrices. We will then formulate a generalized second derivatives test for a real-valued function $z=f(x_1,x_2,...,x_n)$ of n variables with continuous partial derivatives at a critical point $a=(a1,a2,...,an)∈D(f)$ to determine whether $f(a)$ is a local maximum value, local minimum value, or saddle point of $f$

Definition: Let $x=(x1,x2,...,xn)$ and let $z=f(x1,x2,...,xn)=f(x)$ be an n variable real-valued function whose second partial derivatives exist. Then the Hessian Matrix of f is the n×n matrix of second partial derivatives of f denoted $$\mathcal H (\mathbf{x}) = \begin{bmatrix} f_{11} (\mathbf{x}) & f_{12} (\mathbf{x}) & \cdots & f_{1n} (\mathbf{x})\\ f_{21} (\mathbf{x}) & f_{22} (\mathbf{x}) & \cdots & f_{2n} (\mathbf{x})\\ \vdots & \vdots & \ddots & \vdots \\ f_{n1} (\mathbf{x}) & f_{n2} (\mathbf{x}) & \cdots & f_{nn} (\mathbf{x}) \end{bmatrix}$$.

a)$\mathcal H (\mathbf{x})$ is said to be $\textbf{Positive Definite}$ if $D_i>0$ for i=1,2,...,n.

b) $\mathcal H (\mathbf{x})$ is said to be $\textbf{Negative Definite}$ if $D_i<0$ for odd i∈{1,2,...,n} and Di>0 for even i∈{1,2,...,n}.

c) $\mathcal H (\mathbf{x})$ is said to be $\textbf{Indefinite}$ if $det(\mathcal H (\mathbf{x}))=D_n≠0$ and neither a) nor b) hold.

d) If $det(\mathcal H (\mathbf{x}))=D_n=0$, then $\mathcal H (\mathbf{x})$ may be Indefinite or what is known Positive Semidefinite or Negative Semidefinite.

I'm studying a function that has a $det(\mathcal H (\mathbf{x}))=D_n=0$

What is the appropriate way to classify critical points if $det(\mathcal H (\mathbf{x}))=D_n=0$

For example Assume we have this function $$f(x_1,x_2,x_3)=x_1 x_2+x_1 x_3-x_1x_2 x_3 $$ therefore, we found the critical point by solve this system $$\begin{equation} \begin{cases} \frac{df}{dx_1}=x_2+x_3-x_3 x_2=0\\ \frac{df}{dx_2}=x_1-x_1 x_3 =0\\ \frac{df}{dx_3}=x_1-x_1 x_2 =0 \end{cases} \end{equation}$$ we get $c=(0,0,0)$ ,use the Hessian matrix of f $$\mathcal H (\mathbf{f(x_i)})=\left( \begin{array}{ccc} 0 & 1-x_3 & 1-x_2 \\ 1-x_3 & 0 & -x_1 \\ 1-x_2 & -x_1 & 0 \\ \end{array} \right)$$ $det(\mathcal H (\mathbf{f(x_i)}))=D_n=0$

This test fails to determine the type of critical points

Is there another way? Or modification of the solution and correction. Thanks for the help.

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  • $\begingroup$ @user7530 can you help me $\endgroup$ – Emad kareem Jul 2 '17 at 16:38
  • $\begingroup$ $(0,0,0)$ isn’t the only critical point of $f$. There’s an entire curve along which $\nabla f=0$. This should give you a clue as to the nature of these critical points. $\endgroup$ – amd Jul 2 '17 at 21:41
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Compute the eigenvalues of the hessian.

If all the eigenvalues are nonnegative, it is positive semidefinite.

If all the eigenvalues are positive, it is positive definite.

If all the eigenvalues are nonpositive, it is negative semidefinite.

If all the eigenvalues are negative, it is negative definite.

Otherwise, it is indefinite.

Edit:

For that example, you have found $c=(0,0,0)$.

$$H(f(c))=\begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}$$

$$H(f(c))-\lambda I= \begin{bmatrix} -\lambda & 1 & 1 \\ 1 & -\lambda & 0 \\ 1 & 0 & -\lambda\end{bmatrix}$$

\begin{align}\det(H(f(c))-\lambda I)&= \det\left(\begin{bmatrix}1 & 1 \\ -\lambda & 0 \end{bmatrix} \right) -\lambda \det \left( \begin{bmatrix}-\lambda & 1 \\ 1& -\lambda \end{bmatrix} \right) \\&=\lambda-\lambda(\lambda^2-1) \\&=\lambda(2-\lambda^2)\end{align}

Hence, the eigenvalues are $0$, $\sqrt{2}$ and $-\sqrt{2}$. Hence it is indefinite.

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  • $\begingroup$ Please can you solve the example in question, or give me the titles of some references on the subject $\endgroup$ – Emad kareem Jul 2 '17 at 15:42
  • $\begingroup$ Edited the post. $\endgroup$ – Siong Thye Goh Jul 2 '17 at 17:26

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