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Second derivative test, Sylvester’s criterion

TOC

Explanation

Certainly! Let's delve into both topics one by one.

Second Derivative Test in Linear Algebra

The second derivative test is a method used in calculus to determine the local minima and maxima of a function based on its second derivative. However, when we talk about the second derivative test in the context of multivariable calculus or linear algebra, it usually refers to the test for determining whether a critical point of a function of several variables is a local minimum, local maximum, or saddle point.

For a function $f(x, y)$ of two variables, here's how the second derivative test works:

  1. Find the critical points by solving the system of equations given by the first derivatives: $$ \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 0

$$

  1. Evaluate the second partial derivatives at the critical point $(x_0, y_0)$: $$ f_{xx}(x_0, y_0), \quad f_{yy}(x_0, y_0), \quad f_{xy}(x_0, y_0) $$

  2. Calculate the determinant of the Hessian matrix $H$ at the critical point: $$ D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - \left( f_{xy}(x_0, y_0) \right)^2 $$

  3. Use the determinant to classify the critical point:

    • If $D > 0$ and $f_{xx}(x_0, y_0) > 0$, then $(x_0, y_0)$ is a local minimum.
    • If $D > 0$ and $f_{xx}(x_0, y_0) < 0$, then $(x_0, y_0)$ is a local maximum.
    • If $D < 0$, then $(x_0, y_0)$ is a saddle point.
    • If $D = 0$, the test is inconclusive.

Sylvesters Criterion

Sylvester's Criterion is a method used in linear algebra to determine whether a given square matrix is positive definite. A matrix is positive definite if it satisfies the following condition for any non-zero vector $x$:

$$ x^T A x > 0 $$

Here's how Sylvester's Criterion works:

  1. Compute the leading principal minors of the matrix $A$. The $k$-th leading principal minor is the determinant of the top-left $k \times k$ submatrix of $A$.

  2. The matrix $A$ is positive definite if and only if all leading principal minors are positive.

In other words, for a matrix $A$ to be positive definite, the determinants of all upper-left submatrices, including the determinant of the matrix itself, must be positive. This criterion can also be extended to determine if a matrix is positive semidefinite, negative definite, or negative semidefinite by considering the signs of the determinants of the leading principal minors.

  • Positive semidefinite: all leading principal minors are non-negative.
  • Negative definite: the leading principal minors alternate in sign, starting with negative.
  • Negative semidefinite: all leading principal minors are non-positive and alternate in sign, starting with negative.

This criterion is particularly useful because it provides a method to verify definiteness without having to find an eigendecomposition of the matrix. It's worth noting that Sylvester's Criterion only applies to Hermitian (or real symmetric) matrices, which are equal to their own conjugate transpose.

Solution

  1. Find extrema and saddle points for the following function: (10 pts.) $$ f(x,y)=3x^3+y^2+4xy-x+2 $$ Formulate Sylvester’s criterion (2 pts.).

To find the extrema and saddle points for the function $f(x,y) = 3x^3 + y^2 + 4xy - x + 2$, we first need to find the critical points by setting the first partial derivatives equal to zero, and then apply the second derivative test.

Step 1 Find the critical points

We compute the first partial derivatives $f_x$ and $f_y$, and set them equal to zero:

$$ \begin{align*} f_x &= \frac{\partial f}{\partial x} = 9x^2 + 4y - 1 \\ f_y &= \frac{\partial f}{\partial y} = 2y + 4x \end{align*} $$

We need to solve the system of equations given by $f_x = 0$ and $f_y = 0$.

Step 2 Solve for critical points

Solve the equations $9x^2 + 4y - 1 = 0$ and $2y + 4x = 0$ simultaneously.

Step 3 Second derivative test

We then compute the second partial derivatives and evaluate them at the critical points:

$$ \begin{align*} f_{xx} &= \frac{\partial^2 f}{\partial x^2} = 18x \\ f_{yy} &= \frac{\partial^2 f}{\partial y^2} = 2 \\ f_{xy} &= \frac{\partial^2 f}{\partial x \partial y} = 4 \end{align*} $$

The determinant of the Hessian matrix $H$ at a critical point $(x_0, y_0)$ is then calculated as:

$$ D = f_{xx}(x_0, y_0)f_{yy}(x_0, y_0) - \left( f_{xy}(x_0, y_0) \right)^2 $$

We use this determinant to classify each critical point.

Step 4 Sylvesters Criterion

Formulating Sylvester's Criterion involves checking the signs of the leading principal minors of the Hessian matrix of $f$, which is:

$$ H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{xy} & f_{yy} \end{pmatrix} $$

For our function, the leading principal minors are $f_{xx}$ and the determinant $D$ of the whole Hessian matrix.

Let's proceed with the calculations for steps 1 to 3 to find the critical points and use the second derivative test to classify them.

We have found the critical points of the function to be $(-1/9, 2/9)$ and $(1, -2)$.

The second partial derivatives are: $$ f_{xx} = 18x, \quad f_{yy} = 2, \quad f_{xy} = 4 $$

The determinant of the Hessian matrix $H$ as a function of $x$ is: $$ D(x) = 36x - 16 $$

We need to evaluate the second derivatives and the determinant $D$ at each critical point to classify them.

For the critical point $(-1/9, 2/9)$: $$ \begin{align*} f_{xx}(-1/9) &= 18 \cdot (-1/9) = -2 \ f_{yy}(-1/9) &= 2 \ f_{xy}(-1/9) &= 4 \ D(-1/9) &= 36 \cdot (-1/9) - 16 = -4 - 16 = -20 \end{align*} $$ Since $D(-1/9) < 0$, this point is a saddle point.

For the critical point $(1, -2)$: $$ \begin{align*} f_{xx}(1) &= 18 \cdot (1) = 18 \ f_{yy}(1) &= 2 \ f_{xy}(1) &= 4 \ D(1) &= 36 \cdot (1) - 16 = 20 \end{align*} $$ Since $D(1) > 0$ and $f_{xx}(1) > 0$, this point is a local minimum.

Now, let's formulate Sylvester's Criterion based on the Hessian matrix for this function:

Sylvester's Criterion states that a matrix is positive definite if all leading principal minors are positive. The leading principal minors of the Hessian matrix $H$ are:

  1. $f_{xx}$
  2. The determinant $D$

For the Hessian matrix to be positive definite at a point, we must have:

  • $f_{xx} > 0$
  • $D > 0$

In this case, the matrix is not positive definite at the saddle point since $D(-1/9) < 0$, but it is positive definite at the local minimum since both $f_{xx}(1) > 0$ and $D(1) > 0$.