Worksheet 2-2 (with Solutions)

Worksheet 2-2 (with Solutions)#

Download: CMSE382-WS2_2.pdf · Solution PDF

Warning

This is an AI-generated transcript of the worksheet and may contain errors or inaccuracies. Please refer to the original course materials for authoritative content.

Worksheet 2-2: Q1#

For the following symmetric matrices, cross out the classifications it CANNOT be due to the diagonal entries (don’t calculate the eigenvalues).

For each matrix, circle those that remain possible: pos def   pos semidef   indefinite   neg def   neg semidef

  1. \(A_1=\begin{bmatrix} 0 & 1 \\ 1 & 5 \end{bmatrix}\)

    Solution

    The diagonal entry \(5 > 0\) rules out negative definite and negative semidefinite. The diagonal entry \(0\) (not strictly positive) rules out positive definite. Remaining: positive semidefinite or indefinite.

  2. \(A_2=\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}\)

    Solution

    Both diagonal entries are positive, so negative definite and negative semidefinite are ruled out. Remaining: positive definite, positive semidefinite, or indefinite.

  3. \(A_3=\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}\)

    Solution

    One diagonal entry is positive and one is negative, so the only option remaining is indefinite.

  4. \(A_4=\begin{bmatrix} -3 & 1 \\ 6 & -2 \end{bmatrix}\)

    Solution

    Both diagonal entries are negative, so positive definite and positive semidefinite are ruled out. Remaining: negative definite, negative semidefinite, or indefinite.

Worksheet 2-2: Q2#

For the following matrices with their given eigenvalues, what is the correct classification and why?

  1. \(A_1=\begin{bmatrix} 0 & 1 \\ 1 & 5 \end{bmatrix}\); eigenvalues \(= \{-0.193,\; 5.193\}\)

    Solution

    One positive and one negative eigenvalue → indefinite.

  2. \(A_2=\begin{bmatrix} 2 & -1 \\ -1 & 1 \end{bmatrix}\); eigenvalues \(= \{0.382,\; 2.618\}\)

    Solution

    All eigenvalues positive → positive definite.

  3. \(A_3=\begin{bmatrix} -1 & 2 \\ 2 & 1 \end{bmatrix}\); eigenvalues \(= \{-2.236,\; 2.236\}\)

    Solution

    One positive and one negative eigenvalue → indefinite.

  4. \(A_4=\begin{bmatrix} -3 & 1 \\ 6 & -2 \end{bmatrix}\); eigenvalues \(= \{-5,\; 0\}\)

    Solution

    Eigenvalues are either negative or zero → negative semidefinite.

Worksheet 2-2: Q3#

On the last worksheet, we found the stationary points of

\[f(x,y)=6x^{2}y-3x^{3}+2y^{3}-150y\]

were at \(p_1 = (0,5)\), \(p_2 = (0,-5)\), \(p_3 = (4,3)\), \(p_4 = (-4,-3)\).

Use the second optimality condition to classify each stationary point as a local minimum, local maximum, or saddle point. (Visual check: desmos.com/3d/xa4komuwmb.)

Solution

The Hessian is:

\[\begin{split}\nabla^2 f(x,y) = \begin{bmatrix} 12y - 18x & 12x \\ 12x & 12y \end{bmatrix}\end{split}\]

  • \(p_1 = (0,5)\): \(\nabla^2 f = \begin{bmatrix} 60 & 0 \\ 0 & 60 \end{bmatrix}\), eigenvalues \(60, 60\) → positive definite → local minimum.

  • \(p_2 = (0,-5)\): \(\nabla^2 f = \begin{bmatrix} -60 & 0 \\ 0 & -60 \end{bmatrix}\), eigenvalues \(-60, -60\) → negative definite → local maximum.

  • \(p_3 = (4,3)\): \(\nabla^2 f = \begin{bmatrix} -36 & 48 \\ 48 & 36 \end{bmatrix}\), eigenvalues \(60\) and \(-60\) → indefinite → saddle point.

  • \(p_4 = (-4,-3)\): \(\nabla^2 f = \begin{bmatrix} 36 & -48 \\ -48 & -36 \end{bmatrix}\), eigenvalues \(60\) and \(-60\) → indefinite → saddle point.

Worksheet 2-2: Q4#

Find and classify the stationary points of \(f(x,y)=2x^3 + 3y^2 + 3x^2 y - 24y\).

Solution

Gradient:

\[\begin{split}\nabla f(x,y) = \begin{bmatrix} 6x^2 + 6xy \\ 6y + 3x^2 - 24 \end{bmatrix}\end{split}\]

Setting to zero: \(6x(x+y) = 0\) gives \(x=0\) or \(y=-x\); and \(y = -\frac{1}{2}x^2 + 4\).

  • \(x=0\): \(y=4\) → stationary point \((0,4)\).

  • \(y=-x\): solving gives \(x=4 \Rightarrow y=-4\) and \(x=-2 \Rightarrow y=2\).

Stationary points: \((0,4)\), \((4,-4)\), \((-2,2)\).

Hessian: \(\nabla^2 f = \begin{bmatrix} 12x + 6y & 6x \\ 6x & 6 \end{bmatrix}\)

  • \((0,4)\): \(\begin{bmatrix} 24 & 0 \\ 0 & 6 \end{bmatrix}\), eigenvalues \(24, 6\) → positive definite → local minimum.

  • \((4,-4)\): \(\begin{bmatrix} 24 & 24 \\ 24 & 6 \end{bmatrix}\), eigenvalues \(\approx 40.6\) and \(\approx -10.6\) → indefinite → saddle point.

  • \((-2,2)\): \(\begin{bmatrix} -12 & -12 \\ -12 & 6 \end{bmatrix}\), eigenvalues \(\approx -18\) and \(\approx 12\) → indefinite → saddle point.

(Visual check: desmos.com/3d/juzovyr3hd.)

Contents