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Let $A$ be a nonzero symmetric $3\times3$ matrix. Consider the function $f(\textbf x)=\frac{1}{2}(A \textbf x)\cdot \textbf x$.

(a) What is $\nabla f$?

Here's what I did:

Let $A= \left[ \begin{array}{ccc} a & b & c \\ b & e & d \\ c & d & g \end{array} \right]$ and $\textbf x = \left[ \begin{array}{c} x \\ y \\ z \end{array} \right]$.

$ f(\textbf x)=\frac{1}{2}(A \textbf x)\cdot \textbf x = \frac{1}{2}(\left[ \begin{array}{ccc} a & b & c \\ b & e & d \\ c & d & g \end{array} \right] \left[ \begin{array}{c} x \\ y \\ z \end{array} \right])\cdot \textbf[x \space y \space z \textbf] \\= \frac{1}{2}ax^2+\frac{1}{2}ey^2+\frac{1}{2}gz^2+bxy+cxz+dyz=f(x,y,z).$

I basically converted the vectors' dot products into defined matrix multiplication. Then I took the partial derivatives to form $\nabla f=(ax+by+cz,bx+ey+dz,cx+dy+gz)$. In matrix form this reduces to $A \textbf x$. Can someone tell me if I did this correctly?

Here is where I am stuck:

(b) Consider the restriction of $f$ to the unit sphere $S$ in $\mathbb R^3$. We know that $f$ must have a maximum and minimum on $S$. Show that there must be an $\textbf x \in S$ and a $\lambda \neq 0$ such that $A \textbf x = \lambda \textbf x$. (The vector $\textbf x$ is called an eigenvector, while the scalar $\lambda$ is called an eigenvalue.)

I'm not sure how to show this, and I am taking Linear Algebra as well but we haven't covered eigenvectors yet. My instructor says that knowledge of eigenvectors is not necessary to solve this. It is not clear to me what the question is asking, and I don't know where to begin.

Bobby Lee
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  • It wants you to show that there is some vector $\textbf x$ and some scalar $\lambda$ so that transforming $\textbf x$ by $A$ and scaling $\textbf x$ by $\lambda$ produce the same result; that is, $A$ happens to just stretch or compress that particular $\textbf x$. Try just setting up the equations and seeing what happens. (e.g., $\lambda \textbf x = \langle \lambda x_1, \lambda x_2, \lambda x_3\rangle$.) – dfan Mar 01 '14 at 21:50
  • Ok I tried doing what you said, attempted to set up a system of equations $A\textbf x = \lambda \textbf x$, but I could not seem to find a solution similar to that of the back of my book. My solution manual says that "at an $\textbf x$ where $f$ is extreme, there is a $\lambda /2$ such that $A\textbf x = \lambda \textbf x$". This answer makes absolutely no sense to me, and I am wondering if you could lead me to this result so that I can understand it. – Bobby Lee Mar 03 '14 at 23:53

1 Answers1

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Your derivation of $f$ and $\nabla f$ looks good (except for some row vectors that should technically be column vectors).

You have already shown that

$$A \textbf x = \left[ \begin{array}{c} ax + by + cz\\ bx + ey + dz \\ cx + dy + gz \end{array} \right]$$ and you know that $$\lambda \textbf x = \left[ \begin{array}{c}\lambda x \\ \lambda y \\ \lambda z \end{array}\right]$$

and it is given that $$x^2 + y^2 + z^2 = 1$$

So you have a total of four unknowns ($x$, $y$, $z$, and $\lambda$) and four equations (three from $A\textbf x = \lambda \textbf x$ and one from the equation of the sphere). Can you take it from there?

dfan
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