Maximum and minimum ratio of matrix calculation Suppose you have a matrix :
$$A = \begin{pmatrix} 13 & -4 & 2 \\ -4 & 13 & -2 \\ 2 & -2 & 10 \end{pmatrix}.$$
I want to find the maximum and minimum values of the ratio
$$\frac{x'Ax}{x'x}$$ where $x =(x_1,x_2,x_3)$ is nonzero. Is there a way you can figure this out manually?
 A: Since it's a symmetric matrix whose entries are real, it can be diagonalized by an orthogonal matrix $G$. That is the spectral theorem. And remember: an orthogonal matrix is a square matrix $G$ with real entries whose inverse is its transpose, so you have $G'G = I$.  So you get
$$
A= G'\begin{bmatrix} \lambda_1 & 0 & 0 \\ 0 & \lambda_2 & 0 \\ 0 & 0 & \lambda_3 \end{bmatrix}G = G'\Lambda G.
$$
By permuting rows of $G$, you can do this in such a way that $\lambda_1>\lambda_2>\lambda_3$.
Consequently
$$
\frac{x'Ax}{x'x} = \frac{x'(G'\Lambda Gx}{x'(G'G)x} = \frac{(x'G')\Lambda(Gx)}{(x'G')(Gx)} = \frac{(Gx)'\Lambda (Gx)}{(Gx)'(Gx)} = \frac{y'\Lambda y}{y'y}
=\frac{\lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2}{y_1^2+y_2^2+y_3^2}.
$$
If $(y_1,y_2,y_3)=(c,0,0)\ne(0,0,0)$, then the value of the fraction is $\lambda_1$, and similar comments apply to the other two $\lambda$s.  What remains is to show that if more than two coordinates of $(y_1,y_2,y_3)$ are nonzero, then the value of the fraction is a weighted average of the $\lambda$s, so the value is somewhere between what it would be in the opposite extreme cases.
The values of $\lambda_i$, $i=1,2,3$ are eigenvalues, and are therefore the zeros of the characteristic polynomial.
A: I think you may have difficulty calculating the eigenvalues manually.  They happen to be $18$  and $9$ (the latter with multiplicity $2$), so the minimum and maximum values of the ratio are $9$ and $18$. 
EDIT: However, you may notice (well, I didn't until I found the eigenvalues by computer) that there is some symmetry involving the first two coordinates, and this might lead you to find one eigenvector:
$$ A \pmatrix{1\cr 1\cr 0\cr} = 9 \pmatrix{1 \cr 1\cr 0\cr}$$
Thus one eigenvalue is $9$.  By looking at $A$ on the orthogonal complement of this eigenvector, we get a $2 \times 2$ matrix whose eigenvalues are easy to find by hand.
