SOLUTION TO PROBLEM H-10

Set

$$I=\begin{bmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdo... ...dots & \ddots & \vdots \\ 1 & 1 & 1 & \cdots & 1 \end{bmatrix}$$

Then

$$S=U-I\quad\mbox{and}\quad U^2=nU$$

For any real number x, we have

$$\begin{eqnarray*}(U-I)(xU-I)& = & xU^2-(x+1)U+I\\ & = & (xn-(x+1))U+I \end{eqnarray*}$$

If we choose x=1/(n-1) then xn-(x+1)=0. So

$$(U-I)\left (\frac{1}{n-1}U-I\right)=I$$

The inverse of S is

$$\begin{eqnarray*}S^{-1}& =& \left (\frac{1}{n-1}U-I\right) \\ & =& \begin{bmatr... ...{1}{n-1} & \frac{1}{n-1} & \cdots & \frac{2-n}{n-1} \end{bmatrix}\end{eqnarray*}$$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified:   Fri Mar 22 13:11:22 CST 2002