\[A= \left( \begin{array}{ccc} \frac{5}{2} & - \frac{1}{2} & 1 \\ \frac{3}{2} & \frac{1}{2} & -1 \\ 0 & 0 & 0 \end{array} \right) \]
.In general the eigenvalues are complex numbers. Let
\[r_i\]
be the sum of the absolute values of the entries in the ith row except for the entry on the diagonal - entry \[a_{ii}\]
and let \[D_i = \{ z: \| z -a+{ii} \| \le r_i \}\]
be the closed disk centred at \[a_{ii}\]
with radius \[r_i\]
. Gershgorin's Theorem says that the eigenvalues are to be found in the union of the \[D_i\]
's.\[\begin{equation} \begin{aligned} D_1 &= \{ \| z- 5/2 \| \le \| -1/2 \| + \| 1 \| \} \\ &= \{ \| z- 5/2 \| \le 3/2 \} \end{aligned} \end{equation} \]
\[\begin{equation} \begin{aligned} D_2 &= \{ \| z- 1/2 \| \le \| 3/2 \| + \| -1 \| \} \\ &= \{ \| z- 5/2 \| \le 5/2 \} \end{aligned} \end{equation} \]
\[\begin{equation} \begin{aligned} D_3 &= \{ \| z- 0 \| \le \| 0 \| + \| 0 \| \} \\ &= \{ \| z \| \le 0 \} \end{aligned} \end{equation} \]
Solving these inequalities gives
\[1 \le z \le 4\]
\[-2 \le z \le 3\]
\[z=0\]
Hence
\[-2 \le z \le 4\]
Three possible eigenvalues are 0, 1, 2.