Das ist der Fall α=-1
\(\small \left(\begin{array}{rrrr}\lambda=&-1&\left(\begin{array}{rrr}2&2&1\\0&0&1\\0&0&0\\\end{array}\right)&\left(\begin{array}{r}x1\\x2\\x3\\\end{array}\right) = 0\\\lambda=&1&\left(\begin{array}{rrr}0&2&1\\0&-2&1\\0&0&-2\\\end{array}\right)&\left(\begin{array}{r}x1\\x2\\x3\\\end{array}\right) = 0\\\end{array}\right)\)
===>
\(\small EV=\left(\begin{array}{rr}-1&1\\1&0\\0&0\\\end{array}\right)\)
Suche λ=-1 HV ∈ Ker (A-λE)^N mit dim Ker (A-λE)^N = n ∧ HV ¬∈ Ker (A-λE)N-1
===>
\(\small HVKandidaten1 \, := \, \left(\begin{array}{rr}-1&-1\\1&0\\0&1\\\end{array}\right)\to KernHV1 \, := \, \left(\begin{array}{rr}0&-1\\0&1\\0&0\\\end{array}\right)\)
2. Spalten bilden HVs
\(\small T \, := \, \left(\begin{array}{rrr}1&-1&-1\\0&1&0\\0&0&1\\\end{array}\right) \to T^{-1} A \,T = D \, = \, \left(\begin{array}{rrr}1&0&0\\0&-1&1\\0&0&-1\\\end{array}\right)\)