From generalized eigenvector to Jordan form

72 Views Asked by At

I can't figure out the following part of Chen's Linear Systems book. How does he "readily obtain" $Av_2=v_1+\lambda v_2$?


enter image description here

1

There are 1 best solutions below

0
On BEST ANSWER

Note that the fact that we have a "chain of generalized eigenvectors" implies that $(A - \lambda I)v_i = v_{i-1}$. So, we have $$ (A - \lambda I)v_2 = v_1 \implies Av_2 - \lambda v_2 = v_1 \implies Av_2 = v_1 + \lambda v_2. $$