Hi Kyriakov
the procedure is always the same: solve det(A-lambda Id)=0. You will see that if a_12=a-21=0 the two eigenvalues are lambda_1=a_{11} and lambda_2=a_22. So this case is pretty simple.
Best wishes
Professor Ginestra Bianconi
the procedure is always the same: solve det(A-lambda Id)=0. You will see that if a_12=a-21=0 the two eigenvalues are lambda_1=a_{11} and lambda_2=a_22. So this case is pretty simple.
Best wishes
Professor Ginestra Bianconi
