Why $(A -\lambda I)v$ is an eigenvector if $v$ is not?

Question

We have that the eigenvalue is only one real value, and we have a matrix of $2\times 2$, but I fail to see why $(A -\lambda I)v$ , I tried plugging the equation and using that the matrix satisfies its characteristic equation, but that didn't work, and I actually proof that that if we have real values with same magnitude and different sign, then $(A-\lambda ^2)v$ is an eigenvector if v is not. Could it be this is false?, i read that in a book about structures.

Answer 1

I think you're saying that $A$ is a $2\times 2$ matrix with only one eigenvalue, $\lambda$. If so then the characteristic polynomial is $$\det(A-tI)=(t-\lambda)^2,$$ so that $(A-\lambda I)^2=0$. So $$(A-\lambda I)(A-\lambda I)v=0,$$so $(A-\lambda I)v$ is an eigenvector.

Answer 2

The construction you describe is how to create the Jordan Normal Form for this matrix $A.$ You have a column vector $v$ with $(A - \lambda I) v \neq 0.$ Well, make a new column vector $u =(A - \lambda I) v.$ Then the matrix for $A$ in the ordered basis $u,v$ is $$ \left( \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right) $$ which is how Jordan is taught in the U.S. Last week, a guy from Uruguay said they put the extra $1$ below the main diagonal.

Anyway, if $R$ is the 2 by 2 matrix with left column $u$ and right column $v,$ then $R^{-1} AR = J$ is the Jordan form I showed above.