Given a $n\times n$ matrix A,and the matrix's characteristic polynomial is $|\alpha I-A|=(\alpha-a_{1})^{r_{1}}(\alpha-a_{2})^{r_{2}}...(\alpha-a_{p})^{r_{p}}$,and $r_1+r_2+...r_p=n$. Then,as for any eigenvalue $a_i$,do we have $$ rank(A-a_i I)=rank((A-a_iI )^m$$?( m is any positive integer and 1<=i<=p)
That's really confusing me,can someone tell me whether it is true or not? How to prove it?need some help
No, this not true. Take $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ which has eigenvalue 1, its characteristic polynomial is $\det (A-\lambda I) = (\lambda-1)^2$.
You have $$A - I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ which has rank 1, but $$(A-I)^2 = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$$ has rank zero.