rank and eigenvalues Let A be a square matrix of order $14\times 14$. We know that rank(A)=12 and $\lambda=0$ is an eigenvalue with algebraic multiplicity 4. I have to decide which of the following statements is true:


*

*$\text{rank}(A^2)=12$.

*$\text{rank}(A^3)\leq11$.

*There is no matrix satisfying the given conditions.


I tried studying the connection between the Ker and the image of the associate endomorphism, but I don't know how to use the information about the eigenvalue.
Could someone help me?
 A: Since $\dim \ker A = 2$ and $0$ has algebraic multiplicity $4$ we see that there are
two Jordan blocks corresponding to the $0$ eigenvalue. The only possible sizes are
$(1,3)$ and $(2,2)$.
In both cases, $A^2$ will drop rank by at least one, so 1. cannot hold.
Since $\operatorname{rk} A^3 \le \operatorname{rk} A^2 < 12$, we see that 2. is
always true.
Let $A=\begin{bmatrix} 0 & 1 \\
0 & 0 \\
 & & 0 & 1\\
 & & 0 & 0  \\
& & & & I_{12}
\end{bmatrix}$
hence 3. is false.
A: For matrix A, with eigenvalue $\lambda$ with multiplicity $n$ kernels of powers of $(A-\lambda I)^k$ follow the pattern:
\begin{equation*}
ker(A-\lambda I) \subsetneq ker(A-\lambda I)^2 \subsetneq  ... ker(A-\lambda I)^{m} = ker(A-\lambda I)^{m+1} = ...
\end{equation*}
Until the power $m \leq n $,  all the lower kernels are strictly contained in the higher power kernels, and at power $m$, $dim(ker(A- \lambda I)^m) = n$. Also, the dimension difference is non-increasing.
Therefore, for $\lambda = 0$ and $n=4$, we know that the dimension of $ker(A^3) = ker((A-0I)^3)$ can be either 3 or 4. But we also know that $dim(ker(A))=2$ and hence dimension of $ker(A^3)$ is 4. Hence, $rank(A^3)$ is 10.
$ker(A^2) = ker((A-0I)^2)$ can have dimension 4 or 3. So, $rank(A^2)$ can be 10 or 11.
