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?

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.
