rank and eigenvalues

Question

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:

1. $\text{rank}(A^2)=12$.
2. $\text{rank}(A^3)\leq11$.
3. 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?

Answer 1

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.

Answer 2

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.