Proving if a certain matrix exists or not

Question

The question is:

Is there a natural $n$ and $A\in M_{n}(\mathbb C)$ (complex $n\times n$ matrices) such that the conditions

\begin{align} \operatorname{rank}(\,A\,) &= 10\\ \operatorname{rank}(A^2) &= 7\\ \operatorname{rank}(A^3) &= 2 \end{align}

are satisfied? Either provide an example or prove it does not exist.

It's clear that $10\le n$ and because $A$ is not invertible it means $11\le n$. I've tried finding an example using Jordan Blocks: there are two block with the size of $4$ or one block with the size of $5$, in order to answer on the third term. Then I tried filling the matrix with block of 3 but it doesn't work out. I also don't have any idea how to prove such a matrix doesn't exist.

Answer 1

Your idea to use Jordan normal form is a good one. Note that the only way that the rank of your matrix can decrease as you take successive powers is if the matrix contains nilpotent Jordan blocks (corresponding to eigenvectors/generalized eigenvectors with eigenvalue $0$). Call the nilpotent Jordan blocks $N_1,\dots,N_k$. Without loss of generality we may assume that no Jordan block with eigenvalue $0$ is a $1\times 1$ block, as if this is the case then we may restrict ourselves to the nonzero Jordan blocks by disregarding the extraneous blocks. Note that as the rank of $A$ is $10$, by examining the structure of $N_i$ we may see that the rank of $A^2$ is $10-k$; thus we must have exactly $3$ nilpotent Jordan blocks. But now it is easy to see that in general, rank$(A^m)\geq$ rank$(A)-(m-1)k$, so as $k=3$ we must have rank$(A^3)\geq 10-2\cdot3=4$. In particular, no such $A$ can exist.

Answer 2

Each Jordan block $J$ of size $m$ is of the form $\lambda I + N$ where $N$ has $m-1$ ones in its superdiagonal and satisfies $N^m=0$. Then, observing that $I$ commutes with any matrix) we have that

$$J^k= \sum_{i=0}^{\max\{m,k\}} \lambda^{k-i}N^i.$$

In particular, for $\lambda\neq 0$ we have $J^k=\lambda^k\sum_{i=0}^{\max\{m,k\}} \lambda^{-i}N^i$, so that for $l\geq0$ we have

$$J^{m+l}=\lambda^l J^m,$$

which also holds when $\lambda =0$. This makes one think we could try working with different sized Jordan blocks. But how does any one particular Jordan block behave?

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When $\lambda \neq 0$, each Jordan block, as a matrix, has no zero eigenvalues, so its powers all have full rank. Hence, the only way for the rank of $J^k$ to go down is if $J$ is associated to $\lambda =0$.

Now, a Jordan block with $\lambda=0$ is just $N$, and it's easy to see what powers of $N$ look like.
In particular, one can check that $N$ has rank $m-1$ and that moreover each additional multiplication by $N$ further reduces the rank by $1$, until we get to $0$.

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With this in mind, we have that $A$ starts with rank $10$. If $A^2$ has rank $7$, $A$ must have exactly $3$ Jordan blocks associated to $\lambda = 0$.

It follows that another multiplication by $A$ can reduce the rank by at most $3$: this would happen if all of those Jordan blocks had size $3$ or more. Since the requirement is that $A^3$ have rank $2$, a reduction of $5>3$, we see that there are no such $A$ and $n$.