So, I have to show that for all square matrices $A$, $$\newcommand{\rank}{\operatorname{rank}}\rank(A)+\rank(A^3)\geq2\rank(A^2).$$

My attempt at it so far:

So, if I try to use this theory:
Let $A$ be a square matrix similar to a Jordan matrix $J$. Then $J$ is determined by $$a_{\lambda,k} = \dim\ker(A − \lambda I)^k$$ then I know because of this theorem that $$\rank(A − \lambda I)^{k−1} + \rank(A − \lambda I)^{k+1} ≥ 2 \rank(A − \lambda I)^k$$

Then if I know that $\rank(A)+\rank(A^m)\geq m+1$ then $\rank(A)+\rank(A^3)\geq 4$ so then $2\rank(A^2) = 4$, right?

But I'm not exactly sure how to prove all of this because the theorem from above is for the Jordan matrix $J$... Or does this theory not work with this problem at all?

I was just going to say that the eigenvalues were equal to $0$ and prove the above by substituting in the values but I don't think that is a valid solution. Any help on this would be appreciated.

  • 2
    $\begingroup$ You might want to put the inequality in the following form $r(A)-r(A^2)\geqslant r(A^2)-r(A^3)$. The successive differences $r(A^n)-r(A^{n+1})$ are an important invariant. $\endgroup$ – Pedro Tamaroff Feb 13 '15 at 9:26

More generally, one has $dim(\ker(A^{k+1}))-dim(\ker(A^k))\leq dim(\ker(A^k))-dim(\ker(A^{k-1}))$, that is a fundamental result (cf. the Pedro's comment).

Proof: let $e_1,\cdots,e_r$ be a basis of $F$ that satisfies $\ker(A^{k})\oplus F=\ker(A^{k+1})$. Then $Ae_1,\cdots,Ae_r\in\ker(A^k)\setminus\ker(A^{k-1})$. It remains to show that the $Ae_1,\cdots,Ae_r$ are linearly independent. $\sum_i\lambda_iAe_i=0$ implies $A^k(\sum_i\lambda_ie_i)=0$. Then $\sum_i\lambda e_i\in \ker(A^k)\cap F=\{0\}$ and, consequently, the $(\lambda_i)$ are $0$.

Remark: this proof shows how to construct the part of the Jordan basis, associated to the eigenvalue $0$. One step is: take a basis $e_1,\cdots,e_r$ of a complementary of $\ker(A^{k})$ in $\ker(A^{k+1})$ ; complete the free system $Ae_1,\cdots,Ae_r$ in a basis of a complementary of $\ker(A^{k-1})$ in $\ker(A^{k})$.

  • $\begingroup$ Yes this is what I meant in the previous comment, sorry I didn't see your answer before. $\endgroup$ – mathreadler Feb 13 '15 at 15:28
  • $\begingroup$ @ mathreadler , no problem, it is better by saying twice. :) $\endgroup$ – loup blanc Feb 13 '15 at 22:11

Take $A$ similar to $J$, i.e. $A = QJQ^{-1}$. Then

$$A^2 = (QJQ^{-1})^2 = QJQ^{-1}QJQ^{-1} = QJ^2 Q^{-1}$$

and hence $A^2$ is similar to $J^2$, and likewise for $A^3$ and $J^3$. Thus we can replace $A$ with $J$ and the rank does not change. Let us now prove this for a Jordan matrix $J$.

We do not need to worry about those Jordan blocks in $J$ which correspond to non-zero eigenvalues, since taking powers of these does not change the rank. Thus we will consider those blocks which have zeros on the diagonal. Once one squares the matrix $J$ the rank drops by $1$ for each such block of size at least $2$. When one multiplies by $A$ again, the rank drops again for every block of size at least three.

Let $n$ be the dimension of the vector space. Suppose that $b$ of the basis vectors correspond to blocks of nonzero eigenvalue and $n_k$ basis vectors correspond to zero eigenvalue blocks of size $k$.

The rank of $A$ is $b + \sum_{k \geq 1} n_k (k-1)$. The rank of $A^2 = b + \sum_{k \geq 2} n_{k} (k-2)$. The rank of $A^3 = b + \sum _{k \geq 3} n_{k}(k-3)$. Then

$$rank A + rank A^3 = 2b + n_{2} + \sum_{k \geq 3} n_{k}(2k-4)$$


$$\frac{rank A + rank A^3}{2} = b + \frac{n_2}{2} + \sum_{k \geq 3} n_{k}(k- 2)$$

Which is clearly greater than

$$rank A^2 = b + \sum_{k \geq 2} n_{k} (k-2) = b + \sum_{k \geq 3} n_{k}(k-2)$$

And in fact the difference is measured by the number of Jordan blocks with eigenvalue $0$ and size $2$.

  • 1
    $\begingroup$ The realization in the last sentence invites thinking about the difference of $\rank(A^{k+1}) - \rank(A^k)$ and $\rank(A^{k}) - \rank(A^{k-1})$ $\endgroup$ – mathreadler Feb 13 '15 at 15:19

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