2
$\begingroup$

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.

$\endgroup$
  • 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
5
$\begingroup$

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 $\ker(A^{k})$.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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

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

Thus

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

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 $2$.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.