On the rank inequality $\operatorname{rank}(A)+\operatorname{rank}(A^3)\geq2\operatorname{rank}(A^2)$ 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.
 A: 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 $\ker(A^{k})$.
A: 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 $2$.
