Rank of a matrix $A^2$ without calculating the square I have a matrix
$A=\begin{bmatrix}
2 & 0 & 4\\ 
1 & -1 & 3\\ 
2 & 1 & 3
\end{bmatrix}
$
with rank 2.
How do I prove that the matrix $A^2$ has also rank 2 without actually calculating $A^2$.
I know that $rank(AB)\leq min(rank(A), rank(B))$, and so $rank(A^2) \leq 2$, but I still don't have enough information.
Thanks!
 A: If you've shown that the $\operatorname{null}(A) \cap \operatorname{column}(A) = \{ 0 \}$, then  
$$A^2v = A(Av) =  0  \ \Longleftrightarrow \ Av = 0$$
That is $\operatorname{null}(A) = \operatorname{null}(A^2)$ and hence $\operatorname{nullity}(A) = \operatorname{nullity}(A^2)$.
Now as $\dim(\mathbb{R}^3) =  \operatorname{rank}(A) + \operatorname{nullity}(A) =  \operatorname{rank}(A^2) + \operatorname{nullity}(A^2) $, it follows $$\operatorname{rank}(A) = \operatorname{rank}(A^2)$$
A: We can prove generally that if $\operatorname{null}(A)\cap \operatorname{col}(A)=\{0\}$, then $\operatorname{rank}(A^2) = \operatorname{rank}(A)$.  We can do so as follows:
Let $v_1,\dots,v_k$ be a basis for the kernel of $A$.  Extend this to a basis $v_1,\dots,v_n$ of $\Bbb R^n$.
Claim: the vectors $A^2(v_{k+1}),\dots,A^2(v_{n})$ form a basis of the column space of $A^2$.
Proof:  Try it!  It's easy to show that these vectors span the column space; the trick is to show that they are also linearly independent.  Leave a comment if you get stuck and I'll give you another prod along.

Alternatively, use the rank-nullity theorem as the other answer suggests
A: First, the kernel of
$$A=\begin{bmatrix}
2 & 0 & 4\\ 
1 & -1 & 3\\ 
2 & 1 & 3
\end{bmatrix}
$$
is generated by $(2,-1,-1)^t$. 
Now you want to know the number of solutions of
$$\begin{bmatrix}
2 & 0 & 4\\ 
1 & -1 & 3\\ 
2 & 1 & 3
\end{bmatrix}\begin{bmatrix}
x\\ 
y\\ 
z
\end{bmatrix}=\begin{bmatrix}
2\\ 
-1\\ 
-1
\end{bmatrix}
$$
If there are no solution the rank of $A^2$ is the same of the rank of $A$, i.e. 2, if there is a $1-$dimensional space of solution then the rank of $A^2$ is $1$.
Since the rank of
$$\begin{bmatrix}
2 & 0 & 4& 2\\ 
1 & -1 & 3&-1\\ 
2 & 1 & 3&-1
\end{bmatrix}$$
is $3$, the system has no solution, so the rank of $A^2$ is $2$.
In this way you don't need to compute $A^2$, even if probably it would be simpler to do that.
A: in your comment you say that you have proved $image(A) \cup null(A) = {0}.$ that shows $null(A) = null(A^2)$ which in turn by the nullity theorem implies $image(A) = image(A^2).$ so the rank of $A$ and rank of $A^2$ are equal to $2.$
what am i missing?
