Let $A$ be a $7 \times 7$ matrix over a field such that $A^{20}=0$. Show that also $A^{6}=0$ I have the following question:
Let K be a field and A be a 7 by 7 matrix over K such that the 20th power of A is 0. Show that the sixth power of A is also 0.
Which subject in linear algebra does contain this questions? And in here is there any generalization of this question? I need just some hints and reference. Thanks.
 A: Note that for any $n$, $range(A^{n+1}) \subset range(A^n)$. So for all $n$ one has
$$\dim(range(A^{n+1})) \leq \dim(range(A^n))$$
I claim that it is not possible under your assumptions for there to be an $n$ for which
$\dim(range(A^{n+1})) =  \dim(range(A^n) \neq 0$. For if we did, then $range(A^{n+1}) = range(A^n)$ and thus
$$range(A^{n+2}) = A(range(A^{n+1}))$$
$$ = A(range(A^{n}))$$
$$ = range(A^{n+1})$$
Similarly, inductively for all $k \geq n + 1$ one has $range(A^k) = range(A^n)$. This can't happen since $A^{20} = 0$, so $A^{21} = 0,$ $A^{22} = 0...$
So we conclude that if  the dimension of $range(A^n)$ is nonzero, then the dimension of $range(A^{n+1})$ is strictly less than that of $range(A^n)$. Thus starting at $n = 0$, the dimension of $range(A^n)$ drops by at least one each time $n$ increases. Since your space is $7$-dimensional, this means $A^7$ has range $\{0\}$, in other words, $A^7 = 0$. (You don't have to have $A^6 = 0$).
A: The concepts of linear algebra which can help you to deal with this problem is the concept of rank and dimension of a matrix.
Daniel's comment is true. You cannot conclude from what you have that $A^6=0$. For consider the matrix
$$ M = \begin{pmatrix}
     0  &   0  &   0 &    0  &   0  &   0  &   0 \\
     1 &   0  &   0 &    0  &  0   &  0   &  0 \\
     0  &   1  &   0 &    0  &   0  &   0  &   0\\
     0  &   0  &   1 &    0  &   0  &   0  &   0\\
     0  &   0  &   0 &    1  &   0  &   0  &   0\\
     0  &   0  &   0 &    0  &   1  &   0  &   0\\
     0  &   0  &   0 &    0  &   0  &   1  &   0
\end{pmatrix}$$
Then you can compute that $M^{20}=0$ and also $M^7=0$, but $M^6 \neq 0$. 
The hint to get you on the way is these two:


*

*A matrix is zero if and only if $rank(  A) = 0$

*If $A$ is not invertible, then $rank(A^{i+1}) < rank(A^i)$. 

