Question 1136203
.


            Unfortunately,  I don't know, what exactly is your level in Linear Algebra.


            Nevertheless,  I will try to explain the solution in simple terms.



<pre>
You have a linear map (operator) T from a finite-dimensional linear space V to itself.


rank(T)  is the dimension of the image of the space V under this transformation.


{{{T^k}}} are the degrees of the operator T, what you can interpret as sequential iterations of T.


Then it is clear that {{{rank(T^k)}}} can not rise up.  

It only can go down - not necessary strictly down at each step/iteration.

Not necessary in monotonic way down. But not rise up, in any case.



There are two typical examples.



One example is when the matrix of T is zero everywhere, except one diagonal above the major diagonal.

    In this case,  {{{rank(T^k)}}}  decreases monotonically at each step/iteration till 0 (zero) 
    after n iterations, where n = dim(V).


The other example is an operator of projection to a subspace.  Than {{{rank(T^k)}}} stabilizes on some positive value at some step.

    (and this value is, OBVIOUSLY, the dimension of the image of T).



So, after my explanations, n.1 is just proved/explained (using my fingers).



n.2 also becomes evident now, meaning stabilization of values of {{{rank(T^k)}}}.

    This stabilization can be achieved at some positive value of {{{rank(T^k)}}}, or at the zero value 

    (which means then that the operator T and its matrix is/are nilpotent).
</pre>