SOLUTION: suppose that V is finite-dimensional space and T:V->V is a linear operator. 1. prove that rank(T^k+1) <= rank(T^k) 2. Prove that there exists k in N such that rank(T^k+1)=rank(

Question 1136203: suppose that V is finite-dimensional space and T:V->V is a linear operator.
1. prove that rank(T^k+1) <= rank(T^k)
2. Prove that there exists k in N such that rank(T^k+1)=rank(T^k)
I know that R(T^k+1) ⊆ R(T^k) for any k. We can choose v ∈ R(T^k+1).Then for some c, (T^k+1)c = v. But T^k(Tc) = v, so v ∈ R(T^k). How would I proceed with this? any help is appreciated.
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

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

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

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.


 are the degrees of the operator T, what you can interpret as sequential iterations of T.


Then it is clear that  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,    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  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 .

    This stabilization can be achieved at some positive value of , or at the zero value 

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

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