# Many little linear questions

1. Jul 9, 2007

### FlashStorm

1)let A be a matrix of n*n above K. let P be a polynom of K[x] such that the polynom P = 0 when substituting A into that polynom. What can we say about the eigenvalues of A? What can we say about matrices that has one or more of these eigenvalues? can we say that this polynom is zero when substituting one of these eigenvalues? (well , I am 99& sure about it since I read this proof).

2)(Test question) Let T be a linear transformation from V to V , when dimV is finite. ||T*(V)||<=||T(V)|| for all v E V, I need to proof that T is normal. Now , obviously we need to proof that ||T*(V)||=||T(V)|| because then we get (after playing with both sides of the equation: (v,TT*v)<=(v,T*Tv).

Now I don't think that its possible that T* can be such transformation that makes the vector shorter each times (But its probably just intuition I developed from this question). I tried to come up with a sentence or something logical for it but I failed.

Well that's for now, Its kinda simple and basic I guess, but I am stuck and I really need help.

Thanks,
Aviv

2. Jul 9, 2007

### mathwonk

if A is a matrix over k, there is aunique k alkgebra map k[X]-->Mat(n,k) taking X to A. By dimension count, this map has a non trivial kernel generated by a unique monic polynomial f, which thus divides another k polynomial P if and only if P(A) = 0.

This f is called the minimal polynomial of A.

If c is an eigenvaklue of A, there is a basis for the space with the corresponding eigenvector as the first entry. in this basis the matrix for A will have c in the upper left entry and all other entries in the first column are zeroes.

When f is applied to this matrix the uppr left entry is f(c). since the whole matrix is zero, so is f(c). I.e. the minimal polynomial f of A, and hence lso any multiple of it, i.e. any polynomial which annihilates A, will have c as a root.

3. Jul 9, 2007

### FlashStorm

Nice :) . But if The Polynom doesn't go zero when substituting A? Can i deduce that all of A eigenvalues aren't roots? something tells It isn't right. But still if that's the case or not, what can I say about the eigenvalues?

And my second questions is still a problem :( ( There I played with the value
used at first the fact that ||T*(v)+T(v)||<||T*(V)||+||T(V)|| and then transformed it into inner product formula which led me to : (Tv,T*v)+(T*v,Tv)<=0 ____
then Re(Tv,T*v)<=0 with the (a,b)= (b,a).

now from cauchy-"schwarzenegger" I deduced that : |(Tv,T*v)|^2<=||Tv||*||T*v||. But I can't reach anything that can help me. )

Last edited: Jul 9, 2007