Existence of Annihilating Polynomial for a Linear Operator over a fine dimensional vector space.

Question

How do I show the existence of Annihilating Polynomial for a Linear Operator $T:V\to V$ over a finite dimensional vector space $V$.

I had an idea that $\mathbb{L}(V,V)$ is a vector space of dimension $n^2$. Hence considering the set $I,T,T^2,T^3, \dots \dots T^{n^2}$ which has $n^2+1$ elements, must be dependent. Hence we got the annihilating polynomial.

But my Professor wasn't happy, He said that some set had dimension $n$ and considering the set $I,T,T^2,T^3, \dots \dots T^n$ which has $n+1$ elements.

This gives us two cases.

case $1$ If All $T^i$ are distinct, we get a dependent set, hence we can conclude. Otherwise

case $2$ we must have $T^i=T^j$ for some $0\leq i<j\leq n$ Hence it is also dependent. Can someone help me with this?

Have you discussed the "Cayley-Hamilton theorem" in this class? — Ben Grossmann, May 27 '19 at 15:49
the proof you gave is certainly the simplest, most elementary one. If you want a polynomial of degree $n$, then one possibility is to resort to the Cayley Hamilton theorem, since the characteristic polynomial certainly annihilates $T$ — peek-a-boo, May 27 '19 at 15:49
It sounds like your professor was talking about cyclic subspaces — Ben Grossmann, May 27 '19 at 15:51
If you haven't discussed Cayley-Hamitlon, this post should be helpful — Ben Grossmann, May 27 '19 at 15:57
No, we haven't not done cavley-hamilton theorem yet, Its just the begining. He was actually trying to prove that the mapping $\sigma:R[x] \to L\mathbb{T}(V)$ defined by $ \sigma(p(x))=p(T)$ has nontrivial kernel. — GraduateStudent, May 27 '19 at 15:57

Existence of Annihilating Polynomial for a Linear Operator over a fine dimensional vector space.

0 Answers0