Minimal Polynomial of linear map induced by quotients space divides the minimal polynomial of itself

Question

Suppose U is a subspace of V invariant under a linear transformation T : V → V . Prove that T induces a linear map$ \bar T : V /U → V /U $of quotients given by $\bar T(v + U) = T(v) + U$. Prove that the minimal polynomial of $\bar T$ divides the minimal polynomial of T.

For the minimal polynomial: The minimal polynomial of T is the monic polynomial$ m(x)∈[x]$ of least degree such that $m(T)=0$ on V.Therefore, $m(T|_u)=0$

Minimal polynomial of restriction to invariant subspace divides minimal polynomial

Should it to prove the minimal polynomial of the invariant subspace divides the minimal polynomial of T first and then continue?

Your definition of "minimal polynomial is fine, but the crucial property is this: let $m$ be the minlmal polynomial of a linear transformation $\alpha$ on $V$; then $(\forall f)( m|f \iff f(\alpha)=0)$ — ancient mathematician, May 23 '17 at 16:15
Well you have a lin transfm $T|_U$ and a polynomial $m$ that kills it, so .... — ancient mathematician, May 23 '17 at 18:50
And if you look again you'll see that $m(\bar{T})=0$ as well. So the minimal polynomials of $T_{U}$ and $\bar{T}$ both divide the minimal polynomial of $T$. — ancient mathematician, May 25 '17 at 10:23

score 2 · Answer 1 · answered May 25 '17 at 11:02

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$$ m(\bar{T})(v+U)=m(T)v+U=0+U $$ where of course $0+U$ is the zero element of $V/U$.

So $m(\bar{T})=0$, and then by the basic property of minimal polynomials, the minimal polynomial of $\bar{T}$ must divide $m$.

answered May 25 '17 at 11:02

ancient mathematician

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Minimal Polynomial of linear map induced by quotients space divides the minimal polynomial of itself

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