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I am trying to understand that if $f(x)$ is any polynomial over a field $F$ such that $f(T) = 0$ then $m(x)\mid f(x).$ $T$ is a linear transformation from $V$ to $V$ and $m(x)$ is the minimal polynomial of $T$.

My book says apply the division algorithm to $f(x)$:

$f(x) = m(x)Q(x) + R(x)$ where either $R(x) = 0$ or $\deg R(x) < r = \deg m(x)$

Rewriting for $R(T)$ we have:

$R(T) = f(T) - m(T)Q(T) = 0$

Now my book claims that the above statement holds for $x$ and hence $R(x)=0$. I do not understand this implication. Why does this imply that $R(x)=0$?

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You mention something called $m(x)$ but leave us to guess what it's supposed to be. Here's my guess: $m(x)$ is the minimal polynomial of $T$. –  Michael Hardy Dec 6 '12 at 20:41
    
haha good guess! I made the appropriate edit. –  CodeKingPlusPlus Dec 6 '12 at 20:43

2 Answers 2

up vote 1 down vote accepted

If $f(x)=m(x)Q(x)+R(x)$ and $f(T)=0$ and $m(T)=0$, then $R(T)=0$. But since $\deg R(x)<\deg m(x)$, one cannot have $R(T)=0$ unless $R$ is the identically $0$ polynomial.

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Because $R(x)$ is a polynomial of smaller degree than $m(x)$ and $R(T)=0$. But since $m(x)$ is a minimal non-zero polynomial such that $m(T)=0$, that means $R(x)$ can not be a non-zero polynomial, so $R(x)=0$.

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