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so I am trying to show that $f(x)$ divides $g(x)$ for all polynomials $g(x)$ satisfying that $g(A)=0$ where $f(x)$ is the minimal polynomial of a square matrix $A$.

I know from my professor that $f(x)$ is the minimal polynomial of $A$ so $f(A)=0$ so then $g(A)=0$. Therefore, $f(x)$ divides $g(x)$, i.e. $g(x)=f(x)h(x)$ where $h(x)$ is some other polynomial.

My prof told us to solve this question we need to let $b(x)=gcd((f(x),g(x))$ and by using Bezout's Identity, we need to show that $b(A)=0$. I am just having a little trouble trying to prove this. I know that this identity says:

For two polynomials $f_1(x)$ and $f_2(x)$, there exist polynomials $g_1(x)$ and $g_2(x)$ such that $g_1(x)f_1(x) + g_2(x)f_2(x) = > gcd(f_1(x), f_2(x))$.

And the uniqueness of minimal polynomial theorem is:

For a square matrix $A$, if $f_1(A) = f_2(A) = 0$ for two polynomials $f_1(x)$ and $f_2(x)$, then $g(A) = 0$ for $g(x) = gcd(f_1(x), > f_2(x))$. So the minimal polynomial of $A$ is unique up to a scalar.

So since $f(x)$ is already the minimal polynomial, I know that the $deg(b(x))=deg(f(x))$. I know how to prove the Bezout identity using the Euclidean algorithm, I'm just not exactly sure how I would show all of this using the notation he gave us of $b(x)=gcd((f(x),g(x))$. Any help at all would be appreciated.

Sofia June
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5 Answers5

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To me it's unclear what you are actually asking, but if you have defined $f$ to be the monic polynomial of minimal degree such that $f(A) = 0$, then you simply argue like this. Take any $g$ such that $g(A) = 0$. Divide $g$ by $f$, so $$ g = q f + r, \qquad\text{where either $r = 0$, or $r \ne 0$ has degree smaller than that of $f$}. $$ Evaluate in $A$ to get $$ 0 = g(A) = q(A) f(A) + r(A) = q(A) \cdot 0 + r(A) = r(A). $$ Now if $r \ne 0$, you get a contradiction to the fact that $f$ has minimal degree among the polynomials vanishing on $A$.

Andreas Caranti
  • 68,873
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Hint $\ $ The set $\,S\,$ of all $f$ such that $\,f(A)=0\ $ is $ $ closed under mod (remainder), $ $ since if $\,f(A)=0=g(A)\,$ then so too for $\,g\ {\rm mod}\ f = g - qf.\,$ So if $\,0\ne f\in S\,$ has minimal degree then $\,f\,$ divides every $\,g\in S\,$ (else $\,0\neq g\ {\rm mod}\ f \in S\,$ has degree $\,< \deg f).\,$

So if $\,f,\,\bar f\in S\,$ both have min degree then $\,f\mid \bar f\mid f,\ $ so $\,\bar f = cf\,$ for a constant $\,c\ne 0$.

Remark $\ $ The above descent uses the same idea as the classical Euclidean algorithm: $ $ namely, if $\,a\nmid b\,$ then we can obtain a smaller element $\,b\ {\rm mod}\ a = b-qa\in a\Bbb Z + b\Bbb Z.\,$ In every domain $\,D\,$ enjoying (Euclidean) division with smaller remainder this works to show that every ideal $\,I\ne 0\,$ is principal $\,I = aD\,$ generated by any "smallest" element $\,0\neq a\in I,\,$ i.e. Euclidean domains are PIDs. In $\,\Bbb Z\,$ we measure smallness by magnitude $\,|n|;\,$ in $\,F[x]\,$ we use $\,\deg f.$

The idea generalizes to any PID (Dedekind-Hasse criterion): a domain $\rm\,D\,$ is a PID iff given any nonzero $\rm\:a, b \in D,\:$ either $\rm\:a\:|\:b\:$ or some D-linear combination $\rm\:a\,d+b\,c\:$ is "smaller" than $\rm\,a.\,$

Community
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Bill Dubuque
  • 272,048
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One way to approach this problem is using the Euclidean algorithm applied to polynomials: since $f\,\mid\,g$ you can write $g(x) = q(x)f(x) + r(x)$ where $r$ is the zero polynomial or $\textrm{deg}r < \textrm{deg}f$. But then \begin{equation*} r(x) = g(x) - q(x)f(x)\quad\Rightarrow\quad r(A) = g(A) - q(A)f(A) = 0. \end{equation*} By minimality of $f$, $r$ must be the zero polynomial.

rogerl
  • 22,399
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Let A be the matrix of the operator $\varphi$. Let $\nu_A:C[X]\rightarrow M_n$ be the evaluation function $P(x)\mapsto P(A)$.

We know from algebra that $\ker(\nu_A) \trianglelefteq C[x]$. Since $C[X]$ is a PID (Principal ideal domain), every ideal is generated by an element of $C[X]$. This means that $\ker(\nu_a)=(\lambda_A)$. If we define the minimal polynomial to be $\lambda_A$, this becomes trivial, because every polynomial that is annihilated in $A$ must be a multiple of $\lambda_A$.

Lonidard
  • 4,253
0

(*)The mapping $\psi : F[x] \to L(V) = \lbrace A: V \to V \ | \ A $ linear$\rbrace$ defined by $\psi(f) = f(A)$ is a ring homomorphism.

Pf:

$\psi(f+g)= (f+g)(A) = f(A) + g(A) = \psi (f) + \psi (g)$ and

$\psi(f \cdot g) = (f \cdot g)(A) = f(A)g(A) = \psi(f)\psi(g)$

$\psi(1) = 1$

(**)$Ker(\psi)= \lbrace f \in F[x] \ | \ \psi(f) = f(A) = 0\rbrace $ is an ideal in $F[x]$

Pf:

$\forall f,g \in Ker(\psi), h \in F[x],$

$\psi(f+g) = f(A)+g(A) = 0 \implies (f+g) \in Ker(\psi)$ and

$\psi(hf) = h(A)f(A) = h(A)0 = 0 \implies (hf) \in Ker(\psi)$

Now, (***) $F[x]$ is a principal ideal domain:

Pf:

Suppose $I$ is a non-trivial ideal in $F[x]$. Let $g$ be a polynomial of minimal degree in $I$.

By Euclidean Div. Alg. $\forall f \in I, \exists q,r \in F[x]$ so that

$f = qg+r$ with $0 \leq \deg(r) < deg(g)$.

Then $f-qg=r \in I$ which contradicts $g$ having minimal degree in $I$ if $r$ is not the 0 polynomial.

So $r=0 \implies f = qg$, and $f \in I$ was arbitrary so we have that $I = gF[x]$.

$I$ an arbitrary ideal so every ideal in $F[x]$ is principal $\implies$ $F[x]$ a PID.

Thus we know that $Ker( \psi )$ must be generated by an element $\phi \in Ker(\psi)$ with minimal degree. $\phi$ is defined as the minimal polynomial of $A$.

In other words: $Ker(\psi) = \lbrace f \cdot \phi \ | \ f \in F[x] \rbrace$

So for any $g \in Ker(\psi), g = f \cdot \phi$ for some $f \in F[x]$ which shows that $\phi | g$.

ie. the minimal polynomial ($\phi$) divides any other polynomial $g$ for which $g(A) = 0$