If $A$ is a linear transformation with minimal polynomial $p(x)$, and if $q(x)$ is a polynomial such that $q(A)=0$, then show that $q$ is divisible by $p$.
120619
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 $g_1(x)f_1(x) + g_2(x)f_2(x) = &gt; gcd(f_1(x), f_2(x))$.   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 $g(x) = gcd(f_1(x), &gt; f_2(x))$. 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.
48431
Let $P(z)$ be a complex non-constant polynomial. Let $G$ be a connected component of open set $\{z : |p(z)|&lt;1 \}$. How to prove that $G$ contains a zero of $P$?   I have no idea how to even start; maybe I don't know the required theorem? Can anybody help?
40285
I'm trying to show that the polynomial $X^n -2$ ($n \in \mathbb{N}$) is irreducible in $\mathbb{Q}[X]$ but am a bit stuck. Methods I know to show irreducibility:   Gauss' lemma - which says that if I can show it is irreducible in $\mathbb{Z}[X]$ then it will be irreducible in $\mathbb{Q}[X]$. This would allow me to check reduction modulo a prime.    However this doesn't work in this case, because if I reduce mod 2, then the polynomial is just $X^n$ which is reducible.  I'm also aware of Eisensteins criterion where I can check that if a prime divides all the coefficients but its square doesn't divide the constant coefficient, then it is irreducible.  None of these methods are working for this polynomial so help would be much appreciated! Also are there any general methods to look out for when trying to show polynomials are irreducible?  Thanks
26512
Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'.
16355
$P,Q,R,S$ are polynomials such that: $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$   I thought a lot on this but no result!! By the way,one idea is to insert some values for $x$ and try to produce a system of equations for the given polynomials,but I'm not sure it works.
95679
When a polynomial   $$P(x)=x^4- 6x^3 +16x^2 -25x + 10$$  is divided by another polynomial   $$Q(x)=x^2 - 2x +k,$$  then the remainder is   $$x+a.$$  I have to find the values of $a$ and $k$.   Can somebody tell me shortest way to get these values? Which theorem should be applied here?
75159
Question:  If $p(x) \in F[x]$ is of degree $3$, and $p(x)=a_0+a_1x+a_2x^2+a_3x^3$, show that $p(x)$ is irreducible over $F$ if there is no element $r\in F$ such that $a_0+a_1r+a_2r^2+a_3r^3 =0$.  My understanding: If $p(x)$ is reducible, then there exists $ax + b$ such that $a, b \in F$ and $a\ne 0$. And $p(x) = (ax + b)(cx^2 + dx + e)$. Then an $r$ exists such that $p(r) = 0$. Is that how you're supposed to tackle it? I understand that I am supposed to first explain the argument for a polynomial of degree 1, but I do not understand how to go about that, and why that helps.
50384
If $K$ and $k$ are fields, $K\supset k$ is a field extension and $\alpha \in K$ is algebraic over $k$, then we denote by $k[\alpha]$ the set of elements of $K$ which can be obtained as polynomial expressions of $\alpha$.   $$k[\alpha] = \left\{P(\alpha): \quad P\in k[X]  \right\} $$  Also, we denote by $k(\alpha)$ the smallest subfield of $K$ containing both $k$ and $\{\alpha\}$. This is easily seen to be equal to the set of fractions of polynomial expressions in $\alpha$ (just thinking about the way the "smallest" subfield must be generated):  $$k(\alpha) = \left\{\dfrac{P(\alpha)}{Q(\alpha)}: \quad P,Q\in k[X] \text{  and } Q(\alpha)\neq 0 \right\} $$  If $M$ is the minimal polynomial of $\alpha$, it is easily shown that $k[\alpha]\simeq k[X]/\langle M\rangle$ is a field  and therefore a subfield of $k(\alpha)$ containing both $k$ and $\{\alpha\}$. Therefore $k(\alpha)=k[\alpha]$, because $k(\alpha)$ is the smallest.   It follows that all quotients $\dfrac{P(\alpha)}{Q(\alpha)}$ are equal to $H(\alpha)$ for some polynomial $H\in k[X]$.   What I want to see is a more direct proof of this fact. Given $P$ and $Q$, how do you produce this $H$?
23947
I have come across a question about determining possible minimal polynomials for a rank one linear operator and I am wondering if I am using the correct proof method.  I think that the facts needed to solve this problem come from the section on Nilpotent operators from Hoffman and Kunze's "Linear Algebra".     Question: Let $V$ be a vector space of dimension $n$ over the field $F$ and consider a linear operator $T$ on $V$ such that $\mathrm{rank}(T) = 1$.  List all possible minimal polynomials for $T$.   Sketch of Proof:   If $T$ is nilpotent then the minimal polynomial of $T$ is $x^k$ for some $k\leq n$.  So suppose $T$ is not nilpotent, then we can argue that $T$ is diagonalizable based on the fact that $T$ must have one nonzero eigenvalue otherwise it would be nilpotent (I am leaving details of the proof of diagonalization but it is the observation that the characteristic space of the nonzero eigenvalue is the range of T and has dimension $1$).  Thus the minimal polynomial of $T$ is just a linear term $x-c$.    Did I make a mistake assuming that $T$ can have only one nonzero eigenvalue?  Thanks for your help
120366
this is a really easy question but I cant find an answer;  In need to see if $x^4+x^2+x+1$ is an irreducible polynomial over $\mathbb{Q}$
9127
Let $p(x)$ be a polynomial of degree $n$ satisfying $p(x)\geq 0$ for all $x$. That is, for all $x$, $p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \geq 0$, $a_n\neq 0$.  Show that $p(x)+p&#39;(x)+p&#39;&#39;(x)+\cdots+p^{(n)}(x)\geq 0$ for all $x$ where $p^{(i)}(x)$ is the $i^\text{th}$  derivative.  My interest: I know that, we can rewrite the LDE as follows  $$ p(x)+p&#39;(x)+p&#39;&#39;(x)+\cdots+p^{(n)}(x) = Lp(x) $$  where $L := I + D + D^2  + \cdots + D^{(n)}$. Can we say anything about a linear operator of this kind so that it does not change the sign of the input it takes? I can try to solve the question by writing for all the derivatives and factoring them using $p(x)$, but I think there should be a clever way of showing this by the properties of $L$. Can I figure out a solution by just looking at $L$ and the sign of $p(x)$ as in the question? Where should I look for that?
46247
I saw it's been answered before but I could not understand the answers as it delved into material that I did not study yet.  My question is as follows:  Suppose $p \in F[x]$ is the minimal polynomial of some matrix $A \in M_n(F)$ and $F \subseteq K$  I need to show that $p$ is also the minimal polynomial of $A$ over the field $K$. In other words: Show that the minimal polynomial remains the same when we move to a bigger field.  What I thought of:  Obviously $p$ is irreducible in $F$, suppose it is reducible in $K$, which means there are $q,t \in K, q,t \notin F$ such that $p=qt$, now we just need to show that $q(A) \neq 0$ and $t(A) \neq 0$...But why is that so?  Edit: Another way of presenting this question is "Show that if $p$ is the minimal polynomial of $A$ over the field $K$ while the entries of$~A$ are contained in a subfield $F \subseteq K$, then all the coefficients of $p$ are elements of $F$".
2916
Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$.    Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.
85504
Let $n\in\mathbb{N}$ and $A\in M_n(\mathbb{F})$. Assume that the minimal polynomial of the square matrix $A$ is defined as the minimal polynomial of the linear operator $x\mapsto Ax$ for all $x\in\mathbb{F}^n$, on $\mathbb{F}^n$. Now, assume that the minimal polynomial of $A$ splits over $\mathbb{F}$. Show that $A$ and $A^T$ are similar.  Two matrices $A$ and $B$ similar if there exists an invertible matrix $S$ such that $B=SAS^{-1}$. Finding such an $S$ for $B=A^T$ will solve the problem. Since the minimal polynomial of $A$ splits over $\mathbb{F}$, let us denote it by $$m_A(t) = (t-\lambda_1)^{m_1}...(t-\lambda_k)^{m_k}$$ for some $k\in\mathbb{N}$. We know that $m_A(A)$ corresponds to the zero linear operator, i.e. $m_A(A)x=0$, $\forall x\in \mathbb{F}^n$. I quickly noticed that if matrices $P_1,P_2,...P_q$ commute, then $(P_1P_2...P_k)^T = P_1P_2...P_k$. It may be useful to replace each $P_i$ with $(A-\lambda_i)^{m_i}$? I'm not sure how to proceed from here. I'd appreciate any hints!
102659
Given $A\in M_{n \times n}(\mathbb{F})$ and $p_{A}(x)=\det(xI-A)$ why saying that $\det(AI-A)=0$ is not valid?
123230
I am looking at   and trying understand the proof for the theorem that states:  Given a nonzero ideal I in $P(F)$, there is a monic polynomial $p(t)$ such that   $$I=\left\{q(t)p(t) | q(t) \in P(F) \right\}$$  Proof goes like  Since $I$ contains at least one nonzero element, we can let $p(t)$ be a nonzero element of $I$ of minimum degree. We can show that every element of $I$ is a multiple of $p(t)$. Let $f(t) \in I$. Then there exist $q(t),r(t) \in P(F)$ such that  $$f(t) = q(t)p(t) + r(t)$$ and deg $r(t)$ &lt; deg $p(t)$. Since $p(t) \in I$, so is $q(t)p(t)$, and therefore so is $$r(t) = f(t) -q(t)p(t)$$.  By the choice of $p(t),r(t) = 0$, so $f(t)$ is a multiple of $p(t)$.  What I am having trouble understanding is why $r(t)$ must equal $0$. If someone can explain why it is, I'd appreciate it.
92713
Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots. EDIT: ( Over $\Bbb C $ ) though it is obvious i am working over $\Bbb C $ as one of my statements is not true over $ \Bbb R $   I would like a better proof of this result what i did is below, there is the same question on here somewhere but only has an answer to one direction im looking for both.  I proved the result by using a different equivalent statement to diagonalizable looking for any complete proof that is shorter.  We notice that it is equivalent to prove that V has a basis consisting of eigenvectors of T iff the minimal polynomial of T has no repeated roots by theorem.  $(\Rightarrow ) $ First suppose that there is a basis $\beta = (v_1,\cdots , v_n ) $ consisting of eigenvectors of T. let $\lambda_1 , \cdots , \lambda_m $ be distinct eigenvalues of T. Then for each $ v_i $ there exists a $\lambda_k $ with $(T- \lambda_k I) v_i =0 $ it then follows that $(T- \lambda_1 I) \cdots (T- \lambda_mI) v_i =0 $ for each i as we can commute the operators. Since an operator that sends each vector in a basis to the $0$ vector is the $0$ operator we have that $(T- \lambda_1 I) \cdots (T- \lambda_mI) =0 $ Thus the polynomial $(z-\lambda_1) \cdots (z-\lambda_m ) $ when applied to T gives 0. but by theorem we know that that the minimal polynomial of T is a divisor of $(z-\lambda_1) \cdots (z-\lambda_m ) $ which has no repeated roots so the minimal polynomial cannot possibly have repeated roots the result follows.  $(\Leftarrow ) $ Let us assume that the minimal polynomial has no repeated roots; if we let $ \lambda_1 \cdots \lambda_m $ denote the distinct eigenvalues of T, this means the minimal polynomial of T is $(z-\lambda_1) \cdots (z-\lambda_m ) $ It follows that $(T- \lambda_1 I) \cdots (T- \lambda_mI) =0 $ Let $U_m $ be the subspace of a generalized eigenvectors corresponding to the eigenvalue $\lambda_m $. Since $ U_m $ is invariant under T by theorem we consider $ v\in U_m $ let $u= (T- \lambda_m I) v $ it follows that $u\in U_m $  Hence $$ (T|_{U_m} - \lambda_1 I ) \cdots (T|_{U_m} - \lambda_{m-1}I) u =   (T- \lambda_1 I) \cdots (T- \lambda_mI) v =0 $$  by theorem we have that $( T- \lambda_m I )|_{U_m} $ is nilpotent by previous question we have that 0 is the only eigenvalue of $( T- \lambda_m I )|_{U_m} $. Thus $T|_{U_m} - \lambda_jI $ is an invertable operator on $U_m $ for $j= 1, \cdots , m-1 $ it then follows by $$ (T|_{U_m} - \lambda_1 I ) \cdots (T|_{U_m} - \lambda_{m-1}I) u =   (T- \lambda_1 I) \cdots (T- \lambda_mI) v =0 $$ that $u=0$ in other words, $v$ is an eigenvector of T!   We have shown that every generalized eigenvector of T corresponding to the eigenvalue $\lambda_m $ is an eigenvector of T. However we choose $ \lambda_m $ arbitrarily we could of just of easily relabeled the eigenvalues so that any of them was called $ \lambda_m $. Therefore we have that every generalized eigenvector of T is actually an eigenvector of T. By theorem we have that there is a basis for V consisting of generalized eigenvectors of T but by above we have that there is a basis of V consisting of eigenvectors of T the desired result.