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I've seen in my linear algebra textbook that one can prove that the irreducible factors of a characteristic polynomial and minimal polynomial are the same using Primary Decomposition Theorem, but I have no idea how this happens.

So the fact that irreducible factors of a minimal polynomial are irreducible factors of a characteristic polynomial is trivial due to Cayley-Hamilton Theorem, however the converse does not seem so easy. I think I found a solution using field extensions and eigen values, but I want to try to prove this using Primary Decompostion Theorem.

I've found Showing that minimal polynomial has the same irreducible factors as characteristic polynomial this question, it says,

Let the minimal polynomial of a linear transfomation $T:V \to V$ over field F be $m = {f_1}^{m_1} ... {f_n}^{m_n}$. Now restrict $T$ to $ker({f_i}^{m_i})$, and use the Cayley-Hamilton Theorem, and the fact that $f_i$ is irreducible to prove the statement.

How ever I don't see how this happens, if I let $W_i = ker({f_i}^{m_i})$, I can show that the minimal polynomial of $T|_{W_i}$ is ${f_i}^{m_i}$ but using Cayley-Hamilton only gives that the characteristic polynomial of $T|_{W_i}$is a multiple of ${f_i}^{m_i}$, and I don't think this really helps.

I'm thinking that I have to use some unique properties of a characteristic polynomial other than the fact that it's an annihilator, since not every annihilators have the same irreuducible factors as minimal polynomials. Can you help me with this?

+) Okay so to be exact, there's a Lemma on my Textbook:

Let $V$ be a vector space over field $F$, and suppose $T$ is a linear transformation from vector space $V$ to $V$, and let $f_1(t), ...f_k(t)$ be mutually relatively prime monic polynomials over field $F$. if $f_1(t)f_2(t)...f_k(t)$ is an annihilator of $T$, $V$ can be decomposed into direct sums of $kerf_i(T)$.

and there's a exercise that says to prove the following using the Lemma above:

Let $V$ be a vector space over field $F$, and suppose $T$ is a linear transformation from $V$ to $V$, then the 'set of monic irreducible divisor over field F' is the same for the characteristic polynomial of $T$ and the minimal polynomial of $T$.

I have no knowledge about modules, and I know characteristic polynomial as $det(tI-T)$ and minimal polynomial as the minimal annihilator of T.

This isn't my homework or anything, I just got stuck while studying with this textbook on myself :(( Any help?

Ieung
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  • You are right that you need something about the char poly. I think that you can adapt the argument of this question, checking out that it is impossible for any of the $m_i$ to be $0$. https://math.stackexchange.com/questions/369630/primary-decomposition-theorem-what-are-the-characteristic-polynomials-of-the-ma?rq=1 – ancient mathematician Apr 16 '18 at 06:50
  • I find the same problem with the answer below; isn't it possible for $W_i$, or $ker(T_i)$ to be {0}? And I don't think minimal polynomial can be properly defined if we restrict it to {0} – Ieung Apr 18 '18 at 02:32
  • The minimal polynomial of $S$ is a monic generator of the ideal of polynomials $f$ with $f(S)=0$, so the min ploy of the only linear transformation from ${0}$ to itself is $1$. – ancient mathematician Apr 18 '18 at 06:56
  • I have given two answers to the question you linked to. Why did you only consider the (unclear) argument given in the question, but nothing that was mentioned in its answers? – Marc van Leeuwen Apr 22 '18 at 19:23
  • By the way, my proof using invariant factors is probably probably what you are at with using the primary decomposition theorem. See here for how the primary factors refine the invariant factor decomposition, but there is no need to factor into irreducibles since just proving that the characteristic polynomial divides a power of the minimal polynomial suffices. – Marc van Leeuwen Apr 22 '18 at 19:50
  • Well... first of all the question I linked was asking for a solution that does not use primary decomposition theorem, and I was looking for a solution that does use the theorem, so I didn't take a careful look into the answers... Sorry for that. And also even after I read your answers and the wikipedia link you gave me, I've just learned only brief linear algebra, and don't know much about field theory or modules, So I'm having difficulties understanding why this is so :( – Ieung Apr 24 '18 at 00:55
  • My textbook said that I can see this statement as a result of primary decomposition theorem, so I was hoping for a simple logic to prove this, with the primary decomposition theorem I know, which is decomposing a vector space into kernels of polynomials, or T-invariant spaces. I guess I'll have to wait until I get a more general understanding of algebra :( – Ieung Apr 24 '18 at 01:17

1 Answers1

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You did the hardest step.

Assume that the characteristic polynomial of $T_{W_i}$ is $f_i^{t_i}g_i$ where $g_i$ is prime to $f_i$.

Then $W_i=\ker(f_i^{t_i}(T))\oplus \ker(g_i(T))$. It is easy to see (for example by writing down a representative matrix in a basis adapted to this direct sum), that the minimal polynomial of $T_{W_i}$ is the lcm of the minimal polynomial of $T_{\ker(f_i^{t_i}(T))}$ and of the minimal polynomial of $T_{\ker(g_i(T))}$.

Thie first one is a power of $f_i$. The second divides $g_i$, so is prime to $f_i$. Hence their lcm is the product of the two.

We then get that the minimal polynomial $T_{W_i}$ is NOT a power of $f_i$, unless $g_i=1$, and this concludes your proof.

To finish, I would like to give you an alternative approach to prove the theorem you are interested in, which only use polynomials.

Let $u:E\to E$ be an endomorphism, where $E$ has dimension $n$, with minimal polynomial $\mu_u$ and characteristic polynomial $\chi_u$. Then $\mu_u\mid \chi_u\mid \mu_u^n$.

Caylet-Hamilton implies $\mu_u\mid\chi_u$.

Now notice that for all $m\geq 1,$ we have $$ X^m-Y^m =(X-Y)(X^{m-1}+X^{m-2}Y+\cdots+XY^{m-2}+Y^{m-1})\in K(X)[Y].$$ If $d$ is the degree of $\mu_u,$ we deduce easily that there exist polynomials $P_0,\ldots,P_{d-1}\in K[X]$ such that
$$\mu_u(X)-\mu_u(Y)=(X-Y)(P_0+P_1Y+\cdots +P_{d-1}Y^{d-1})\in K(X)[Y].$$

Let $M$ be the representative matrix of $u$ in a fixed basis of $E.$ Then$$\mu_M(M)=0\in M_n(K)\subset M_n\big(K(X)\big).$$ Considering $M$ as an element of $M_n\big(K(X)\big)$, this implies $$\mu_u(X)I_n=(XI_n-M)Q\in M_n\big(K(X)\big),$$ où $Q=(P_0I_n+P_1M+\cdots +P_{d-1}M^{d-1})\in M_n\big(K(X)\big),$ since $\mu_M(M)=0$. Taking determinants , we get $$\mu_u^n=\chi_u \det(Q) \in K(X).$$ Notice now that $Q\in M_n(K[X]).$ Thus $\det(Q)\in K[X],$ hence the result.

Now the divisibilty relations $\mu_u\mid\chi_u\mid \mu_u^n$ easily implies that $\mu_u$ and $\chi_u$ have the same irreducible factors.

GreginGre
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    The second proof looks very cool, proving the theorem without any advanced theory. However I have some questions with the first proof; what if ker(gi) = {0}? Doesn't that make gi not necessarily 1? – Ieung Apr 18 '18 at 01:32
  • Sorry it's a misprint. You have to read $\ker(f_i^{t_i}(T))\oplus\ker(g_i(T))$ And if the second kernel is trivial, what you can say is that $g_i(T)$ is injective. This means that $g_i$ is prime to the minimal polynomial of $T_{W_i}$if I remember well ,even it needs a proof, which is more or less as long as the proof I gave you... – GreginGre Apr 18 '18 at 08:34
  • Your first proof misses the main point, namely to show that $f_i$ is a factor of the minimal polynomial. To that end you need to show that the power of it that arises is not the power $0$, which is equivalent to showing that $f_i[T]$ cannot be injective (whenever $f_i$ is a factor of $\chi_T$, i.e., when $t_i>0$). – Marc van Leeuwen Apr 22 '18 at 19:38
  • How can I prove that the fact $g_i(T)$ is injective implies $g_i$ is prime to the minimal polynomial? I've came up to this point but failed to prove this :( – Ieung Apr 24 '18 at 01:30