4

The question that I have (modified from the original problem) is the following:

Say $ V =\mathbb{C}[x]/(x-2) \oplus \mathbb{C}[x]/(x^2)$. What is the minimal and characteristic polynomial of the corresponding linear transformation $T: V \to V$, where $ T $ is multiplication by $ x $?

I have problem converting the $ \mathbb{C}[x] $-module structure of $V$ back to $T$. In particular, I have trouble understanding or seeing what $ T $ is like, or how does $ V$ look as a $ \mathbb{C} $-vector space. It is very confusing to me.

Any help is appreciated.

user26857
  • 53,190
  • You didn't explicitly say what $T$ does. Presumably, you have $T(p(x),q(x)) = (xp(x),xq(x))$. – Ben Grossmann Mar 29 '17 at 03:59
  • Yes, it is multiplication by $ x $ as per usual. –  Mar 29 '17 at 04:00
  • "as per usual" in what context? – Ben Grossmann Mar 29 '17 at 04:14
  • If $ V $ is a $F[x]$-module, then it determines a linear operator $ T: V \to V $ where $ T $ is the action of $ x $ on $ V $ (Dummit and Foote). –  Mar 29 '17 at 04:16
  • Aha! It's "the usual" in abstract algebra (or at least in Dummit and Foote). You're directing this question to the linear algebra people (going off of your tags), so you should be ready to give that sort of context. – Ben Grossmann Mar 29 '17 at 04:19
  • Noted, I have edited the question. Sorry for the confusion. –  Mar 29 '17 at 04:20
  • 2
    Didn't mean to make such a big deal out of it; obviously, I was able to guess what you meant with my first comment. Still, I think the change is for the better. – Ben Grossmann Mar 29 '17 at 04:23

3 Answers3

4

Note that $V$ is a vector space consisting of polynomials of the form $(p(x),q(x))$ modulo the appropriate relations. So really, the elements look like $(p(x) + (x-2),q(x) + (x^2))$ (I leave off the ideals for convenience). $V$ has the basis $$ \mathcal B = \{(1,0),(0,x),(0,1)\} =: \{v_1,v_2,v_3\} $$ To see that this is really a basis, note that each element of $V$ can be uniquely expressed as $$ (p(x),q(x)) = (a_0, b_1x + b_0) = a_0(1,0) + b_1 (0,x) + b_0(0,1) $$ We find that $$ T(v_1) = x(1,0) = (x,0) = (2,0) = 2v_1\\ T(v_2) = x(0,x) = (0,x^2) = (0,0) = 0\\ T(v_3) = x(0,1) = (0,x) = v_2\\ $$ With that, we find that the matrix representation of $V$ with respect to this basis is $$ \pmatrix{2&0&0\\0&0&1\\0&0&0} $$ Hopefully that helps.

The minimal and characteristic polynomial will both be $x^2(x-2)$.

Ben Grossmann
  • 234,171
  • 12
  • 184
  • 355
  • Thank you. Your answer is very helpful. Just a quick question: can you give a minimal example similar to the problem where the minimal and characteristic polynomials are different? –  Mar 29 '17 at 04:25
  • Sure: any direct sum with a common factor in the ideal will do. For example, take $\Bbb C[x]/(x - 1) \oplus \Bbb C[x]/(x - 1)$. Minimal polynomial is $(x-1)$, characteristic polynomial is $(x-1)^2$. If you prefer, consider $\Bbb C[x]/(x - 1) \oplus \Bbb C[x]/(x(x - 1))$. – Ben Grossmann Mar 29 '17 at 04:28
  • In that case, the min. poly. is $ x(x-1) $ and the char. poly. is $ x(x-1)^2 $? –  Mar 29 '17 at 04:32
  • That's exactly right. – Ben Grossmann Mar 29 '17 at 04:33
  • Nice answer (+1). What is the easiest way to see that the minimal polynomial of the matrix in your answer is $x^2(x-2)$? Here is one way: we know that characteristic polynomial is $x^2(x-2)$. Since minimal polynomial divides characteristic polynomial (Cayley-Hamilton Theorem) and these two polynomials share the same irreducible factors (some general fact), we know that the minimal polynomial is either $x(x-2)$ or $x^2(x-2)$. I guess one can directly check that $x(x-2)$ does not annihilate the matrix, so the minimal polynomial has to be $x^2(x-2)$. Do you have any easier/alternative way? – Prism Mar 29 '17 at 15:01
  • 1
    @Prism For a matrix in Jordan form, the minimal polynomial is $\prod_i(x - \lambda_i)^{d_i}$, where each $d_i$ is the size of the largest Jordan block. Alternatively: if you have a rational canonical form matrix (which I would if I had switched $v_2,v_3$), the minimal polynomial is the LCM of the polynomials corresponding to each diagonal block. – Ben Grossmann Mar 29 '17 at 15:10
3

By the CRT, $\mathbb{C}[x] / (x - 2) \oplus \mathbb{C}[x] / (x^2) \cong \mathbb{C}[x] / (x^2(x - 2))$. The min and char polynomials are therefore $x^2(x - 2)$.

Flowers
  • 805
2

For every summand $\def\C{\Bbb C}\C[x]/P$ with $P\in\C[x]$ monic, the matrix on the basis of (images of) monomials $x^i$ for $0\leq i<\deg P$ of multiplication by (the image of) $x$ is the companion matrix for$~P$. It has minimal and characteristic polynomials both equal to$~P$. All you need to do it take the product of these factors$~P$ for the characteristic polynomial, and their least common multiple for the minimal polynomial. In the example the two polynomials are relatively prime, so it gives the same result in both cases.

Community
  • 1
Marc van Leeuwen
  • 119,547