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I was studying minimal polynomials in a Linear Algebra course. I am using the book Linear Algebra by Stephen H Friedberg, Insel, and Spence for this purpose.

I was doing a problem but while reading a solution for the same, I came across a concept about which I need a little clarification.

Suppose $A$ is a $6\times 6$ real matrix and $(x-1)^2(x+1)^2(x^2+x+1)$ is the characteristic polynomial of $A.$ If $p(t)$ is a minimal polynomial of $A$ then the possibilities of $p(t)$ are as follows:

  • $(x+1)(x-1)(x^2+x+1)$
  • $(x+1)^2(x-1)(x^2+x+1)$
  • $(x+1)(x-1)^2(x^2+x+1)$
  • $(x-1)^2(x+1)^2(x^2+x+1)$

In all these possibilities I see, $(x^2+x+1)$ is present as a factor of $p(x).$

I assume this is because of the theorem:

Let $T$ be a linear operator on a finite dimensional vector space $V$ over a field $F.$ If $p(t)$ is the minimal polynomial of $T$ then a scalar $\lambda$ is an eigenvalue of $T$ iff $\lambda$ is a root of $p(t).$ This means that the minimal polynomial and characteristic polynomial have the same set of roots in $F.$

Here's my question: If $A$ is a real matrix as mentioned above then the eigenvalues of $A$ must all be real if they exist. So, the eigenvalues of $A$ are precisely $1$ and $-1$ respectively. This means $1$ and $-1$ must also be a root of the minimal polynomial of $A$ which is, $p(t).$ So, $(x^2-1)|p(t)$ by the application of the theorem above. However, there is no reason to conclude from this theorem that $(x^2+x+1)|p(x)$ since the theorem only states that every eigenvalue of $A$ is a root of the minimal polynomial and vice versa, i.e the characteristic and minimal polynomials have the same set of roots over the field $\Bbb R.$ But $x^2+x+1$ has no real root and so the roots of $x^2+x+1$ are not eigenvalues of $A$ which implies they are also not a root of $p(x).$

This is the reason why, I do not get why are $(x^2+x+1)$ included as a factor in every possibility of $p(t)?$ Any help regarding this will be greatly appreciated.

A rural reader
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Thomas Finley
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    the roots of $x^2 + x + 1$ are distinct, each must be a root of the minimal p. So $(x-r_1),(x-r_2)$ both divide minimal p. Finally, they have GCD 1 in the Euclidean ring $F(x)$ where $F$ is the complexes. Then the product $(x-r_1)(x-r_2) = x^2 + x + 1$ divides the minimal polynomial – Will Jagy Jun 07 '24 at 16:32
  • @WillJagy Sorry, but can you please elaborate a bit? Why each of the roots of $(x^2+x+1)$ must be a root of $p(x)$? – Thomas Finley Jun 07 '24 at 16:32
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    $x^2+x+1$ has complex roots so it is irreducible over the real numbers. You can also check that $x^2-1$ is a factor in all these options a well but they decomposed it into $(x-1)(x+1)$ since it has real roots. – CyclotomicField Jun 07 '24 at 16:51
  • @CyclotomicField Yes and? – Thomas Finley Jun 07 '24 at 17:08
  • @ThomasFinley since the characteristic polynomial and minimal polynomial have the same roots they all have the same factor of $x^2+x+1$. This is just like how in $\mathbb{Z}$ that $5$ is prime but in $\mathbb{Z}[i]$ it's not because $5=(2+i)(2-i)$. When we're discussing $\mathbb{Z}$ you would just leave it as $5$ rather than decomposing it into its factors just like how for a matrix in $\mathbb{R}$ we don't break $x^2+x+1$ into its roots over $\mathbb{C}$. – CyclotomicField Jun 07 '24 at 17:33
  • @CyclotomicField "since the characteristic polynomial and minimal polynomial have the same roots..." If you are saying this statement using the theorem I mentioned in the post, then it seems that they both have same roots over the field given in the problem. In case it was $\Bbb R.$ But that doesn't necessarily mean $(x^2+x+1)$ is root of $p(x)$. – Thomas Finley Jun 07 '24 at 17:37
  • I don't understand why you ask about the factor $x^2+x+1$. The other two factors ($x+1$ and $x-1$) also are included in every possibility of $p(t)$. – jjagmath Jun 07 '24 at 18:18
  • @ThomasFinley You're reversed the implication. It's true that the minimal and characteristic polynomials will have the same roots over the base field but they also have the same roots over the field's algebraic completion. The real roots are a subset of the eigenvalues but it's not all of them. – CyclotomicField Jun 07 '24 at 23:16
  • @CyclotomicField The real roots are the only eigenvalues and they are all of them, if the field is $\Bbb R.$ – Thomas Finley Jun 08 '24 at 03:26
  • @jjagmath That's because $(x+1)$ and $(x-1)$ have real roots and so their roots forms the full set of eigenvalues of $A$ over $\Bbb R$ but this is not the case $x^2+x+1$ – Thomas Finley Jun 08 '24 at 03:28
  • @ThomasFinley that's not true. Consider the matrix $\begin{bmatrix}0 & 1 \ -1 & 0\end{bmatrix}$ which has characteristic equation $x^2+1$ so the eigenvalues are $\pm i$ Even real matrices can have complex eigenvalues. Any rotation matrix will have complex eigenvalues. – CyclotomicField Jun 08 '24 at 03:30
  • @CyclotomicField Wait, but you never specified with which field are we working upon in that example. If it's a complex field of course the eigenvalues exist and they are the ones you mention. But if it's $\Bbb R$ upon which we are working on, then eigenvalues of the matrix doesn't exist. Eigenvalues of matrix defined over a field $ F$ are (by definition of eigenvalues) from the field $F$ as well – Thomas Finley Jun 08 '24 at 03:33
  • @ThomasFinley the eigenvalues do exist, they're complex numbers. It's the main motivations for using the complex numbers as eigenvalues are extremely useful in applications. Like they're pervasive in quantum mechanics because wave equation and rotations are intimately linked. – CyclotomicField Jun 08 '24 at 03:35
  • @CyclotomicField Then I think our definitions do not match. – Thomas Finley Jun 08 '24 at 03:35
  • @ThomasFinley I'm using the standard definition. You'll find that they satisfy exactly the same equations that the real eigenvalues do. Review the proof and convince yourself it applies to all roots, even if they're not in the base field. – CyclotomicField Jun 08 '24 at 03:37
  • @CyclotomicField My definition for eigenvalues: Let $A\in M_{n\times n}(F)$. If $\exists v(\neq 0)\in F^n$ such that $Av=\lambda v$ for some $\lambda\in F$ then, $\lambda$ is called an eigenvalue of $A$ corresponding to the eigenvector $v.$ – Thomas Finley Jun 08 '24 at 03:40
  • If you're using the wrong definition I can under why it's caused confusion. The eigenvalues should be in the algebraic completion of $F$, not in $F$. – CyclotomicField Jun 08 '24 at 04:06
  • @CyclotomicField But this is the definition that's given in Friedberg! – Thomas Finley Jun 08 '24 at 04:15
  • @ThomasFinley then it's probably an error. It would be highly unusual for them to use a nonstandard definition. – CyclotomicField Jun 08 '24 at 04:46

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The minimal polynomial over $\mathbb{R}$ and the minimal polynomial over $\mathbb{C}$ coincide for real matrices. This is because taking the real and imaginary parts of the minimal polynomial over $\mathbb{C}$ gives two real polynomials which annihilate the matrix. So they are both divisible by the real minimal polynomial. And therefore the real minimal polynomial divides the complex minimal polynomial. On the other hand, the real minimal polynomial is a complex polynomial annihilating the matrix, so it is divisible by the complex minimal polynomial. Two monic polynomials which divide eachother are equal.

So, you use the theorem in your post applied with the field $\mathbb{C}$ to show that the roots of $x^2+x+1$ are roots of the minimal polynomial, and therefore $x^2+x+1$ is a factor.

Joshua Tilley
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  • Thank you for your suggestion! However, I didn't get what you meant by "real and imaginary parts of the minimal polynomial over $\Bbb C$" in your answer. – Thomas Finley Jun 11 '24 at 08:51
  • By the real part of a polynomial, I mean a polynomial whose coefficients are the real part of the original polynomial's coefficients. – Joshua Tilley Jun 11 '24 at 11:53
  • One can say more generally that the minimal polynomial of a matrix cannot change under field extensions, because of the more basic fact that a set of linearly independent vectors over $F$ in some space $F^n$ remain independent when $F$ is embedded into a larger field: either argue that some non-vanishing minor ensures the independence or that Gaussian elimination always detects existing dependencies. Because the minimal polynomial of $A$ by definition exhibits the first linear dependence between powers of $A$. – Marc van Leeuwen Jun 13 '24 at 11:19