3

Example 4: In Example $2$, the operator $T$ also had the characteristic polynomial $f=(x-1)(x-2)^2$. But, this $T$ is not diagonalizable, so we don’t know that the minimal polynomial is $(x-1)(x-2)$. What do we know about the minimal polynomial in this case? From Theorem 3 we know that its roots are $1$ and $2$, with some multiplicities allowed. Thus we search for $p$ among polynomials of the form $(x-l)^k(x-2)^l$, $k\geq 1$, $l\geq 1$. Try $(x-1) (x-2)$: $$(A-I)(A-2I)=\begin{bmatrix} 2&0&-1 \\ 2&0&-1 \\ 4&0&-2\\ \end{bmatrix}$$ Thus, the minimal polynomial has degree at least $3$. So, next we should try either $(x-1)^2(x-2)$ or $(x-1)(x-2)^2$. The second, being the characteristic polynomial, would seem a less random choice. One can readily compute that $(A - I)(A - 2I)^2 = 0$. Thus the minimal polynomial for $T$ is its characteristic polynomial.

Que: Why didn’t author consider possibility of minimal polynomial may have irreducible (over $\Bbb{R}$) factor, let say $x^2+1$?

Also we can’t use minimal polynomial divides characteristic polynomial results to confine minimal polynomial possibility to $(x-2)^2$, $(x-1)(x-2)$ and $(x-1)(x-2)^2$, because Cayley-Hamilton theorem is not yet proved in book.

user264745
  • 4,143

2 Answers2

1

Because the roots of the minimal polynomial and the characteristic polynomial have the same roots. Since the only roots of the characteristic polynomial are $1$ and $2$, then minimal polynomial can have no other roots. Any irreducible factor would have complex roots.

José Carlos Santos
  • 427,504
  • Thank you so much for the answer. I think, roots of a polynomial function depends on what field we’re working on. In above example, characteristic and minimal polynomial are map from $\Bbb{R}$ into $\Bbb{R}$. So “complex root” are not roots of characteristic and minimal polynomial. In proof of roots of minimal polynomial and characteristic polynomial have same roots, we assume roots in field $F$, not in extension of field $F$. – user264745 Jan 05 '23 at 09:41
  • The assertion “the roots of the minimal polynomial and the characteristic polynomial have the same roots” are valid for any field for which our operator makes sense. So, since the roots of the characteristic polynomial are $1$ and $2$, then the minimal polynomial can have no roots other than $1$ and $2$, even if we are working on $\Bbb Q\left(\sqrt2\right)$, on the field of algebraic numbers, or on $\Bbb C$. – José Carlos Santos Jan 05 '23 at 09:55
  • Let say $i\in \Bbb{C}$ is root of minimal polynomial. $i$ is also root of characteristic polynomial. We know $i$ is root of characteristic polynomial if and only if $i$ is eigenvalue of $T$. So $\exists 0\neq (x,y)\in \Bbb{R}^2$ such that $T(x,y)=i(x,y)\in \Bbb{C}$. Which is absurd since $T(x,y)\in \Bbb{R}$. – user264745 Jan 05 '23 at 10:12
1

The question seems to be about minimal and characteristic polynomials in field extensions.

$\mathbf{Claim:}$ Every irreducible factor of the characteristic polynomial of $T$ is also an irreducible factor of the minimal polynomial of $T$ and vice versa.

Let $k$ be the base field. Choose the splitting field $K$ of the minimal polynomial $p(x)$. We will suppose here that the polynomial has only one irreducible factor $r(x)$ (We can generalize to how many ever we want later). We can say that the irreducible factor $r(x)$ in $K[x]$ will be the product of its Galois conjugates, i.e., $r(x)=(x-c)(x-\sigma_1c) \cdots (x-\sigma_nc)$.

$\mathbf{Fact\ 1:}$ Some $(x-\sigma_ic)$ must divide the characteristic polynomial.

Why? The minimal polynomial of $T$ as a matrix with entries in $K$ must contain at least one of the $(x-\sigma_ic)$ as factor. Otherwise, $q(x)\in k[x]$ would be a minimal polynomial for $T$ as a matrix with entries in $k$.

$\mathbf{Fact\ 2:}$ Since $(x-\sigma_ic)$ divides the characteristic polynomial, $r(x)$ must divide it as the characteristic polynomial is in $k[x]$.

Combining these two facts shows us that the irreducible factors of the minimal polynomial must also be irreducible factors for the characteristic polynomial.

Now, for the other direction, if $\alpha$ is a root of the characteristic polynomial $c(x)$, then it is a factor of the minimal polynomial of $T$ over $K$, which must divide the minimal polynomial of $T$ over $k$. Therefore, every other Galois conjugate of $\alpha$ must also be a root of the minimal polynomial of $T$ over $k$. So, $r(x)$ divides the minimal polynomial of $T$ over $k$.

Umesh Shankar
  • 2,787
  • Yup. I saw “minimal and characteristic polynomials of a linear operator have the same irreducible factors” in this article by Keith Conrad (page 6 & corollary 4.10). – user264745 Jan 05 '23 at 10:24
  • 1
    https://math.stackexchange.com/questions/66443/a-square-matrix-has-the-same-minimal-polynomial-over-its-base-field-as-it-has-ov#66916 – Umesh Shankar Jan 05 '23 at 10:49