3
$\begingroup$

Let $T_\mathbb{R}$ be a real $n \times n$ matrix, and let $T_\mathbb{C}$ be the same matrix but considered as a complex one. What is the easiest way to see that $\min_{T_\mathbb{C}} \in \mathbb{C}[t]$ equals the image of $\min_{T_\mathbb{R}} \in \mathbb{R}[t]$ under the tautological map $\mathbb{R}[t] \to \mathbb{C}[t]$?

$\endgroup$
3
  • 1
    $\begingroup$ If $p = \operatorname{Re} \operatorname{min}_{T_{\mathbb{C}}}$, what is $p(T)$? $\endgroup$ – Daniel Fischer Sep 8 '15 at 19:14
  • $\begingroup$ The linked question is not posed in the same terms, but amounts to the same issue: since the minimal polynomial over $\Bbb C$ of a matrix with real entries is the tautological image of the minimal polynomial over $\Bbb R$, it must have real entries. $\endgroup$ – Marc van Leeuwen Sep 9 '15 at 15:13
  • $\begingroup$ And note that (at least) one of the answers shows that this is true in any field extension; it is not necessary to use anything particular about the extension $\Bbb C/\Bbb R$, such as complex conjugation. $\endgroup$ – Marc van Leeuwen Sep 9 '15 at 15:22
0
$\begingroup$

Set $T = T_\mathbb{C}$, and let $p = \min_T$. Without loss of generality, $p$ is monic. Since $\mathbb{C}$ is algebraically closed, it follows that we can decompose$$V = \bigoplus_\lambda V^{(\lambda)}.$$Now, the minimal polynomial is given by$$p(t) = \prod_{\lambda \in \text{Spec}(T)} (t - \lambda)^{m_\lambda},$$where $m_\lambda$ is the smallest integer such that $T^{m_\lambda}$ annihilates $V^{(\lambda)}$.

We now prove that $p$ consists of real coefficients. To do so, it suffices to prove that $m_\lambda = m_{\overline{\lambda}}$ for any $\lambda \in \mathbb{C}$. But $\overline{T} = T$ as $T \in \mathbb{R}^{n \times n}$, and the claim follows immediately upon observing that$$(T - \lambda)^k(v) = 0 \iff (T - \overline{\lambda})^k(\overline{v}) = \overline{0} = 0.$$Hence, the minimal $k$ to kill a $\lambda$-eigenvector is the same as the minimal $k$ to kill a $\overline{\lambda}$-eigenvector.

Alternatively, one can use the observation that if $1, T_\mathbb{R}, \dots, T_\mathbb{R}^{n-1}$ are linearly independent, then so are $1, T_\mathbb{C}, \dots, T_\mathbb{C}^{n-1}$.

$\endgroup$
1
$\begingroup$

Let $p_{\mathbb C}(T)$ be the minimal polynomial of $A$ over $\mathbb C$ ; by definition, this is the monic polynomial of least degree in $\mathbb C[T]$ for which evaluating at $A$ gives zero. Take the basis $\{1,i\}$ for $\mathbb C$ as a real vector space. Let $I$ be the $n \times n$ identity matrix and consider the (left) $\mathrm{Mat}_{n \times n}(\mathbb R)$-module $\mathrm{Mat}_{n \times n}(\mathbb C)$. In this module, the elements $1 \cdot I$ and $i \cdot I$ are $\mathrm{Mat}_{n \times n}(\mathbb R)$-linearly independent (make the check coefficient-wise).

Now write $$ p_{\mathbb C}(T) = T^k + \alpha_1 T^{k-1} + \cdots + \alpha_k. $$ We can write $\alpha_j = \beta_j + i \gamma_j$, from which we can define $$ p_{\mathbb C}^1(T) = T^k + \beta_1 T^{k-1} + \cdots + \beta_k, \quad p_{\mathbb C}^i(T) = \gamma_1 T^{k-1} + \cdots + \gamma_k, \quad \Rightarrow \quad p_{\mathbb C}(T) = p_{\mathbb C}^1(T) + i p_{\mathbb C}^i(T). $$ Plugging in $A$, we see that since $p_{\mathbb C}(A) = 0$, by the linear independence of $1 \cdot I$ and $i \cdot I$, we have $p_{\mathbb C}^1(A) = p_{\mathbb C}^i(A) = 0$. Since $\deg p_{\mathbb C}^i < \deg p_{\mathbb C}$, by definition of the complex minimal polynomial, $p_{\mathbb C}^i = 0$. Therefore $p_{\mathbb C} = p_{\mathbb C}^1 \in \mathbb R[T]$.

I wrote the argument this way because in fact this generalizes : let $K/F$ be a finite field extension and $A \in \mathrm{Mat}_{n \times n}(F)$. Then if $p_K(T)$ (resp. $p_F(T)$) denotes the minimal polynomial of $A$ over $K$ (resp. $F$), we have $p_K(T) = p_F(T)$ ; the proof is the same as above, just replace the basis $\{1,i\}$ of the $\mathbb C/\mathbb R$ case by an $F$-basis of $K$ which contains $1$.

(Note that I didn't use the fact that $K$ is closed, nor that $K/F$ Galois, as in the case of $\mathbb C/\mathbb R$ ; all you need is that the extension is finite to allow the use of linear algebra.)

Hope that helps,

$\endgroup$