Polynomials

Polynomials

$ T \in \mathcal{L}(V) $	$ \ q \in \mathcal{P}(\mathbf{F}) $	Minimal polynomial
$ T/U $	$ q(T/U) = q(T)/U $	$ p_{T/U} \mid p_{T} $
$ T \vert_U $	$ q(T \vert_U) = q(T) \vert_U $	$ p_{T \vert_U} \mid p_{T} $
		$ p_T \mid p_{T \vert_U} \cdot p_{T/U} $
$ \mathbf{F} = \mathbb{R} $, $ T_{\mathbb{C}} $	$ q(T_{\mathbb{C}}) = (q(T))_{\mathbb{C}} $ ($q$ has real coefficients)	$ p_{T_{\mathbb{C}}} = p_T $
$ T’ $	$ q(T’) = (q(T))’ $	$ p_{T’} = p_T $
$ S \in \mathcal{L}(V) $ is invertible, $ STS^{-1} $	$ q(STS^{-1}) = q(T) $	$ p_{STS^{-1}} = p_T $

Complexification

$V_{\mathbb{C}} = V \oplus iV$ with elements $u + iw$ ($u, w \in V$), and $T_{\mathbb{C}}(u + iw) = Tu + iTw$.

Write $p = p_T$ and $q = p_{T_{\mathbb{C}}}$.

Direction 1: $q \mid p$

$p$ has real coefficients and $p(T) = 0$, so $p(T_{\mathbb{C}}) = 0$. Thus $p$ annihilates $T_{\mathbb{C}}$, giving $q \mid p$. In particular $\deg q \le \deg p$.

Direction 2: $p \mid q$

$q$ a priori has complex coefficients, so split $q$ into real and imaginary parts.

Write $q(z) = g(z) + i\,h(z)$, where $g, h \in \mathcal{P}(\mathbb{R})$ are obtained by taking the real and imaginary parts of each coefficient of $q$. Since $g, h$ have real coefficients:

\[0 = q(T_{\mathbb{C}}) = g(T_{\mathbb{C}}) + i\,h(T_{\mathbb{C}}) = (g(T))_{\mathbb{C}} + i\,(h(T))_{\mathbb{C}}.\]

Evaluate at $u + i\cdot 0 = u, \ \forall u \in V$: \((g(T))_{\mathbb{C}}u = g(T)u \in V\) and $i(h(T))_{\mathbb{C}}u = i\,h(T)u \in iV$. These lie in the complementary summands $V$ and $iV$, so both must vanish:

\[g(T)u = 0 \quad\text{and}\quad h(T)u = 0 \quad\text{for all } u,\]

i.e. $g(T) = 0$ and $h(T) = 0$.

Now $q$ is monic, so its leading coefficient is $1 = 1 + i\cdot 0$; hence $g$ is monic of degree $\deg q$, while $\deg h < \deg q$. Since $g(T) = 0$ and $g$ is a monic real annihilator of $T$, minimality gives $p \mid g. \blacksquare$

Bézout identity

Suppose $ p, q \in \mathcal{P}(\mathbb{C}) $ are nonconstant polynomials with no zeros in common. Let $ m = \deg p $ and $ n = \deg q $. There exist \(r \in \mathcal{P}_{n - 1}(\mathbb{C})\) and \(s \in \mathcal{P}_{m - 1}(\mathbb{C})\) such that

\[rp + sq = 1.\]

Define \(T: \mathcal{P}_{n - 1}(\mathbb{C}) \times \mathcal{P}_{m - 1}(\mathbb{C}) \to \mathcal{P}_{m + n - 1}(\mathbb{C})\) by

\(T(r,s) = rp + sq\).

$T$ is a square map:

\[\dim\big(\mathcal{P}_{n-1}(\mathbb{C}) \times \mathcal{P}_{m-1}(\mathbb{C})\big) = n + m,\] \[\dim \mathcal{P}_{m+n-1}(\mathbb{C}) = m+n.\]

Equal. And $T$ is linear: $T(r,s) = rp + sq$ is linear in $(r,s)$ since $p, q$ are fixed.

Next, show $ T $ is injective. Suppose $T(r,s) = 0$, i.e.

\[rp + sq = 0, \qquad\text{so}\qquad rp = -sq. \tag{$\ast$}\]

We want to force $r = 0$ and $s = 0$. The no-common-zeros hypothesis enters here, through unique factorization / the divisibility structure of $\mathbb{C}[z]$.

From $(\ast)$, $q$ divides $rp$. Now list the zeros of $q$: by the FTA, $q$ factors as $q(z) = c\prod_{i}(z - \mu_i)$ over its roots $\mu_i$ (with multiplicity). Each root $\mu_i$ of $q$ is a zero of the left side $rp$, hence a zero of $r$ or of $p$. But $p$ and $q$ share no zeros, so $\mu_i$ is not a zero of $p$ — therefore $\mu_i$ must be a zero of $r$, and by matching multiplicities (a root of $q$ of multiplicity $t$ is not absorbed by $p$ at all, so all $t$ copies must come from $r$), the full factor $q$ divides $r$:

\[q \mid r.\]

But now degrees: $r \in \mathcal{P}_{n-1}(\mathbb{C})$ so $\deg r \le n - 1 < n = \deg q$. The only multiple of $q$ with degree below $\deg q$ is the zero polynomial. Hence

\[r = 0.\]

Plugging back into $(\ast)$: $sq = 0$ with $q \neq 0$, so $s = 0$. Thus $\ker T = {0}$ and $T$ is injective.

It’s easy to show $T$ is invertible since $T$ is an operator.

The constant polynomial $1$ lives in the codomain \(\mathcal{P}_{m+n-1}(\mathbb{C})\) (its degree $0$ is $\le m + n - 1$, using that $p, q$ nonconstant gives $m, n \ge 1$, so $m + n - 1 \ge 1 \ge 0$). By surjectivity from (b), $1$ is hit: there exist \(r \in \mathcal{P}_{n-1}(\mathbb{C})\) and $s \in \mathcal{P}_{m-1}(\mathbb{C})$ with

\[T(r,s) = rp + sq = 1. \qquad\blacksquare\]

“no common zeros” is the $\mathbb{C}[z]$-analogue of “coprime,” and $rp + sq = 1$ is exactly the statement that the gcd is a unit.

Suppose $ p \in \mathcal{P}(\mathbb{C}) $ has degree $ m $.

$ p $ has $ m $ distinct zeros $ \iff p $ and its derivative $ p’ $ have no zeros in common $ \iff $ The greatest common divisor of $ p $ and $ p’ $ is the constant polynomial $1$.

Minimal Polynomial

Suppose $ V $ is finite-dimensional and $ T \in \mathcal{L}(V) $. The minimal polynomial of $ T $ is the unique monic polynomial $ p \in \mathcal{P}(\mathbf{F}) $ of smallest degree such that $ p(T) = 0 $.

$ \deg p \le \dim V $.

Computation ($O((\dim V)^2)$)

Find the smallest positive integer $m such that the equation

\[c_0I + c_1T + \dots + c_{m-1}T^{m-1} = -T^m\]

has a solution $c_0, c_1, \dots, c_{m-1} \in \mathbf{F}$.

Pick a basis of $V$ and replace $T$ in the equation above with the matrix of $T$, then the equation above can be thought of as a system of $(\dim V)^2$ linear equations in the $m$ unknowns $c_0, c_1, \dots, c_{m-1} \in \mathbf{F}$.

Use Gaussian elimination or another fast method of solving systems of linear equations can tell us whether a solution exists, testing successive values $m = 1, 2, \dots, \dim V $ until a solution exists.

A usually faster way ($O((\dim V))$):

Pick $v \in V$ with $v \ne 0$ and consider the equation

to check whether the following system of $\dim V$ linear equations has a unique solution:

\[c_0v + c_1Tv + \dots + c_{\dim V-1}T^{\dim V-1}v = -T^{\dim V}v.\]

Use a basis of $V$ to convert the equation above to a system of $\dim V$ linear equations in $\dim V$ unknowns $c_0, c_1, \dots, c_{\dim V -1} $.

If this system of equations has a unique solution $\dim V$ unknowns $c_0, c_1, \dots, c_{\dim V -1} $ (as happens most of the time), then the scalars $\dim V$ unknowns $c_0, c_1, \dots, c_{\dim V -1}, 1 $ are the coefficients of the minimal polynomial of $T$.

Every monic polynomial is the minimal polynomial of some operator.

See

• https://en.wikipedia.org/wiki/Companion_matrix
• https://mathworld.wolfram.com/CompanionMatrix.html

For every polynomial $q$,

\[T\, q(ST) \;=\; q(TS)\, T.\]

First for monomials: $T(ST)^k = (TS)^k T$, by induction on $k$. The case $k=0$ is $T = T$. Assuming it for $k$,

\[T(ST)^{k+1} = \big(T S\big) T (ST)^{k} = (TS)\,(TS)^k T = (TS)^{k+1}T,\]

where the first step just regroups $T(ST)(ST)^k$. Both sides of the identity are linear in $q$, so it extends from monomials to all polynomials. $\square$

If $q$ annihilates $ST$, then $z\,q(z)$ annihilates $TS$.

Suppose $q(ST) = 0$. By the identity, $q(TS)\,T = T\,q(ST) = 0$. Multiply on the right by $S$:

\[q(TS)\,TS = 0.\]

Since $q(TS)$ is a polynomial in $TS$, it commutes with $TS$, so this says exactly that the polynomial $z\,q(z)$ evaluated at $TS$ is zero. $\square$

Suppose $ V$ is finite-dimensional and $ T \in \mathcal{L}(V) $. Let $ \mathcal{E} $ be the subspace of $ \mathcal{L}(V) $ defined by

\(\mathcal{E} = \{ q(T) : q \in \mathcal{P}(\mathbf{F}) \}\).

Then the list $I, T, T^2, \dots, T^{m-1}$ is a basis of $\mathcal{E}$, where $m = \deg p_T$.

$\mathcal{E}$ is indeed a subspace: it’s closed under addition and scalar multiplication because $q_1(T) + q_2(T) = (q_1 + q_2)(T)$ and $c\,q(T) = (cq)(T)$. In fact $\mathcal{E}$ is the image of the linear map $\mathcal{P}(\mathbf{F}) \to \mathcal{L}(V)$, $q \mapsto q(T)$.

Consider the linear map $\Phi : \mathcal{P}(\mathbf{F}) \to \mathcal{L}(V)$, $\Phi(q) = q(T)$. Then $\operatorname{range}\Phi = \mathcal{E}$ and $\operatorname{null}\Phi = {q : q(T) = 0}$ — exactly the multiples of $p$. So $\mathcal{E} \cong \mathcal{P}(\mathbf{F})/\langle p\rangle$, and the quotient has the remainders of degree $< m$ as canonical representatives — $m$ dimensions’ worth.

Local Minimal Polynomial

Suppose $V$ is finite-dimensional, $ T \in \mathcal{L}(V) $, and $ v \in V $. Then

\[\operatorname{span}(v, Tv, \dots, T^m v) = \operatorname{span}(v, Tv, \dots, T^{\dim V - 1}v)\]

for all integers $ m \ge \dim V - 1 $.

Write $n = \dim V$ and

\[U_m := \operatorname{span}(v, Tv, \dots, T^m v).\]

Since each list extends the previous one, the chain is increasing:

\[U_0 \subseteq U_1 \subseteq U_2 \subseteq \dots\]

The goal is: $U_m = U_{n-1}$ for all $m \geq n-1$.

Claim 1 (stabilizing chain). If $U_k = U_{k-1}$ for some $k \geq 1$, then $U_m = U_{k-1}$ for all $m \geq k-1$.

Proof. It suffices to show $U_{k+1} = U_k$, then induct.

$U_k = U_{k-1}$ says $T^k v \in U_{k-1} = \operatorname{span}(v, Tv, \dots, T^{k-1}v)$, so write

\[T^k v = a_0 v + a_1 Tv + \dots + a_{k-1}T^{k-1}v.\]

Apply $T$ to both sides:

\[T^{k+1}v = a_0 Tv + a_1 T^2 v + \dots + a_{k-1}T^{k}v \in U_k.\]

So the one new vector in the list for $U_{k+1}$ already lies in $U_k$, giving $U_{k+1} \subseteq U_k$, hence $U_{k+1} = U_k$. Induction extends this to all $m \geq k$. $\square$

Claim 2. There exists $k$ with $1 \leq k \leq n$ and $U_k = U_{k-1}$. (Treat the case $v = 0$ separately: then every $U_m = {0}$ and the exercise is trivial. So assume $v \neq 0$.)

Proof. The list $v, Tv, \dots, T^n v$ has $n+1$ vectors in the $n$-dimensional space $V$, so it is linearly dependent. Therefore, some vector in the list lies in the span of the ones preceding it: there is $k$ with $0 \le k \leq n$ and

\[T^k v \in \operatorname{span}(v, Tv, \dots, T^{k-1}v) = U_{k-1}.\]

Since $v \ne 0$, we have $k \geq 1$. And $T^kv \in U_{k-1}$ gives $U_k \subseteq U_{k-1}$, i.e. $U_k = U_{k-1}$. $\square$

Take the $k \leq n$ from Claim 2. By Claim 1, $U_m = U_{k-1}$ for all $m \geq k-1$. Since $k - 1 \leq n-1$, the index $n-1$ is itself in that range, so $U_{n-1} = U_{k-1}$. Therefore, for every $m \geq n-1 \; (\geq k-1)$,

\[U_m = U_{k-1} = U_{n-1},\]

which is exactly

\[\operatorname{span}(v, Tv, \dots, T^m v) = \operatorname{span}(v, Tv, \dots, T^{\dim V - 1}v). \qquad \blacksquare\]

$U_{n-1}$ is $T$-invariant. Since $T(U_{n-1}) \subseteq U_n = U_{n-1}$. In fact $U_{n-1}$ is the smallest $T$-invariant subspace containing $v$ — any such subspace must contain all $T^jv$. So, closing $v$ up under $T$ never requires more than $\dim V$ terms.

Local minimal polynomial

Suppose $ V $ is finite-dimensional, $ T \in \mathcal{L}(V) $, and $ v \in V $. Then there exists a unique monic polynomial $ p_v $ of smallest degree such that $ p_v(T)v = 0 $.

$ p_v $ is the minimal polynomial of the smallets $T$-invariant subspace $U$ containing $v$, and $ \deg p_v = k = \dim U $.

In Claim 2, $T^kv = \sum_{j<k} a_j T^j v$, rearranges to $q(T)v = 0$ with $q(z) = z^k - a_{k-1}z^{k-1} - \dots - a_0$ monic of degree $k$ — and minimality of $k$ makes $q$ the least-degree monic polynomial with $q(T)v = 0$.

Suppose $ V $ is finite-dimensional, $ T \in \mathcal{L}(V) $, $ q \in \mathcal{P}(\mathbf{F}) $ and $ q(T) = 0 $, then

\[p_v \mid p_T \mid q.\]

If $p_u$ and $p_w$ are coprime, then $p_{u+w} = p_u p_w$.

First, $(p_up_w)(T)(u+w) = p_w(T)p_u(T)u + p_u(T)p_w(T)w = 0$, so $p_{u+w} \mid p_up_w$.

Conversely suppose $r(T)(u+w) = 0$. Applying $p_w(T)$ and using $p_w(T)w = 0$: \((p_w r)(T)u = p_w(T)r(T)u + p_w(T)r(T)w = p_w(T)r(T)(u+w) = 0,\) so $p_u \mid p_w r$, and coprimality gives $p_u \mid r$. Symmetrically $p_w \mid r$, hence $p_up_w \mid r$. Taking $r = p_{u+w}$ finishes it. $\square$

There exists $v \in V$ with $p_v = p$.

Factor $p = q_1^{m_1}\dots q_k^{m_k}$ into powers of distinct monic irreducibles. Fix $i$.

Since $\deg(p/q_i) < \deg p$, minimality of $p$ gives $(p/q_i)(T) \neq 0$, so choose $u_i$ with $(p/q_i)(T)u_i \neq 0$. Set \(w_i = \big(p/q_i^{m_i}\big)(T)\,u_i.\)

Then $q_i^{m_i}(T)w_i = p(T)u_i = 0$, so $p_{w_i} \mid q_i^{m_i}$; and $q_i^{m_i-1}(T)w_i = (p/q_i)(T)u_i \neq 0$, so $p_{w_i} \nmid q_i^{m_i-1}$. As $q_i$ is irreducible, the only possibility is $p_{w_i} = q_i^{m_i}$.

The polynomials $q_1^{m_1},\dots,q_k^{m_k}$ are pairwise coprime, so $v = w_1 + \dots + w_k$ gives

\[p_v = p_{w_1}\dots p_{w_k} = q_1^{m_1}\dots q_k^{m_k} = p. \qquad \blacksquare\]

Over $\mathbb{C}$ this is the Jordan statement in disguise. There $q_i = z - \lambda_i$, and $\deg p = \sum m_i$ while $n = \sum \dim G(\lambda_i, T)$. Since always $m_i \le \dim G(\lambda_i,T)$, the hypothesis $\deg p = n$ forces $m_i = \dim G(\lambda_i,T)$ for every $i$ — that is, one Jordan block per eigenvalue. The $w_i$ above is precisely a vector at the top of the $i$-th block’s chain, and $v$ is their sum.

If $p_v = p$ and $\deg p = n$, then $v, Tv, \dots, T^{n-1}v$ is a basis.

Suppose $a_0 v + a_1 Tv + \dots + a_{n-1}T^{n-1}v = 0$ with the $a_j$ not all zero. Then $q(z) = a_0 + a_1 z + \dots + a_{n-1}z^{n-1}$ is a nonzero polynomial with $q(T)v = 0$, so $p_v \mid q$ — impossible, since $\deg q < n = \deg p_v$. So the list is linearly independent, and $n$ independent vectors in an $n$-dimensional space form a basis. $\square$