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3.5 Minimal Polynomial

Annihilating Polynomials

Assume \(k\) filed, \(V\) be \(k-vs\) and \(\dim_kV=m\) and \(\dim_kW=n\)

Let \(f:V\to V\) be \(k\)-linear map and \(\dim_kL(V,W)=m\times n\)

Notation: \(L(V,W)=\text{Hom}_k(V,W)\), thus \(\dim_{k}\text{Hom}(V,V)=m^{2}\)

Main Theorem

Prove that there exists a monic polynomial \(p_0(x)\) of minimal degree such that \(p_0(f) = 0\) and \(p(x)\mid t(x)\) with \(t(f) = 0\).

Main idea: consider set of polynomial and degree, then WOP

Strategy: Consider a linearly dependent list, because it guarantees constructing a nonzero polynomial \(p(x)\) such that \(p(f)=0\).

Note \(p_0(x)\) cannot be zero polynomial

Proof

Consider \(p(x) = a_{0} + a_{1}x + \dots + a_{m}x^{m^2}\), then since \(\underbrace{\{Id_{v},f,...,f^{m^2}\}}_{m^2+1\text{ elements}}\subseteq\text{Hom} (V,V)\) is linearly dependent

Thus \(\exists a_0,a_1,...,a_{m^2}\in k\) not all \(0\) such that \(a_{0}\cdot Id+a_{1}\cdot f+\cdots+a_{m^2}f^{m^2}=0\) (this \(0\) is linear map)

Thus \(p(f) = a_0 + a_1 f + \dots + a_m f^{m^2} = 0\)

Then consider set \(A = \{ t(x) \in k[x] : t(f) = 0\}\). Since \(p(x) \in A\), then \(A \neq \emptyset\).

And consider set \(D_A = \{ \deg(t(x)) : t(x) \in A \}\). Since \(\deg(p(x)) \in D_A\), then \(D_A \neq \emptyset\).

Thus, by WOP, take \(p_0(x)\) is monic and \(\deg(p_0(x))\) is the minimal degree \(m_0\) such that \(p_0(f) = 0\).

Then \(m_0>0\), because if \(m_0=0\), then constant linear map \(a_0=0\) is linearly independent!

Let \(t(x)\in A\), using the division algorithm, we find polynomials \(q(x),r(x)\in k[x]:t(x)=p_0(x)q(x)+r(x)\) with \(r(x)=0\) or \(dgr(x)<dgp_0(x)\)

Then \(t(f)=p_0(f)\circ q(f)+r(f)\) here \(t(f)=0\) since \(t(f)=0\) and \(p_0(f)=0\)

Then \(0=r(f)\). Then \(r(x)=0\) because if \(r(x)\neq 0\), contradicts with the minimality of \(dgp_0(x)\)

Thus \(t(x)=p_0(x)\cdot q(x)\), thus \(p_0(x)\) is a division of \(t(x)\)

Thus \(A=\{p_0(x)\cdot q(x):q(x)\in k[x]\}=\lang p_0(x)\rang\)

Remark

\(\empty\neq A\subseteq k[x]\), \(A\) is an ideal of the ring \(k[x]\)

If \(t_{1}(x)\in A,t_{2}(x)\in A\), then \(t_1(x)+t_2(x)\in A\)

If \(t(x)\in A,s(x)\in k[x]\), then \(s(x)\cdot t(x)\in A\)

Remark

Minimal Polynomial \(p_0(x)\) is unique with the properties:

  1. \(p_0(f)=0\)​

  2. \(p_0(x)\) is monic

  3. \(p_0(x)\) is of minimal degree

Unique because if there exists \(\gamma(x)\in k[x]\) satisfying 1 2 3

Then \(p_{0}(x)\mid \gamma(x)\) because \(\gamma(f)=0\)

Also we can show \(\gamma(x)\mid p_{0}(x)\). Moreover, \(p_0(x)\) and \(\gamma(x)\) are monic, thus \(\gamma(x)=p_0(x)\) because

\(p_{0}(x)\mid \gamma(x)\implies\exists q_{1}\left(x\right)\) such that \(\gamma(x)=p_0(x)\cdot q_1(x)\)

\(\gamma(x)\mid p_{0}(x)\implies\exists q_{2}\left(x\right)\) such that \(p_{0}(x)=\gamma(x)\cdot q_{2}(x)\)

Then \(p_0(x)=p_0(x)q_1(x)q_2(x)\), thus \(p_0(x)(1-q_1(x)q_2(x))=0\)

Since \(p_0(x)\neq 0\), we get \(q_1(x)q_2(x)=1\), thus \(q_1(x),q_2(x)\) are constants

And \(p_0(x)=\gamma(x)q_2\implies q_2=1\implies q_1=1\). Then \(p_0(x)=\gamma(x)\)

Definition

\(p_0(x)\) is called the minimal polynomial of \(f:V\to V\). Notation: We will denote it by \(m_f(x)\)

Remark

Assume \(\dim_kM_m(k)=m^2\). In a similar way, given a matrix \(M\in M_{m}(k)\) one can define the minimal polynomial of \(M\) denoted \(m_M(x)\) since there is a bijection from linear map to matrix

Then given a matrix \(M\in M_m(k)\) and an invertible matrix \(C\in M_m(k)\) where \(CMC^{-1}\) exists, then \(m_M (x) = m_{CMC^{-1}} (x)\)

Proof

To prove the minimal polynomial, we need to use properties "p(x) | t(x)".

Thus, NTP \(m_{M}(x)\mid m_{CMC^{-1}}(x)\) and \(m_{CMC^{-1}}(x) \mid m_M(x)\) .

This is equivalent to proving \(m_M(CMC^{-1}) = 0\) and \(m_{CMC^{-1}}(M) = 0\) .

Let's \(m_M(x) = a_0 + a_1 x + \ldots + x^r\) , then \(m_{M}(CMC^{-1})=a_{0}CM^{0}C+a_{1}CMC^{-1}+a_{2}CMC^{-1}CMC^{-1}+\ldots+(CMC^{-1} )^{r}\)

\(=a_{0}I+a_{1}CMC^{-1}+a_{2}CM^{2}C^{-1}+\ldots+CM^{r}C^{-1}=Cm_{M}(M)C^{-1}\)​.

Since \(m_M(M) = 0\), then \(m_M(CMC^{-1}) = 0\)

Similarly, \(m_{CMC^{-1}}(M) = \ldots\) , then \(m_{CMC^{-1}}(CMC^{-1}) = \ldots = C m_{CMC^{-1}}(M) C^{-1}\)

Since \(m_{CMC^{-1}}(M) = 0\), then \(m_{CMC^{-1}}(M) = 0\) .

Thus \(m_M(x) \mid m_{CMC^{-1}}(x)\) and \(m_{CMC^{-1}}(x) \mid m_M(x)\) by minimal polynomial.

Thus \(m_M(x) = m_{CMC^{-1}}(x)\) .

This means: if a matrix \(M\) can change basis into another matrix \(D=CMC^{-1}\)(two matrix are similar), then they have the same polynomial

Theorem

Let \(A\in M_m(k)\), \(\chi_A(x)\) and \(m_A(x)\) have the same roots

Proof

Main idea: the root of \(\chi_A(x)\) means a eigenvalue, the root of \(m_A(x)\) means a factor of polynomial, sometimes properties of minimal polynomial are useful

Given \(c\in k\), we have to prove that \(\chi_A(c)=0\iff m_A(c)=0\)

\(\Rightarrow\)) Idea: Consider \(m_A(x)\) and put definition into it

Now suppose that \(\chi_A(c)=0\), that is \(c\) is an eigenvalue of \(A\), we want to prove \(m_A(c)=0\)

We know that \(\exists v\neq 0\in k^m\) such that \(Av=cv\), moreover \(0=m_A(A)\cdot v=m_A(c)\cdot v\) lecture 1

Thus \(m_{A}(c)\cdot v=0_{V}\in k^{m}\). Thus \(m_A(c)=0\)

\(\Leftarrow\)) Idea: Factor \((x-c)\) and argue \(q(x)\), use \(A-cId\)

Suppose that \(m_A(c)=0\). We have to prove that \(c\) is an eigenvalue of \(A\), that is \(A-cId=0\)

Since \(m_A(c)=0\), then \(c\) is a root of \(m_A(x)\), then \(x-c\mid m_A(x)\), that is \(\exists q(x)\in k[x]\) such that \(m_A(x)=(x-c)q(x)\). Here \(\deg q(x) < \deg m_{A}(x)\)

Then \(m_{A}(A)\left(v\right)=(A-cId)q(A)(v)=0\) .

Since \(\deg q(x) < \deg m_{A}(x)\), then \(q(A)\neq0\)​, since minimal polynomial.

Then \((A-cId)(v)=0\Rightarrow v\) is an eigenvector corresponding to eigenvalue \(c\). Then \(\chi_A(c) = 0\)

Theorem

Suppose that \(f:V\to V\) is a diagonalizable linear map (\(\dim_kV=n\)) \(\iff\)​\(m_{f}(x)=(x-c_{1})...(x-c_{r})=\prod_{i=1}^{r}\left(x-c_{i}\right)\) where \(c_1,...,c_r\) all the different eigenvalue of \(f\),

Proof

\(\Rightarrow\)) Main idea: To prove \(p(x)=m_f(x)\) we can prove \(p(x)\mid m_f(x)\) and \(m_f(x)\mid p(x)\) since we are in polynomials

Let \(p(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)\). We want to prove that \(p(x)=m_f(x)\)

Let's prove \(m_{f}(x)\mid p(x)\) firstly, we know \(m_f(f)=0\), thus we only need to prove \(p(f)=0\), that is \(p(f)(v)=0,\forall v\in V\)​

Clearly, \(p(x)\) is monic. Given an eigenvector \(v\) of eigenvalue \(c_j\) for some \(j,1\leq j\leq r\)

Then \(p(f)(v)=\prod_{i=1,i\neq j}^{r}(f-c_{i}Id)\circ(f-c_{j}Id)(v)=0\)

Since \(f\) is diagonalizable, it has a basis \(\mathcal{B}\) of eigenvectors

Given any \(w\in \mathcal{B}\), \(p(f)(w)=0\). Thus we get that \(p(f)=0\) (so \(m_{f}(x)\mid p(x)\))

On the other hand, by previous theorem: \(c_1,...,c_r\) are the roots of \(m_f(x)\), so \(x-c_{i}\mid m_{f}(x),\forall i,1\leq i\leq r\)

Since \(x-c_1,...,x-c_r\) are coprime (\(c_i\neq c_j,\forall i\neq j\)), we have that \(\prod_{i=1}^{r}\left(x-c_{i}\right)\mid m_{f}(x)\)

Thus \(p(x)=m_f(x)\)

\(\Leftarrow\)) Main idea: Use Lemma since we have the same hypothesis and we need eigenvalue. Then consider minimal polynomial and factor out \(x-c\)

Trick: Consider \(q(x)-q(c_{j})\in k[x]\)

Let \(W\) be spanned by eigenvectors of \(f\). We want to prove that \(W=V\)

Suppose \(W\subsetneq V\), by lemma there exists \(v\in V\setminus W\) and an eigenvalue \(c_j\) for some \(j\), \(1\leq j\leq r\) such that \(\underbrace{(f-c_{j}Id_{V})(v)}_{w}\in W\)

We know that \(m_f(x)=(x-c_j)q(x)\) then \(m_{f}(f)(v)=0_{v}=(f-c_jId_V)(q(f)(v))\)​

Thus \(q(f)(v)\in W\) because \(q(f)(v)\in \ker(m_f)\), then \(q(f)(v)\) is an eigenvector, then in \(W\) since \(W\) is spanned by eigenvectors

Here \(q(f)(v)\in W\) but \(v\in W\), here we cannot get a contradiction, but if it is just a polynomial about \(v\), we will get

Thus consider \(q(x)-q(c_{j})\in k[x]\)​. Since \(q(c_j)-q(c_j)=0\), then it is divisible by \(x-c_j\).

Then \(q(x)-q(c_j)=(x-c_j)h(x)\) for some \(h(x)\in k[x]\)

Since \(q(c_j)-q(c_j)=0\), then it is divisible by \(x-c_j\). Then \(q(x)-q(c_j)=(x-c_j)h(x)\) for some \(h(x)\in k[x]\)

Thus \(q(f)(v)-q(c_j)(v)=h(f)(f-c_jId)(v)\in W\)(*) since \((f-c_jId)(v)\in W\) and by lemma

Since \(q(f)(v)\in W\), then \(q(c_{j})(v)=q(f)(v)-h(f)(w)\in W\)

Since \(q(c_j)\neq 0\) and \(q(c_j)(v)\in W\), then \(v\in W\) which contradiction to \(v\in V\setminus W\)

Thus \(W=V\), that is \(V\) has a basis of eigenvectors

EXAMPLES

  1. Let \(A = \begin{bmatrix} 5 & -6 & -6 \\ -1 & 4 & 2 \\ 3 & -6 & -4 \end{bmatrix} \in \mathbb{R}^{3 \times 3}\), \(\chi_A(x) = (x-2)^2 (x-1)\)

    So we know that \(m_A(x) = (x-2)^i (x-1)^j\) for some \(i, j \in \mathbb{N}\).

    Doing the computations, one can verify that \(A\) is diagonalizable, so \(m_A(x) = (x-2)(x-1)\)​

  2. \(A = \begin{bmatrix} 2 & 0 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \in \mathbb{R}^{3 \times 3}\), \(\chi_A(x) = (x-2)^2 (x-3)\)

    Again \(m_A(x) = (x-2)^i (x-3)^j\) for some \(i, j \in \mathbb{N}\).

    Here \(A\) is not diagonalizable, then we compute \((A - 3I)(A - 2I) \neq 0\) \((A - 3I)(A - 2I)^2 = 0\)

    So \(m_A(x) = (x - 3)(x - 2)^2\)

Theorem(CAYLEY-HAMILTON)

Let \(\dim_kV=m\), \(f:V\to V\) be \(k\)-linear map, then \(\chi_f(f)=0\), equivalent \(m_{f}(x)\mid\chi_{f}(x)\), then \(\deg m_{f}(x)\leq \deg\chi_{f}(x)\), so in particular \(\deg m_{f}(x)\leq m\)

Exercise

Find \(A\in \R^{3\times 3}\) such that \(m_A(x)=x^2\)

\(\chi_f(x)=x^3\), then \(A^2=0\). Assume \(A= \begin{pmatrix} 0 & a & b \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\) and compute