3.12 Theorems of minimal polynomial

Theorem

Let \(V\neq\{0_V\}\) be a k-vs and \(\dim V<\infty\), \(f\in End(V)\) is \(k\)-linear map

\(\exists\) a basis \(\mathcal{B}\) of \(V\) such that \([f]_{\mathcal{B}}\) is upper triangular\(\iff\)\(m_{f}(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)^{m_i}\) for some \(c_1,...,c_r\in k:c_i\neq c_j\) if \(i\neq j\) and \(m_i\geq 1\)

Proof

Main idea: Recall Lemma since we need eigenvalue and hypothesis are same and choose vector in \(W\) inductively from \(0\)

\(\Leftarrow\)) Suppose that \(m_{f}(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)^{m_i},c_{1},...,c_{r}\in k,c_{i}\neq c_{j},if\, i\neq j\)

Let \(W_{0}=\{0_{V}\}\subsetneq V\). By the previous lemma, \(\exists v_{1}\notin W_{0}\) and some eigenvalue \(c\) of \(f\) such that \((f-c\cdot Id)(v_{1})\in W_{0}\) that is \((f-cId)(v_1)=0\) then \(f(v_{1})=cv_{1}\).

Note that \(v_1\) is an eigenvector of \(f\) of eigenvalue \(c\).

Let \(W_1=\lang v_1\rang\), if \(W_1=V\), we are done taking \(\mathcal{B}=\{v_1\}\). If \(W_1\subsetneq V\), again, using the lemma, \(\exists v_2\in V\setminus W_1\) and eigenvalue \(c'\) of \(f\) such that \((f-c^{\prime}Id_{V})(v_{2})\in W_{1}=\{v_{1}\}\), that is \(f(v_2)=c'\cdot v_2+\alpha_1v_1\) for some \(\alpha_1\in k\)

Note that \(\{v_1,v_2\}\) is linear independent since \(v_{2}\notin\lang v_1\rang=W\)

Thus basis will start with \(\{v_1,v_2\}\), up to our \([f]_{\mathcal{B}}=\)

Now, set \(W_2=\lang v_1,v_2\rang\). Again if \(W_2=V\), we are done. If not, that is \(W_2\subsetneq V\) we have to continue

In this way, suppose we have obtained a subspace \(W_i\) where \(i\geq 2\) such that \(W_i\) has basis \(\{v_1,v_2,...,v_i\}\) and \(\forall j,1\leq j\leq i,f(v_{j})\in \lang v_{1},...,v_{j-1}\rangle\)

If \(W_i=V\), we are done. If not, \(W_i\subsetneq V\), using the lemma again, we find a vector \(v_{i+1}\notin W_i\) and an eigenvalue, \(c_{l_{i+1}}\) eigenvalue of \(f\) such that \(f(v_{i+1})=c_{l_{i+1}} v_{i+1}+\beta_1v_1+...+\beta_iv_i\) for some \(\beta_1,...,\beta_i\in k\)

Note that \(\{0_{V}\}=W_{0}\subsetneq W_{1}\subsetneq W_{2}\subsetneq\ldots\subsetneq V\) and \(0=\dim W_{0}<\dim W_{1}<\ldots<\dim V\) finite

So the produce must stop at some \(l=\dim V\) and thus we get a basis \(\mathcal{B}=\{v_1,...,v_l\}\) such that \([f]_\mathcal{B}\) is upper triangular

\(\Rightarrow\)) If \(\exists\) a bsis \(\mathcal{B}\) of \(V\) such that \([f]_{\mathcal{B}}= \begin{pmatrix} a_{1} & \placeholder{} & \star \\ \placeholder{} & \ddots & \placeholder{} \\ 0 & \placeholder{} & a_{l} \end{pmatrix}\), so \(\chi_{f}(x)=\prod_{i=1}^{l}\left(x-a_{i}\right)=\prod_{i=1}^{r}\left(x-c_{i}\right )^{m_{i}},c_{i}\neq c_{j},if\,i\neq j\) and \(m_i\geq 1,\forall i\)

Then by C-H theorem, \(m_f(x)\mid \chi_f(x)\), so \(m_{f}(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)^{n_{i}},m_{i}\geq n_{i}\geq1\) since the roots of \(m_f(x)\) and \(\chi_{f}(x)\) are same

Corollary

If the field \(k\) is algebraically closed, that is \(\forall p[x]\in k[x]\), \(p(x)\) has all its roots in \(k\)

\(\forall\) finite dimensional \(k\)-vector space \(V\), \(\forall k\) linear map \(f:V\to V\), \(\exists\) a basis \(\mathcal{B}\) of \(V\) such that \([f]_\mathcal{B}\) is upper triangular

Theorem

Let \(V\) be a finite dimension k-vs, \(f\in End(V)\) is \(k\)-linear map

Then \(f\) is diagonalizable\(\iff\)\(m_{f}(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)\) with \(c_i\neq c_j\) if \(i\neq j\)

Proof

See 3.5 notes

Example

\(A,B\in M_2(\R)\), \(A= \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix},B= \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\)

Verify that \(A\cdot B=B\cdot A\), also: \(\chi_A(x)=x(x-2)=m_A(x)\Rightarrow\) \(A\) is diagonalizable

\(\chi_B(x)=(x+2)(x-4)=m_B(x)\Rightarrow B\) is diagonalizable

For \(A\): \(V_{0}^{A}=\lang (1,-1)\rangle\), \(V_{2}^{A}=\langle(1,1)\rangle\)

For \(B\): \(V_{-2}^{B}=\langle(1,-1)\rangle\), \(V_{4}^{B}=\langle(1,1)\rangle\)

Fix \(\mathcal{B}=\{\left(1,-1\right)=v_{1},\left(1,1\right)=v_{2}\}\), then \(\exists C_{1}\in Gl_{2}(\mathbb{R})\) and \(C_1\) is invertible such that \(C_{1}AC_{1}^{-1}= \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}\)

Notation: \(\{c\in M_m(k):c\text{ is invertible}\}=Gl_{m}(k)\subsetneq M_{m}(k)\)

\(\exists C_{2}\in Gl_{2}(\mathbb{R})\) and \(C_{2}\) is invertible such that \(C_{2}BC_{2}^{-1}= \begin{pmatrix} -2 & 0 \\ 0 & 4 \end{pmatrix}\)

\(C_1=C_2\) is the change of bases matrix from \(E\) since we have the common basis

Analogously, defining the linear maps \(f_A:\R^2\to \R^2\) such that \([f_A]_E=A\) and \(f_{B}:\R^{2}\to \R^{2}\) such that \([f_{B}]_{E}=B\)

We get that both \(f_A\) and \(f_B\) are diagonalizable in the same basis \(\mathcal{B}=\{v_1,v_2\}\)

Then \([f_{A}]_{\mathcal{B}}= \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix},[f_{B}]_{\mathcal{B}}= \begin{pmatrix} -2 & 0 \\ 0 & 4 \end{pmatrix}\)

Theorem

Let \(V\) a finite dimensional k-vs. Let \(\mathcal{F}\) be a set/family of \(k\)-linear maps from \(V\) to \(V\) such that they commute and they are diagonalizable

Then \(\exists\) a basis \(\mathcal{B}\) of \(V\) such that \([f]_\mathcal{B}\) is diagonal, \(\forall f\in \mathcal{F}\)

Proof

By induction on \(\dim V=n\). If \(n=1\), it is evident

Assume that the result is true \(\forall\)k-vector space of \(\dim <n=\dim V\)

If \(\mathcal{F}\) contains just scalar multiplies of \(Id_V\) that is \(\forall f\in \mathcal{F},\exists \lambda\in k:f=\lambda Id_V\)

(If \(f=\lambda Id_V\), then for all basis \(\mathcal{B}\), \([f]_{\mathcal{B}}= \begin{pmatrix} \lambda & \placeholder{} & \placeholder{} \\ \placeholder{} & \ddots & \placeholder{} \\ \placeholder{} & \placeholder{} & \lambda \end{pmatrix}\))

Suppose \(\exists f\in \mathcal{F}\) such that \(f\neq \lambda \cdot Id_{V},\forall \lambda\in k\)

\(f\) is diagonalizable by hypothesis. Let \(c_1,...,c_r\) be all the different eigenvalues

We know that \(r\geq 2\) since \(f\neq Id_V\) \(\forall \lambda\). Let \(W_i\) be the subspace spanned by the eigenvectors of eigenvalue \(c_i\).

Then \(\dim W_i<\dim V=n\) \(\forall i,1\leq i\leq r\)

\(\forall h\in \mathcal{F},h\circ f=f\circ h\), \(W_i\) is \(f\)-invariant \(\forall i\), so \(W_i\) is \(h\)-invariant \(\forall i,\forall h\in\mathcal{F}\), we will prove this

Given \(v\in W_i\), we want to prove that \(h(v)\in W_i\), then \(f(h(v))=f\circ h(v)=h\circ f(v)=h(c_i\cdot v)=c_i \cdot h(v)\)

Thus \(h(v)\in W_i\)

Let \(\mathcal{F_{i}}=\left\lbrace h|_{W_i}:\forall h\in\mathcal{F}\right\rbrace\) again any pair of elements in \(\mathcal{F_{i}}\) commute and \(\dim W_i<n\) and \(h|_{W_i}\) is diagonalizable, \(\forall h\in \mathcal{{F}}\) because \(m_{h|_{w_i}}\mid m_h(x)\) and \(m_h(x)\) is a product of different factors of dg1, then \(m_{h|_{w_i}}\) also is a product of some different factors of dg1

Then by theorem, \(h|_{W_i}\) is diagonalizable

Now, we use the inductive hypothesis for \(W_i\) and \(\mathcal{F_i}\) and we find a basis \(\mathcal{B}_i\) of \(W_i\) such that \([h|_{W_i}]_\mathcal{B_i}\) is diagonal \(\forall h\in \mathcal{F}\)

Since \(V=W_{1}\oplus...\oplus W_{r}\), for each \(W_i\), we get a basis \(\mathcal{B_i}\) and then we get a basis \(\bigcup_{i=1}^{r}\mathcal{B}_{i}=\mathcal{B}\) of \(V\) and \(\forall h\in\mathcal{F},[h]_{\mathcal{B}}=\)

Summary: We want to prove any diagonalizable and commutative linear map can find a common basis

Since finite dimension, we can consider induction on dimension

We assume it's true for dimension less than \(n\), then we consider a eigenspace \(W_i\) of \(f\)-invariant

Then we want to prove \(W_i\) is \(h\) invariant \(\forall h\), then we can consider a smaller linear map \(h|_{W_i}\)

Then prove \(h|_{W_i}\) is commutative and diagonalizable, then use hypothesis.

Then since direct sum, combine its into a basis

Additional knowledge

\(V=S\oplus T\) where \(S,T\) subspaces \(S\neq \{0_{V}\}\neq T\), then \(\dim V=\dim S+\dim T\) and \(S\cap T=\{0\}\)

Unicity: \(\forall v\in V,\exists s\in S,t\in T:v=s+t,v=s'+t'\), then \(s=s',t=t'\)
Define \(p:V\to V\) given \(v\in V,p(v)=s\) (\(\exists !s\in S,t\in T:v=s+t\))

\(p\) is k-linear map and \(p^2=p\) since \(p(p(v))=p(s+0)=p(s)=p\left(v\right)=s\)
Definition: \(p:V\to V\) is a projection iff \(p^2=p\)

Example
1. \(lmp=S\), \(\ker p=T\) and \(p^2=p\iff p^2-p=0\)
  
  Taking \(h(x)=x^{2}-x\) and since \(h(p)=p^2-p=0\), we have that \(m_{p}(x)\mid x^{2}-x=x\left(x-1\right)=h\left(x\right)\)
  
  If \(m_{p}(x)=x\Rightarrow m_{p}\left(p\right)=p=0\) not possible since \(lmp=S\neq \{0\}\)
  
  If \(m_{p}(x)=x-1\Rightarrow m_{p}\left(p\right)=p-1=0\Rightarrow p=Id_{V}\) not possible since \(\ker p=T\neq\{0\}\)
  
  So \(m_p(x)=x(x-1)\) 2. Trivial example: \(p=Id_{V}(Imp=V,\ker p=\{0_{V}\})\), \(p=0_{V}(Imp=\left\lbrace0\right\rbrace,\ker p=V)\)

Remark

For example3(a), i find it's useful to consider a polynomial and evaluate on linear map \(p\) to get a minimal polynomial, it utilize the property of minimal polynomial