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3.31 Admissible subspaces

Suppose moreover, \(f\) is nilpotent, then \(\chi_f(x)=x^n\) (by definition of nilpotent, 0 is the only eigenvalue), then \(C_{f}= \begin{pmatrix} 0 & \cdots & 0 & 0 \\ 1 & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \cdots & 1 & 0 \end{pmatrix}\)

When is \(\chi_f(x)=m_f(x)\)? When the index of nilpotency of \(f\) is \(n\)

In this case \(\exists v\in V:V=\lang v\rang_f=\lang v,f(v),...,f^{n-1}(v)\rang\) and \(f^n(v)=0\)

Let \(h:V\to V\) be a k-linear map such \(\chi_h(x)=m_h(x)=(x-\lambda)^n\) for some \(\lambda\in k,\lambda \neq0\)

Let \(g:V\to V\) be the k-linear map, \(g=h-\lambda Id_V\).

Let's find a Jordan basis of \(h\)

\(\chi_{g}(x)=\det[xId_{V}-g]_\mathcal{B}\) and since \(g=h-\lambda Id\), then \(\chi_{g}(x)=\det[xId_{V}-h+\lambda Id_{V}]=\det\left\lbrack\left(x+\lambda\right )Id_{V}-h\right\rbrack_{\mathcal{B}}=x^{n}\)

So \(g\) is nilpotent. By theorem \(m_g(x)=x^n\), if not \(m_g(g)=g^k=0\) contradicts to definition of nilpotent

So \(\exists v\in V:V=\lang v\rang_g\) and if \(\mathcal{B}=\{v,g(v),...,g^{n-1}(v)\}\), then \([g]_{\mathcal{B}}= \begin{pmatrix} 0 & \cdots & 0 & 0 \\ 1 & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \cdots & 1 & 0 \end{pmatrix}\)

\([h]_{\mathcal{B}}=[g+\lambda Id_{V}]_{\mathcal{B}}=\left\lbrack g\right\rbrack_{\mathcal{B}} +\lambda Id= \begin{pmatrix} \lambda & \cdots & 0 & 0 \\ 1 & \ddots & 0 & \vdots \\ \vdots & \ddots & \ddots & \vdots \\ 0 & \cdots & 1 & \lambda \end{pmatrix}\) where \(g=h-\lambda Id_V\Rightarrow h=g+\lambda Id_V\).

This \([h]_\mathcal{B}\) is a Jordan form

\(f:V\to V,W\) is \(f\)-invariant subspace. Does it exists a subspace \(W'\) such that

  • \(W'\) is \(f\)-invariant
  • \(V=W\oplus W'\)​

Not in general

Example

\(V=k[x]_{\leq7}=\left\lbrace p\left(x\right)\in k[x]:p(x)=0\text{ or }\deg p(x)\leq 7\right\rbrace\), \(f:V\to V\) where \(p(x)\mapsto p'(x)\)

Then take \(W=k[x]_{\leq 5}\) is \(f\)-invariant. Suppose \(\exists\) subspace \(S\) of \(V\) such that \(S\) is \(f\)-invariant, \(V=W\oplus S\)

We can deduce that \(S\) contains some polynomials of \(\deg 6\) and \(\deg 7\)

Let \(p(x)\in S:\deg p(x)=6\), since \(S\) is \(f\)-invariant, \(f(p(x))\in S\)

But \(f(p(x))=p'(x),\deg p'(x)\leq 5\)

Suppose \(char k\neq 2,3\), then \(\deg p'(x)=5\)

So \(p'(x)\in W\)

If \(char k=2,3\), take \(W_{1}=k[x]_{\leq6}\) and a polynomial \(p(x)\) of \(\deg =7\)

The characteristic of a field is \(\min\left\lbrace n\in\mathbb{N}:n\cdot1_{k}=0\text{ in }k\right\rbrace\) and it is \(0\) if the set: min... is empty

Definition

Let \(f:V\to V\) be \(k\)-linear map and let \(W\) be a subspace.

\(W\) is \(f\)-admissible\(\iff\)1. \(W\) is \(f-\)invariant 2. If given \(v\in V\), \(\exists p(x)\in k[x]:p(f)(v)\in W\), then \(\exists w\in W:p(f)(v)=p(f)(w)\) (In other words, if \(p(f)(v)\in W\), then \(p(f)(v)\in Imp(f)|_{W}\))

Proposition

Let \(W\) be a subspace of \(V\). Suppose \(W\) is \(f\)-invariant, then \(W\) has an \(f\)-invariant complement\(\iff\)​\(W\) is \(f\)-admissible

Proof

\(\Rightarrow\)) Suppose \(\exists W'\) \(f\)-invariant subspace such that \(V=W\oplus W'\)

Given \(v\in V\), \(\exists\) unique \(w\in W,w'\in W':v=w+w'\)

Given \(p(x)\in k[x],p(f)(v)=\underbrace{p(f)(w)}_{\in W}+\underbrace{p(f)(w')}_{\in W'}\) since \(W,W'\) are \(f\)-invariant

So if \(p(f)(v)\in W\), we must have \(p(f)(w')=0_V\)

So \(p(f)(v)=p(f)(w)\)​

\(\Leftarrow\)) Suppose that \(W\) is \(f\)-admissible. If \(W=\{0\}\), then it's complement is \(V\), which is \(f-\)invariant.

If \(W=V\), then its complement is \(\{0_V\}\), which is also \(f\)-invariant.

Now given \(\{0\}\subsetneq W\subsetneq V\), \(W\) is \(f\)-admissible. By Cyclic decomposition theorem, \(\exists z_{1},...,z_{r}\in V\) such that \(V=W\oplus\langle z_{1}\rangle_{f}\oplus\ldots\oplus\langle z_{r}\rangle_{f}\) and \(\langle z_{1}\rangle_{f}\oplus\ldots\oplus\langle z_{r}\rangle_{f}=W'\) is \(f\)-invariant

Remark

In previous example, of course, \(W\) is not \(f\)-admissible

Example

Let \(f:V\to V\) be a projector (\(f^2=f\)). Let \(W=Imf\), then \(W\) is \(f\)-admissible

  1. \(W\) is \(f\)-invariant (if \(v\in Imf\), then \(f(v)\in Im f\)​)

  2. Let \(v\in V\), \(p(x)\in k[x]\) such that \(p(f)(v)\in W=Imf\)

    \(p(f)(v)=\sum_{i=0}^{n}a_{i}f^{i}\left(v\right)=a_{0}v+a_{1}f\left(v\right)+\ldots +a_{n}f^{n}\left(v\right)\)​

    Since \(f\) is a projector, then \(f|_{Imf}=Id_{Imf}\), then \(f(p(f)(v))=p(f)(v)\)​

    Then \(f\circ p(f)(v)=p(f)(f(v))=p(f)(v)\), then take \(w=f(v)\), we have \(w\in W\) and \(p(f)(v)=p(f)(w)\)

Definition

Let \(V\) finite dimensional, \(f:V\to V\) linear. \(S\) is \(f\)-invariant subspace of \(V\), \(v\in V\), consider \(\{p(x)\in k[x]:p(f)(v)\in S\}\neq \empty\) since \(0\in S(f,v)\)

It contains at least one polynomial of positive \(\deg\), since \(m(f)(x)\in S(f,v)\)

So \(\exists\) a unique polynomial of minimal \(\deg\) of \(q(x)\) such that \(q(f)(v)\in S\)

This polynomial is called the \(f\)-conductor of \(v\) into \(S\)

Let's denote it \(c_{f,v,S}(x)\)

Example

Take \(S=\{0_V\}\), \(v\in V\), \(c_{f,v,\{0\}}=m_{f,v}(x)\)

Remark

Let \(S,T\) be \(f\)-invariant subspace such that \(S\subseteq T\). Let \(v\in V,c_{f,v,S}(x)\)

We know \(c_{f,v,S}(f)(v)\in S\), also since \(S\subseteq T\), then \(c_{f,v,S}(f)(v)\in T\)

And we know \(c_{f,v,T}(f)(v)\in T\) and \(c_{f,v,T}(f)(v)\) is the minimality

Then \(c_{f,v,T}(x)\mid c_{f,v,S}(x)\)